
Local structure of convex surfaces
A. Yu. Plakhov^{ab}^{document.write(decode_email('DMEBCAHEGJHEGMGFDNCHEFCNGNGBGJGMDKCAHAGMGBGLGIGPHGGBGMGFHIGBGOGEGFHCDAEAGHGNGBGJGMCOGDGPGNCHCAGDGMGBHDHDDNFDEMGJGOGLCAEIFCEFEGDNCCGNGBGJGMHEGPDKHAGMGBGLGIGPHGGBGMGFHIGBGOGEGFHCDAEAGHGNGBGJGMCOGDGPGNCCDODMGJGNGHCAHDHEHJGMGFDNCCGNGBHCGHGJGOCNGMGFGGHEDKDDHAHICCCAGBGMGJGHGODNCCGBGCHDGNGJGEGEGMGFCCCAHDHCGDDNCCCPGGHEGJGDGPGOHDCPGFGNGBGJGMGJGDGPDBCOGKHAGHCCCAHHGJGEHEGIDNCCDBDIHAHICCCAGCGPHCGEGFHCDNCCDACCCPDODMCPEBDO'));email} ^{a} CIDMA, Department of Mathematics, University of Aveiro, Aveiro,
Portugal
^{b} Institute for Information Transmission Problems of the Russian Academy of Sciences (Kharkevich Institute), Moscow, Russia
Abstract:
A point on the surface of a convex body and a supporting plane to the body at this point are under consideration. A plane parallel to this supporting plane and cutting off part of the surface is drawn. The limiting behaviour of the cutoff part of the surface as the cutting plane approaches the point in question is investigated. More precisely, the limiting behavior of the appropriately normalized surface area measure in $S^2$ generated by this part of the surface is studied. The cases when the point is regular and singular (a conical or a ridge point) are considered. The supporting plane can be positioned in different ways with respect to the tangent cone at the point: its intersection with the cone can be a vertex, a line (if a ridge point is considered), a plane angle
(which can degenerate into a ray or a halfplane), or a plane (if the point is regular and, correspondingly, the cone degenerates into a halfspace). In the case when the intersection is a ray, the plane can be tangent (in a one or twosided manner) or not tangent to the cone.
It turns out that the limiting behaviour of the measure can be different. In the case when the intersection of the supporting plane and the cone is a vertex or in the case of a (one or twosided) tangency, the weak limit always exists and is uniquely determined by the plane and the cone. In the case when the intersection is a line or a ray with no tangency, there may be no limit at all. In this case all possible weak partial limits are characterized.
Bibliography: 13 titles.
Keywords:
convex geometry, conical and ridge singular points, surface area measure of convex bodies, tangent cone, Newton's aerodynamic problem.
Received: 10.04.2023 and 02.10.2023
§ 1. Introduction We consider a convex compact set $C$ with nonempty interior in Euclidean space $\mathbb R^d$, $d=3$ or $d=2$, a point $r_0$ on its boundary, $r_0 \in \partial C$, and a supporting plane (or line, in the twodimensional case) $\Pi$ to $C$ at $r_0$. We consider a part of the surface $\partial C$ containing $r_0$ and bounded by some plane parallel to $\Pi$. We are interested in the limiting properties of this part of the surface as the boundary plane tends to $\Pi$. A point $r_0 \in \partial C$ is said to be regular if the supporting plane to $C$ at this point is unique, and it is singular otherwise. It is known that regular points form a set of full measure in $\partial C$. In this paper we study mainly the case $d=3$. The twodimensional case $d=2$ is very simple and exhausted by Remark 2. We leave consideration of the case of higher dimension for the future. In what follows a convex compact set with nonempty interior is called a convex body. We let $e$ denote the outer unit normal to $\Pi$ and $\Pi_t$ denote the plane parallel to $\Pi$ and lying at distance $t>0$ from it on the side opposite to the normal. Thus, the plane $\Pi=\Pi_0$ has the equation $\langle rr_0, e \rangle=0$, while $\Pi_t$ has the equation $\langle rr_0, e \rangle= t$. The body $C$ is contained in the closed halfspace $ \{r \colon \langle rr_0, e\rangle \leqslant 0\}$. In what follows $\langle \cdot\,{,}\,\cdot \rangle$ denotes the scalar product. For each $t>0$ we consider the convex body
$$
\begin{equation*}
C_t=C \cap \bigl\{{r} \colon \langle rr_0, e\rangle \geqslant t\bigr\}.
\end{equation*}
\notag
$$
In other words, $C_t$ is the part of $C$ cut off from it by the plane $\Pi_t$. The boundary of $C_t$ is the union of the plane convex set
$$
\begin{equation}
B_t=C \cap \bigl\{{r} \colon \langle rr_0, e\rangle=t\bigr\}
\end{equation}
\tag{1}
$$
and the convex surface
$$
\begin{equation}
S_t=\partial C \cap \bigl\{{r} \colon \langle rr_0, e\rangle \geqslant t\bigr\};
\end{equation}
\tag{2}
$$
so $\partial C_t=B_t \cup S_t$. From now on, we let $\mathcal{A}$ denote the twodimensional Lebesgue measure of the Borel set $\mathcal{A}$ in the plane or in the convex surface $\partial C$. In particular, $\square ABCD$ denotes the area of the quadrilateral $ABCD$. Slightly misusing the notation, we denote similarly the length of a line segment or a curve; for example, $MN$ denotes the length of the line segment $MN$; in what follows the measure we mean is always clear from the context. We let ${n}_{r}$ denote the outer unit normal to $C$ at a regular boundary point ${r} \in \partial C$. A surface area measure of a convex body $C$ is a Borel measure $\nu_C$ in $S^2$ satisfying the equality
$$
\begin{equation}
\nu_C(\mathcal{A}) :=\{{r} \in \partial C \colon n_{r} \in \mathcal{A}\},
\end{equation}
\tag{3}
$$
where $\mathcal{A} \subset S^2$ is an arbitrary Borel set. (From here to Remark 1 we replace $S^2$ by $S^1$ in the twodimensional case.) The above definition of a surface area measure appeared apparently in A. D. Aleksandrov’s studies (see [1], § 2). A review of known results concerning the surface area measure and the corresponding references, as well as some new results, can be found in [2]. The following relation for the surface area measure (3) is well known (see [1], § 3, Lemma 2):
$$
\begin{equation}
\int_{S^2} n\, d\nu_C(n)=\vec 0.
\end{equation}
\tag{4}
$$
We let $\nu_t=\nu_{C,r_0,e,t}$ denote the normalized measure induced by the surface $S_t$. More precisely, by definition,
$$
\begin{equation}
\nu_t(\mathcal{A}) :=\frac{1}{B_t} \bigl\{r\in S_t \colon n_r\in\mathcal{A}\}\bigr
\end{equation}
\tag{5}
$$
for each Borel set $\mathcal{A} \subset S^2$. The surface area measure of the convex body $C_t$ is $\nu_{C_t}=B_t \delta_{e}+B_t \nu_t$; therefore,
$$
\begin{equation*}
\int_{S^2} n\, d\nu_{C_t}(n)=B_t \biggl(e+\int_{S^2} n\, d\nu_t(n)\biggr).
\end{equation*}
\notag
$$
From now on, $\delta_{e}$ denotes the unit atom concentrated at ${e}$. Applying (4) to $C_t$ we obtain
$$
\begin{equation}
\int_{S^2} n\, d\nu_t(n)=e.
\end{equation}
\tag{6}
$$
The measure $\nu_t$ is said to converge weakly to $\nu_*$ as $t \to 0$, which we denote by $\lim_{t\to0}\nu_t=\nu_*$, if
$$
\begin{equation*}
\lim_{t\to0} \int_{S^2} f(n)\, d\nu_t(n)=\int_{S^2} f(n)\, d\nu_*(n)
\end{equation*}
\notag
$$
for each continuous function $f$ on $S^2$. In a similar way, $\nu_*$ is a weak partial limit of $\nu_t$ if there exists a sequence of positive numbers $t_i, i \in \mathbb{N}$, converging to zero such that
$$
\begin{equation*}
\lim_{i\to\infty} \int_{S^2} f(n)\, d\nu_{t_i}(n)=\int_{S^2} f(n)\, d\nu_*(n)
\end{equation*}
\notag
$$
for each continuous function $f$ on $S^2$. In this paper we are going to study the limiting properties of the measure $\nu_t$ as ${t \to 0}$. One property of this kind can be deduced immediately. Let $\nu_*$ be the weak limit or a weak partial limit of the measures $\nu_t$. Passing to the limit as $t \to 0$ in (6) or as $i \to \infty$, we arrive at the relation
$$
\begin{equation}
\int_{S^2} n\, d\nu_*(n)=e.
\end{equation}
\tag{7}
$$
Remark 1. This research has been prompted by extremal problems in classes of convex bodies, in particular, by Newton’s problem of a convex body with minimum resistance (see [3]–[9]). It is natural to try to develop a method of small variations for such problems; the simplest variation of this kind is perhaps to cut off a small part of the body by a plane. This approach has turned out to be rather efficient in the case of Newton’s problem (see [4]):  find the least value of the functional
$$
\begin{equation}
\iint_\Omega \frac{1}{1+\nabla u(x,y)^2}\, dx\,dy
\end{equation}
\tag{8}
$$
in the class of concave functions $u\colon \Omega \!\to\! \mathbb R$ satisfying the inequality ${0 \!\leqslant\! u \!\leqslant\! M}$, where $\Omega \subset \mathbb R^2$ is a plane convex body and $M>0$. The results of this paper were announced in [10] for the special case when $r_0$ is a ridge singular point; and then they were used to prove the following assertion. If $u$ minimizes the functional (8) and the maximum level set $L_M=\{(x,y)$: $u(x,y)=M\}$ has a nonempty interior, then the slope of the lateral part of the graph of $u$ is $1$ at almost all boundary points of $L_M$. In other words, for almost all $(x, y) \in \partial L_M$
$$
\begin{equation*}
\lim_{\stackrel{(x',y')\to(x, y)}{u(x',y')<M}}\nabla u(x',y')=1.
\end{equation*}
\notag
$$
This is a partial proof of a conjecture stated (with no restriction on the maximum level set) in the mid1990s. Remark 2. The limiting behaviour of $\nu_t$ (see (5)) is rather simple for $d=2$. To be precise, assume that the tangent angle to $C \subset \mathbb R^2$ at $r_0$ (that is, the least angle with vertex at $r_0$ containing $C$) is specified by the inequalities
$$
\begin{equation*}
\langle rr_0, e_1 \rangle\leqslant0\quad\text{and} \quad \langle rr_0, e_2 \rangle\leqslant0, \qquad e_1=1, \quad e_2=1,
\end{equation*}
\notag
$$
and $e$ is specified by the formula
$$
\begin{equation*}
e=\lambda_1 e_1+\lambda_2 e_2, \qquad \lambda_1 \geqslant 0, \quad \lambda_2 \geqslant 0, \quad e=1.
\end{equation*}
\notag
$$
Hence $e_1$ and $e_2$ are the outer normal vectors to the sides of the angle and $e$ is the outer normal to a supporting line at $r_0$. Then it is true that
$$
\begin{equation*}
\lim_{t\to 0} \nu_t=\lambda_1 \delta_{e_1}+\lambda_2 \delta_{e_2}.
\end{equation*}
\notag
$$
In the special case when $r_0$ is a regular point we have $e_1=e_2=e$. In the case when the point is singular, $e_1 \ne e_2$, it can turn out that $e$ coincides with one of the vectors $e_1$ and $e_2$. In both cases the limit measure is an atom:
$$
\begin{equation*}
\lim_{t\to 0} \nu_t=\delta_{e}.
\end{equation*}
\notag
$$
The proof of these assertions is simple and is left to the reader. In what follows we focus on the threedimensional case $d=3$. Convergence is always understood as weak convergence, which is not specifically indicated. By definition, the tangent cone to a convex body $C$ at a point $r_0 \in \partial C$ is the least closed cone with vertex $r_0$ containing $C$. In other words, it is the closure of the union of rays with vertex $r_0$ the intersection of which with the interior of $C$ is nonempty. Clearly, the tangent cone is a convex set. If the tangent cone is a halfspace, then the point $r_0$ is regular. If the tangent cone is a dihedral angle, then $r_0$ is called a ridge singular point. If the tangent cone is proper, that is, does not contain lines, then $r_0$ is called a conical singular point (see, for example, [11]). In particular, if $C$ is a polyhedron, then all interior points of its faces are regular points, interior points of its edges are ridge points, and all vertices are conical points. We let $N(r_0)$ denote the set of outer unit normals to the supporting planes at $r_0$. If $r_0$ is a regular point, then $N(r_0)$ is a onepoint set in $S^2$. If it is a ridge point and the outer normals to the corresponding dihedral angle are $e_1$ and $e_2$, then the dihedral angle has the form
$$
\begin{equation}
\langle rr_0, e_1 \rangle \leqslant 0, \qquad \langle rr_0, e_2 \rangle \leqslant 0,
\end{equation}
\tag{9}
$$
and $N(r_0)$ is the minor arc of the great circle in $S^2$ with endpoints at $e_1$ and $e_2$. If $r_0$ is a conical point, then $N(r_0)$ is a spherically convex set^{1}^{[x]}^{1}A set $N \subset S^2$ is called spherically convex, if $\{ \lambda x : \lambda \geqslant 0,\, x \in N \} \subset \mathbb R^3$ is convex. in $S^2$ with nonempty relative interior. We let $\partial N(r_0)$ denote the relative boundary of $N(r_0)$ in $S^2$. If $r_0$ is a regular or a ridge point, then $\partial N(r_0)=N(r_0)$. A point in $N(r_0)$ is said to be extreme if it is not an interior point of an arc of a great circle remove which is entirely contained in $N(r_0)$. All extreme points are clearly on $\partial N(r_0)$. It is clear that $e$ is in $N(r_0)$. Three cases (i)–(iii) are possible. (i) $e$ is an extreme point of $N(r_0)$. This is true when one of the following holds: Case (a) means that $\Pi$ is the tangent plane. Case (b) means that the tangent cone is a dihedral angle and $\Pi$ contains a face of this angle. Case (c) means that the tangent cone is proper and the plane $\Pi$ is tangent to the cone (the tangency can be one or twosided). (ii) $e$ lies in the interior of $N(r_0)$. In this case $r_0$ is a conical singular point. The tangent cone is proper, and the plane $\Pi$ intersects the cone at its vertex. (iii) $e$ is in $\partial N(r_0)$ but is not an extreme point. This is true when one of the following holds: In case (b) the points $e_1$ and $e_2$ are uniquely defined up to interchanging. The plane $\Pi$ intersects the cone in a line (in case (a)) or in a ray (in case (b)), but it is not tangent to the cone. It turns out that the limiting behaviour of the measure $\nu_t$ is different in these three cases. This behaviour is described in detail in Theorems 1–5 below. In case (i) the limit always exists and is equal to $\delta_e$. In case (ii) the limit of the measure also exists and is uniquely defined: it is proportional to the measure generated by the cutoff lateral surface of the tangent cone. Case (iii) is the most complicated and interesting one. In this case $e$ lies in $\partial N(r_0)$ and is an interior point of an arc $\Gamma$ of a great circle whose endpoints $e_1$ and $e_2$ are extreme points of $N(r_0)$. It is possible that $\nu_t$ has no limit at all; however, the set of limit points is nonempty and each limit point is a measure whose support lies on $\Gamma$ and contains $e_1$ and $e_2$. In addition, the following statement is true, which is converse in a certain sense: assume that $e_1 \ne \pm e_2$; then any closed set containing $e_1$ and $e_2$ and lying in the minor arc of the great circle with endpoints $e_1$ and $e_2$ is the support of the limit of some family of measures $\nu_t$. We consider several simple examples. Example 1. Let $C$ be a convex polyhedron, and let the intersection of the supporting plane $\Pi$ and $C$ be a vertex $r_0$ of $C$. In this situation we have case (ii); the measure $\nu_t$ does not depend on $t$ and is a linear combination of atoms $\delta_{e_i}$, where the $e_i$ denote the outer normals to the faces incident to $r_0$. Example 2. Let $C$ be a unit ball in $\mathbb R^3$, and let $r_0$ be a point on its surface. In this situation we have case (i), (a); $e$ coincides with $r_0$, and the measure $\nu_t$ is uniformly distributed over a spherical cap of opening angle $\arccos(1t)$ with centre $e$. The measure $\nu_t$ converges weakly to $\delta_e$ as $t \to 0$. Example 3. Let $C$ be a convex polyhedron, $r_0$ be a point in one of its (closed) faces, and $e$ be the outer normal to this face. If $r_0$ is an interior point of the face, then we have case (i), (a); if $r_0$ is an interior point of an edge of this face, then this is case (i), (b); and if $r_0$ is a vertex, then this is case (i), (c). We let $i$ mark each face of the polyhedron adjacent to the above face. Then the measure $\nu_t$ is a linear combination of atoms or, more precisely,
$$
\begin{equation*}
\nu_t=c(t) \delta_e+\sum_i c_i(t) \delta_{e_i}.
\end{equation*}
\notag
$$
Here $c(t)$ is the ratio of the area of the face in question to the area of the intersection of $C$ and the plane $\Pi_t$, which tends to 1 as $t \to 0$; each coefficient $c_i(t)$ is the ratio of the area of the cutoff part of the $i$th face to the area of the intersection. For each $i$ we have $c_i(t)=O(t)$ as $t \to 0$. Thus, $\nu_t$ converges to $\delta_e$, not just weakly but also in norm. Example 4. Again, let $C$ be a convex polyhedron and $r_0$ be a point in one of its edges. Let $e_1$ and $e_2$ be the outer normals to the faces containing this edge, and let $e=\lambda_1 e_1+\lambda_2 e_2$, $e=1$, where $\lambda_1>0$ and $\lambda_2>0$. If $r_0$ is an interior point of the edge, then we have case (iii), (a) (Figure 1); if $r_0$ coincides with an endpoint of this edge (a vertex; the point $M$ or $N$ in Figure 1), then we have case (iii), (b). The surface $S_t$ in Figure 1 is the union of the two quadrilaterals $MNBA$ and $MNFD$ and the two triangles $BFN$ and $ADM$. The vectors $e_1$ and $e_2$ are the outer normals to $MNBA$ and $MNFD$, respectively; the area of each quadrilateral is of order $t$, that is, $\square MNBA=c_1 t+ O(t^2)$ and $\Box MNFD=c_2 t+O(t^2)$, $c_1>0$, $c_2>0$. The areas of the triangles $BFN$ and $ADM$ are $O(t^2)$. The plane set $B_t$ is the quadrilateral $ABFD$; the normal vector to it is $e$, and its area is of order $t$: $\square ABFD=c_0 t+O(t^2)$. Therefore, the corresponding measure $\nu_t$ converges to the measure $\nu_*=\frac{c_1}{c_0} \delta_{e_1}+\frac{c_2}{c_0} \delta_{e_2}$, the support of which is the twopoint set $\{e_1, e_2\}$. Using (7) we obtain $\frac{c_1}{c_0}=\lambda_1$ and $\frac{c_2}{c_0}=\lambda_2$; hence
$$
\begin{equation}
\nu_*=\lambda_1 \delta_{e_1}+\lambda_2 \delta_{e_2}.
\end{equation}
\tag{10}
$$
Example 5. Consider the cylinder $C=\{{r}=(x,y,z) \colon x^2+y^2 \leqslant 1,\, 0 \leqslant z \leqslant 1\}$ and the ridge point $r_0=(1,0,0) \in \partial C$ on its boundary. Then we have case (iii), (a). The tangent cone at $r_0$ is the dihedral angle with outer normals $e_1=(1,0,0)$ and $e_2=(0,0,1)$. Let $e= \lambda_1 e_1+\lambda_2 e_2$, where $\lambda_1>0$, $\lambda_2>0$ and $\lambda_1^2+\lambda_2^2=1$ (Figure 2). The surface $S_t$ is part of the surface $\partial C$ bounded by the plane $\langle r r_0, e\rangle=t$; it is the union of part of the cylinder surface $S_t^1$ and part of the lower end face of the cylinder $S_t^2$. The outer normals at points in $S_t^2$ coincide with $e_2$. The outer normals at points in the surface $S_t^1$ belong to some arc with midpoint $e_1$ in $S^2$, the length of which tends to 0 as $t \to 0$. Therefore, $\nu_t$ is the sum of two terms such that one term is proportional to $\delta_{e_2}$ and the other is proportional to a measure weakly converging to $\delta_{e_1}$. Using (7) we conclude that $\nu_t$ converges to the measure $\nu_*$ specified by (10). It can seem that the limit measure always exists in case (iii) and is specified by (10), like in the twodimensional case and Examples 4 and 5. However, it turns out that this is not true. We consider the following example. Example 6. Let $C$ be a part of a cylinder bounded by two planes:
$$
\begin{equation*}
C=\{(x,y,z) \colon z1 \leqslant x \leqslant z+1,\, y^2+z^2 \leqslant 1\}.
\end{equation*}
\notag
$$
Consider the ridge point $r_0=(0,0,1) \in \partial C$. The outer normals to the corresponding dihedral angle are $e_1=\frac{1}{\sqrt{2}}\, (1,0,1)$ and $e_2=\frac{1}{\sqrt{2}} \, (1,0,1)$. Let $e=(0,0,1)=\frac{1}{\sqrt{2}} \, e_1+\frac{1}{\sqrt{2}}\, e_2$ (Figure 3). Here we have $C_t=C \cap \{z \leqslant 1+t\}$, and $B_t$ is the rectangle $t \leqslant x \leqslant t$, $\sqrt{2tt^2} \leqslant y \leqslant \sqrt{2tt^2}$ in the plane $z=1+t$. The surface $S_t$ is the union of three parts: $S_t=S_t^1 \cup S_t^2 \cup S_t^0$, where $S_t^1$ is the plane domain with outer normal $e_1$ that is specified by the relations $x=z1$, $x \geqslant t$ and $(x+1)^2+y^2 \leqslant 1$ and $S_t^2$ is the plane domain with outer normal $e_2$ that is specified by the relations $x=z+1$, $x \leqslant t$ and $(x1)^2+y^2 \leqslant 1$. These surfaces are parts of ellipses lying in the planes $x=z 1$ and $x=z+1$, respectively. The surface $S_t^0$ is the graph of the function $z(x,y)=\sqrt{1y^2}$ with the domain
$$
\begin{equation*}
(1\sqrt{1y^2}) \leqslant x \leqslant 1\sqrt{1y^2},\qquad \sqrt{2tt^2} \leqslant y \leqslant \sqrt{2tt^2}
\end{equation*}
\notag
$$
(Figure 4). The set of outer normals at points in the $S_t^0$ is an arc in $S^2$ with midpoint $e$, whose length tends to 0 as $t \to 0$. It is straightforward to derive the following estimates for the areas of each of these three surfaces:
$$
\begin{equation*}
\begin{gathered} \, B_t=2t\cdot 2\sqrt{2tt^2}=4\sqrt 2 \, t^{3/2}(1+o(1)), \qquad t \to 0, \\ S_t^0=\frac{4\sqrt 2}{3}\, t^{3/2}(1+o(1))\quad\text{and} \quad S_t^1=S_t^2=\frac{8}{3}\, t^{3/2}(1+o(1)), \qquad t \to 0. \end{gathered}
\end{equation*}
\notag
$$
It follows that $\nu_t$ converges to the measure
$$
\begin{equation*}
\nu_*=\frac{\sqrt{2}}{3} \, \delta_{e_1}+\frac{\sqrt{2}}{3} \, \delta_{e_2}+\frac{1}{3} \delta_{e},
\end{equation*}
\notag
$$
whose support is the threepoint set $\{e_1, e_2, e\}$. Thus, the limit measure is not necessarily the sum of two atoms of the form (10). In addition, it turns out that $\nu_t$ can have no limit at all, as the next example shows. Example 7. We define $C$ by
$$
\begin{equation*}
C=\{(x,y,z) \colon z \leqslant x \leqslant z,\ \gamma(y) \leqslant z \leqslant 1\},
\end{equation*}
\notag
$$
where $\gamma \colon \mathbb R \to \mathbb R$ is a convex even function, $\gamma(0)=\gamma'(0)=0$, and $\gamma(y)>0$ for $y \ne 0$. The point $r_0=(0,0,0)$ is a ridge point, and the corresponding tangent cone is specified by inequalities (9), where $e_1=\frac{1}{\sqrt{2}}\, (1,0,1)$ and $e_2= \frac{1}{\sqrt{2}}\, (1,0,1)$. Let $e=(0,0,1)=\frac{1}{\sqrt{2}}\, e_1+\frac{1}{\sqrt{2}}\, e_2$. Then $e_1$, $e_2$ and $e$ are the same as in Example 6. We have $C_t=C \cap \{z \leqslant t\}$, and $B_t$ is the rectangle $t \leqslant x \leqslant t$, $\gamma(y) \leqslant t$ in the plane $z=t$. The surface $S_t$ is a union of three parts: $S_t=S_t^1 \cup S_t^2 \cup S_t^0$, where $S_t^1$ is the plane domain with outer normal $e_1$ specified by $x=z$, $t \leqslant x \leqslant \gamma(y)$, $S_t^2$ is the plane domain with outer normal $e_2$ specified by $x=z$, $\gamma(y) \leqslant x \leqslant t$, and $S_t^0$ is the graph of the function $z(x,y)=\gamma(y)$ with domain $\gamma(y) \leqslant x \leqslant \gamma(y)$, $\gamma(y) \leqslant t$. The outer normals to $S_t^0$ lie in some neighbourhood of $e$ of diameter tending to 0 as $t \to 0$. The projections of $S_t^1$, $S_t^2$, and $S_t^0$ onto the $xy$plane look roughly the same as in Figure 4. If the family of points $\bigl(S_t^/B_t, S_t^+/B_t, S_t^0/B_t \bigr)$ has a partial limit $(b^, b^+, b^0)$ as $t \to 0$, then the family of measures $\nu_t$ has the corresponding partial limit $\nu_*=b^ \delta_{e_1}+b^+\delta_{e_2}+b^0 \delta_{e}$. Assume that the graph of the function $x=\gamma(y)$, $y>0$, is a polygonal line with infinitely many edges. We define its vertices $(x_i, y_i)$, $i \geqslant i_0$, recursively as follows. Fix $0<a<b<1$. The initial values $x_{i_0}>0$ and $y_{i_0}>0$ are chosen arbitrarily. Assume that $(x_i, y_i)$ has already been specified. Then we set $x_{i+1}=x_i/i$; if $i$ is even, then we set $y_{i+1}=ay_i$, whereas if $i$ is odd, then $y_{i+1}=by_i$. The initial value $i_0$ is chosen to be sufficiently large for the resulting function $\gamma$ to be convex. Setting $t=x_i$ we obtain $B_{x_i}=4x_i y_i$; if $i$ is even, then
$$
\begin{equation*}
\begin{gathered} \, S_{x_i}^0=4 \cdot \frac{1a}{2}\, x_i y_i (1+o(1))\quad\text{and}\quad S_{x_i}^1= S_{x_i}^2=2\sqrt 2 \cdot \frac{1+a}{2}\, x_i y_i (1+o(1)), \\ i \to \infty, \end{gathered}
\end{equation*}
\notag
$$
whereas if $i$ is odd, then
$$
\begin{equation*}
\begin{gathered} \, S_{x_i}^0=4 \cdot \frac{1b}{2}\, x_i y_i (1+o(1))\quad\text{and}\quad S_{x_i}^1= S_{x_i}^2=2\sqrt 2 \cdot \frac{1+b}{2}\, x_i y_i (1+o(1)), \\ i \to \infty. \end{gathered}
\end{equation*}
\notag
$$
Therefore, $\nu_t$ has at least two partial limits of the form
$$
\begin{equation*}
\nu_*^1=\lim_{\stackrel{i\text{ is even}}{i\to\infty}} \nu_{x_i}=\frac{1+a}{2\sqrt 2} (\delta_{e_1}+\delta_{e_2})+\frac{1a}{2}\delta_e
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\nu_*^2=\lim_{\stackrel{i\text{ is odd}}{i\to\infty}} \nu_{x_i}=\frac{1+b}{2\sqrt 2} (\delta_{e_1}+\delta_{e_2})+\frac{1b}{2}\delta_e.
\end{equation*}
\notag
$$
We introduce some additional notation. Let $e_1$ and $e_2$ be unit vectors such that $e_1 \ne \pm e_2$. We let $\Gamma_{e_1,e_2}$ denote the minor arc of the great circle in $S^2$ with endpoints $e_1$ and $e_2$. Let $r_0$ be a conical point, and let $e$ be an interior point of the set $N(r_0)$. The planes $\Pi_t$ cut off similar parts from the tangent cone. We choose $t$ so that the intersection of $\Pi_t$ and the tangent cone has area 1. We let $\nu_\star$ denote the surface area measure induced by the cutoff part of the cone surface. It has support in the set $\partial N(r_0)$ and satisfies
$$
\begin{equation*}
\int_{S^2} n\, d\nu_\star(n)=e.
\end{equation*}
\notag
$$
So we understand the local structure of a convex surface as the limiting behaviour of the measure $\nu_t$ induced by the cutoff part of the surface. The cutoff is performed by a plane with outer normal vector $e$. This limiting behaviour depends on the position of $e$ in the set $N(r_0)$, as can be seen from the following theorems. In Theorems 1–3 the behaviour depends on the vector itself and the tangent cone. In Theorem 4 the situation is not that clear. Theorem 1. If $e$ is an extreme point of the set $N(r_0)$, then
$$
\begin{equation}
\lim_{t\to 0} \nu_t=\delta_{e}.
\end{equation}
\tag{11}
$$
Theorem 2. If $e$ is an interior point of the set $N(r_0)$ and thus the point $r_0$ is conical, then
$$
\begin{equation}
\lim_{t\to 0} \nu_t=\nu_\star.
\end{equation}
\tag{12}
$$
Hence, according to Theorems 1 and 2, if $e$ is an extreme or an interior point of the set $N(r_0)$, then the limit measure is defined uniquely. If $e$ is in $\partial N(r_0)$ but it is not an extreme point of $N(r_0)$, then the convergence of the measure $\nu_t$ depends on whether the set $B_t$ at $t=0$ is a nondegenerate segment or degenerates into a point. In the first case the limit exists and is the sum of two atoms; in the second case convergence is not guaranteed and we manage only to describe the properties of the partial limits. An answer is given by Theorems 3 and 4 below. Theorem 3. Assume that $e$ lies on $\partial N(r_0)$ and it is not an extreme point of $N(r_0)$; thus, $e=\lambda_1 e_1+\lambda_2 e_2$, $\lambda_1>0$, $\lambda_2>0$, where $e_1$ and $e_2$ are extreme points of $N(r_0)$. In this case the set $B_0$ is a line segment orthogonal to the vectors $e_1$ and $e_2$ and containing the point $r_0$. If this segment does not degenerate into a point, then
$$
\begin{equation}
\lim_{t\to 0} \nu_t=\lambda_1 \delta_{e_1}+\lambda_2 \delta_{e_2}.
\end{equation}
\tag{13}
$$
Theorem 4. Under the assumptions in Theorem 3, if $B_0$ is $r_0$, then the set of partial limits of $\nu_t$ is nonempty and each partial limit is a measure whose support lies on the arc $\Gamma_{e_1,e_2}$ and contains $e_1$ and $e_2$. Hence Theorem 4 specifies only some property of a partial limit of $\nu_t$. The following theorem is a converse of Theorem 4 in a certain sense: it states that this property is not only necessary but also sufficient for the corresponding limit measure to exist. Theorem 5. Consider unit vectors $e_1$, $e_2$ and $e$ such that $e_1 \ne \pm e_2$ and $e=\lambda_1 e_1+\lambda_2 e_2$ for some $\lambda_1>0$ and $\lambda_2>0$. Thus, $e$ is an interior point of the arc $\Gamma=\Gamma_{e_1,e_2}$. Let $K$ be a closed subset of the arc $\Gamma$ containing its endpoints $e_1$ and $e_2$, that is, $\{e_1, e_2\} \subset K \subset \Gamma$. Then there exists a convex body $C$ and a ridge point $r_0$ on its surface such that the tangent cone to $C$ at $r_0$ is specified by (9) and the measure $\nu_t=\nu_{C,r_0,e,t}$ converges to some measure $\nu_*$ with support $K$ as $t \to 0$. Theorems 1–5 are the main results of this paper. Summarizing and somewhat paraphrasing them, we can state that the limit measure exists and is uniquely determined by the plane and cone in the following cases: If the plane $\Pi$ intersects the cone in a ray or a line, but it is not tangent to the cone and its intersection with $C$ is a point, then we cannot guarantee that the limit measure exists. Partial limits of the measure always exist and satisfy certain conditions. It can also be proved that any measure satisfying these conditions is the limit of measures corresponding to a ridge point of some convex body. Remark 3. We believe that Theorem 5 can be strengthened in the following way. Proposition. Assume that $N \subset S^2$ is a closed spherically convex set lying on a hemisphere, the intersection of $N$ and the great circle bounding this hemisphere is an arc $\Gamma=\Gamma_{e_1,e_2}$, and $e$ is an interior point of $\Gamma$. Assume that a measure $\nu_*$ satisfies (7) and its support lies in $\Gamma$ and contains $e_1$ and $e_2$. Then there exists a convex body $C$ and a point $r_0 \in \partial C$ such that $N(r_0)=N$ and the measure $\nu_t=\nu_{C,r_0,e,t}$ converges to $\nu_*$ as $t \to 0$ . We leave the proof of this proposition for the future. Some particular assertions in Theorems 1–5 concerning the cases when the point $r_0$ is regular or conical and $e$ is an interior point of $N(r_0)$ were proved in [12]. Proofs of Theorems 1–5 are given in §§ 2–6.
§ 2. Proof of Theorem 1 First we prove some simple assertions. Lemma 1. If a plane convex set $D \subset \mathbb R^2$ contains a disc of radius $a$, then its perimeter $P$ and area $A$ satisfy
$$
\begin{equation}
P \leqslant \frac{2}{a} A.
\end{equation}
\tag{14}
$$
Proof. In fact, let $A(d\varphi)$ be the area of the subset of $D$ lying between two rays from the centre of the disc that bound an infinitesimal arc $d\varphi$ on the boundary of $D$, and let $p(d\varphi)$ be the length of this arc (Figure 5).
Then we have
$$
\begin{equation*}
A(d\varphi) \geqslant \frac{a}{2}\, p(d\varphi);
\end{equation*}
\notag
$$
hence
$$
\begin{equation*}
A=\int_0^{2\pi} A(d\varphi) \geqslant \frac{a}{2} \int_0^{2\pi} p(d\varphi)=\frac{a}{2}P.
\end{equation*}
\notag
$$
It follows that (14) holds. The lemma is proved. Lemma 2. Assume that a convex body $C$ in Euclidean space $\mathbb R^3$ with coordinates $(x_1, x_2, z)$ lies between the planes $z=0$ and $z=t$, $t>0$. Let $D$ denote the image of $C$ under the natural projection of $\mathbb R^3$ onto the $x$plane, $(x,z) \mapsto x$, and let $P=\partial D$ be its perimeter. Assume that for some $0<\varphi<\pi/2$ and some domain $\mathcal{U} \subset \partial C$, the outer normal $n_r=(n_{r1}, n_{r2}, n_{r3})$ at any regular point $r \in \mathcal{U}$ satisfies the inequality $n_{r3} \leqslant \cos\varphi$. (In other words, the tangent plane at all regular points of $\mathcal{U}$ forms an angle $\geqslant \varphi$ with the plane $z=0$.) Then
$$
\begin{equation*}
\mathcal{U} \leqslant \frac{2tP}{\sin\varphi}.
\end{equation*}
\notag
$$
Proof. The body $C$ is bounded below by the graph of a convex function $u_1$ and bounded above by the graph of a concave function $u_2$, that is,
$$
\begin{equation*}
C=\{(x,z) \colon u_1(x) \leqslant z \leqslant u_2(x)\}
\end{equation*}
\notag
$$
(Figure 6). Both functions are defined in the domain $D$. We let $\mathcal{U}_i$ $(i=1,2)$ denote the intersection of $\mathcal{U}$ and the graph of $u_i$. Clearly, if $(x, u_i(x))$ is a regular point in $\mathcal{U}_i$, then $\nabla u_i(x) \geqslant \tan\varphi$.
For $0 \leqslant z \leqslant t$ we let $l_z$ denote the length of the curve $\{x\colon u_1(x)=z\}$. We introduce the natural parameter $s \in [0, l_z]$ on each curve of this type. If the corresponding set is empty, then we set $l_z=0$. It is obvious that $l_z \leqslant P$. We let $x(z,s)$ denote the point corresponding to $s$ in this curve. Then the area of $\mathcal{U}_1$ is
$$
\begin{equation*}
\mathcal{U}_1=\int_0^t dz \int_0^{l_z} \sqrt{1+\frac{1}{\nabla u_1(x(z,s))^2}}\, ds \leqslant \int_0^t l_z \sqrt{1+\cot^2 \varphi}\, dz \leqslant \frac{tP}{\sin\varphi}.
\end{equation*}
\notag
$$
The same arguments hold for $\mathcal{U}_2$; therefore, $\mathcal{U}=\mathcal{U}_1+\mathcal{U}_2 \leqslant 2tP/\sin\varphi$. The lemma is proved. Lemma 3. If a convex set in a plane contains two mutually orthogonal line segments of length 1, then it also contains a circle of radius $c$, where $c$ is some positive constant. Proof. We consider two cases.
1. One of the line segments (we call it segment 1) intersects the straight line containing the other (segment 2). This line divides segment 1 into two smaller line segments such that the length of one of them (say, segment $1'$) is at least $1/2$. The second segment is divided by the first into two parts such that the length of one of them (segment $2'$) is at least $1/2$. The convex hull of the segments $1'$ and $2'$ is a rectangular triangle with legs of length $\geqslant 1/2$. This triangle contains a disc of radius $r_1=1/(4+2\sqrt 2)$.
2. The line containing one of the segments does not intersect the other. In an appropriate system of coordinates the segments have the form $[b1, b] \times \{0\}$ and $\{0\} \times[a 1, a]$, where $a> 1$ and $b>1$. Without loss of generality we assume that ${a \geqslant b}$.
We consider a triangle with base $[b1, b] \times \{0\}$ and the vertex $(0, a)$. Its area is $S=a/2$, and the perimeter $p$ satisfies the inequalities $p \leqslant 1+2\sqrt{a^2+b^2} \leqslant 1+ 2\sqrt 2 a$. Hence the radius $r$ of the circle inscribed in this triangle satisfies the relations
$$
\begin{equation*}
r=\frac{2S}{p} \geqslant \frac{a}{1+2\sqrt 2 a}>\frac{1}{1+2\sqrt 2}=r_2.
\end{equation*}
\notag
$$
Since $r_1<r_2$, we conclude that the convex set contains a disc of radius $c=r_1$.
The lemma is proved. Let $K \subset \mathbb R^2$ be a convex body. Recall that the tangent angle to $K$ at a point $p \in \partial K$ is the closure of the union of all rays with vertex $p$ intersecting the interior of $K$. If the point $p$ is regular, then the tangent angle is a halfplane. We let $K_\varepsilon$ denote the set of points inside $K$ whose distance to $\partial K$ is at least $\varepsilon$. In other words, $K_\varepsilon=K \setminus \mathcal{N}_\varepsilon(\partial K)$, where $\mathcal{N}_\varepsilon$ denotes the $\varepsilon$neighbourhood. Let $l$ be a supporting line to $K$ at some point $r^{\mathrm{o}} \in \partial K$, and let $\epsilon^{\mathrm{o}}$ be the outer normal to $l$. Thus,
$$
\begin{equation*}
l=\{r \colon \langle rr^{\mathrm{o}}, \epsilon^{\mathrm{o}} \rangle=0\}.
\end{equation*}
\notag
$$
We let $l_\delta$, $\delta>0$, denote the line parallel to $l$ and passing at distance $\delta$ from it on the side opposite to the vector $\epsilon^{\mathrm{o}}$. Its equation has the form
$$
\begin{equation*}
l_\delta=\{r \colon \langle rr^{\mathrm{o}}, \epsilon^{\mathrm{o}} \rangle=\delta\}.
\end{equation*}
\notag
$$
For sufficiently small $\delta>0$ the line $l_\delta$ intersects $K$. The intersection $l_\delta \,{\cap}\, K_\varepsilon$ is a line segment (possibly degenerating into a point or an empty set). We let $\beta_{\delta,\varepsilon}= l_{\delta} \cap K_\varepsilon$ denote the length of this segment. Lemma 4. Let $l$ contain one of the sides of the tangent angle to $K$ at $r^{\mathrm{o}}$, and let $\kappa>0$. Then $\beta_{(1+\kappa)\varepsilon,\varepsilon}/\varepsilon \to \infty$ as $\varepsilon \to 0$. Proof. In an appropriate system of coordinates $(x_1, x_2)$, $r^{\mathrm{o}}$ coincides with the origin $(0,0)$, the line $l$ is the $x_1$axis, the line $l_\delta$ is specified by the equation $x_2=\delta$, one side of the tangent angle is the ray $x_1 \geqslant 0$, $x_2=0$, and part of the lower boundary of $K$ is specified by a convex nonnegative function $x_2=\varphi(x_1)$ on an interval of the positive halfaxis containing 0 such that $\varphi'_+(0)=0$.
For sufficiently small $\varepsilon>0$ the equation $\varphi(\xi)=\kappa\varepsilon$ has a unique solution $\xi= \xi(\varepsilon)$ such that $\lim_{\varepsilon\to 0} \xi(\varepsilon)/\varepsilon=0$. The line segment $[\xi/2, \xi\varepsilon] \times \{ (1+\kappa)\varepsilon\}$ lies in $K_\varepsilon$ for sufficiently small $\varepsilon$; hence it lies in $l_{(1+\kappa)\varepsilon} \cap K_\varepsilon$ (Figure 7). It follows that $\beta_{(1+\kappa)\varepsilon,\varepsilon} \geqslant \xi/2\varepsilon$ and $\beta_{(1+\kappa)\varepsilon,\varepsilon}/\varepsilon \geqslant \xi/(2\varepsilon)1 \to \infty$ as $\varepsilon \to 0$.
The lemma is proved. We consider a convex body $C$ in $\mathbb R^3$. Throughout the rest of the proof of Lemma 6 we assume that $r_0 \in \partial C$ is a singular conical point and consider a vector $e$ lying in $\partial N(r_0)$ (which is not necessarily an extreme point of $N(r_0)$). A plane $\Pi$ contains the point $r_0$ and is orthogonal to $e$. We also do the following: we take a vector $\widehat e$ in the interior of $N(r_0)$ and let $\widehat\Pi_t$ denote the plane $\langle rr_0, \widehat e \rangle=t$. Thus, $\widehat\Pi_t$ lies at distance $t$ from $r_0$; its intersection with the tangent cone is the plane convex figure $\widehat K_t$, and $\widehat B_t=C \cap \widehat\Pi_t$ is a convex figure contained in $\widehat K_t$. The line $\widehat l_t=\Pi \cap \widehat\Pi_t$ is a supporting line to $\widehat K_t$. Lemma 5. A vector $e$ is an extreme point of $N(r_0)$ if and only if the line $\Pi \cap \widehat\Pi_t$ contains a side of some tangent angle to $\widehat K_t$. The proof is left to the reader. Recall that $\mathcal{A}_\delta$ denotes the set of points in $\mathcal{A}$ lying at least at distance $\delta$ from $\partial\mathcal{A}$, that is, $\mathcal{A}_\delta=\mathcal{A} \setminus \mathcal{N}_\delta(\partial\mathcal{A})$. Lemma 6. There exists a function $\tau=\tau(\delta)$, $\delta>0$, such that
$$
\begin{equation*}
\lim_{\delta\to 0} \frac{\delta}{\tau(\delta)}= 0
\end{equation*}
\notag
$$
and $\widehat B_\tau$ contains $(\widehat K_\tau)_\delta$. Proof. We place the origin at $r_0$; then the dilation of a set $\mathcal{A}$ with centre $r_0$ and coefficient $k$ is $k\mathcal{A}$. We have $\frac{1}{\tau} \widehat K_\tau=\widehat K_1$.
Assume that Lemma 6 is not valid; then there exist sequences of positive numbers $\delta_n$ and $\tau_n$ converging to zero such that $\delta_n/\tau_n$ converges to a positive number $c$ as ${n \to \infty}$ and $\widehat B_{\tau_n}$ does not contain $(\widehat K_{\tau_n})_{\delta_n}$. This means that there is a sequence of points $r_n \in \widehat K_{\tau_n}$ such that $\operatorname{dist}(r_n, \partial\widehat K_{\tau_n}) \geqslant \delta_n$ but $r_n \not\in \widehat B_{\tau_n}$.
The corresponding sequence of points $\frac{1}{\tau_n} r_n$ satisfies the relations
$$
\begin{equation*}
\frac{1}{\tau_n} r_n \in \widehat K_1\quad\text{and} \quad \operatorname{dist}\biggl(\frac{1}{\tau_n} r_n, \partial\widehat K_1\biggr) \geqslant \frac{\delta_n}{\tau_n}.
\end{equation*}
\notag
$$
Choosing a subsequence if necessary, we assume that $\frac{1}{\tau_n} r_n$ converges to a point $r^{\mathrm{o}} \in \widehat K_1$ such that $\operatorname{dist}(r^{\mathrm{o}}, \partial\widehat K_1) \geqslant c$; hence $r^{\mathrm{o}}$ is an interior point of $\widehat K_1$.
We draw a line through $r_n$ in the plane $\widehat\Pi_{\tau_n}$ so that one of the halfplanes bounded by this line does not intersect $\widehat B_{\tau_n}$. Let this halfplane (recall that it lies in $\widehat\Pi_{\tau_n}$) have the form $\langle rr_n, e_n \rangle \geqslant 0$, $e_n=1$. Again, choosing a subsequence if necessary, we can view $e_n$ as converging to some unit vector $e^{\mathrm{o}}$.
We take an arbitrary point $r$ in the intersection of the interior of $\widehat K_1$ and the open halfplane $\langle rr^{\mathrm{o}}, e^{\mathrm{o}} \rangle>0$ in $\widehat\Pi_{1}$. By the convergence $\frac{1}{\tau_n} r_n \to r^{\mathrm{o}}$ and $e_n \to e^{\mathrm{o}}$, we have $\langle r\frac{1}{\tau_n} r_n, e_n \rangle>0$ for sufficiently large $n$; thus, $\langle {\tau_n} rr_n, e_n \rangle>0$. The last inequality means that ${\tau_n} r$ is not in $\widehat B_{\tau_n}$, nor therefore in $C$. It follows that the whole of $\mathbb R^+_0 r$ intersects $C$ only at 0. We have obtained a contradiction with the fact that this ray lies in the interior of the tangent cone.
The lemma is proved. Lemma 7. Let $e$ be an extreme point of $N(r_0)$. Then the set $B_t$ contains two mutually orthogonal line segments of length $\beta_t$, where $\beta_t/t \to \infty$ as $t \to 0$. Proof. (a) Assume that $r_0$ is a regular point; then the set $N(r_0)$ consists of the unique point $e=n_{r_0}$.
The intersection of $B_t$ and any plane containing $r_0$ and orthogonal to $\Pi_t$ is a line segment whose length $\beta_t^\varphi$ satisfies the limit relation $\beta_t^\varphi/t \to \infty$ as $t \to 0$. (Here ${0 \leqslant \varphi<\pi}$ is the angle of rotation of the plane, which is used as a parameter for planes.) Taking two mutually orthogonal planes, for example, the ones corresponding to $\varphi=0$ and $\varphi=\pi/2$, we obtain a proof of the assertion for $\beta_t=\min \{ \beta_t^0,\beta_t^{\pi/2}\}$.
(b) Assume that $r_0$ is a ridge point; thus, $N(r_0)$ is an arc of a great circle and $e$ coincides with $e_1$ or $e_2$. We draw two planes containing $r_0$: one of them is orthogonal to the edge of the dihedral angle, while the other contains this edge and some interior point of the dihedral angle (for example, it is the bisector of this angle). The intersections of these planes with $B_t$ are two mutually orthogonal line segments of lengths $\beta_t^1$ and $\beta_t^2$ satisfying $\beta_t^i/t \to \infty$ as $t \to 0$, $i=1, 2$. We can set $\beta_t=\min \{ \beta_t^1,\beta_t^{2}\}$ again.
(c) Assume that $r_0$ is a conical point. Recall that the plane $\Pi$ contains $r_0$ and is orthogonal to $e$. Its intersection with the tangent cone is an angle with vertex $r_0$, which can degenerate into a ray.
We take the bisector of this angle (the ray $r_0 p$ in Figure 8) and draw a plane $\widetilde\Pi$ through it and some interior point of the tangent cone. The intersections of this plane with $C$, the tangent cone, and the plane $\Pi_t$ are, respectively, a convex plane figure, a tangent angle to it, and a line parallel to the fixed ray $r_0 p$ and passing at a distance proportional to $t$ from it. The intersection of $\widetilde\Pi$ and $B_t$ is a line segment ($b_t$ in Figure 8); let us show that its length $\beta_t^1$ satisfies $\beta_t^1/t \to \infty$ as $t \to 0$.
Indeed, each ray from $r_0$ in $\widetilde\Pi$ which lies in the open tangent angle intersects $C$ in some lie segment one endpoint of which is $r_0$. This means that it also intersects the segment $b_t$ for sufficiently small $t$. Given $\varepsilon>0$ smaller that the opening of the tangent angle, we consider two rays of this type from $r_0$ (the dashed lines in Figure 8) such that one of them is fixed, whereas the other one forms an angle of $\varepsilon$ with the ray $r_0 p$. For sufficiently small $t$, both rays intersect $b_t$ and the distance between the points of intersection is $\frac{ct}{\varepsilon} (1+o(1))$ as $ \varepsilon \to 0$. In view of the fact that the length $\beta_t^1$ of this segment majorizes this distance, letting $\varepsilon$ tend to zero we deduce the required relation. We take a line $l$ orthogonal to the above bisector and draw a plane $\widehat\Pi$ containing it and intersecting the tangent cone in a convex bounded figure (we denote it by $\widehat K$). We let $\widehat e$ denote the unit normal to the plane $\widehat\Pi$ pointing to the same side of it as $r_0$. The plane $\widehat\Pi_\tau$ has the equation $\langle rr_0, \widehat e \rangle=\tau$. Thus, $\widehat\Pi_\tau$ is parallel to $\widehat\Pi$ and lies at distance $\tau$ from $r_0$. Its intersection with the tangent cone is a convex bounded figure, which we denote by $\widehat K_\tau$, while $\widehat B_\tau=C \cap \widehat\Pi_\tau$ is a convex figure contained in $\widehat K_\tau$. The line $\widehat l_\tau=\Pi \cap \widehat\Pi_\tau$ is parallel to $l$ and is a supporting line to $\widehat K_\tau$. By Lemma 5 it contains a side of some tangent angle to $\widehat K_\tau$. For each $t>0$ we fix $\tau=\tau(t/2)$ as indicated in Lemma 6. The intersection of $B_t$ and the plane $\widehat\Pi_\tau$ coincides with the intersection of $\widehat B_\tau$ and the line $\widehat l_{\tau,ct}$ drawn in $\widehat\Pi_\tau$ parallel to $\widehat l_\tau$ and lying on the inner side at distance $ct$ from it. Here $c \geqslant 1$ is the parameter equal to the reciprocal of the sine of the angle between $e$ and $\widehat e$, that is, $c=1/\sqrt{1\langle e, \widehat e\, \rangle^2}$. By Lemma 6, $\widehat B_\tau$ contains $(\widehat K_\tau)_{t/2}$. Taking the parameters $\varepsilon=t/2$ and $\kappa=2c1$ in Lemma 4 we conclude that the intersection of $(\widehat K_\tau)_{t/2}$ and the line $\widehat l_{\tau,ct}$ is a line segment whose length satisfies the relation $\widehat K_\tau)_{t/2} \cap \widehat l_{\tau,ct}/t \to \infty$ as $t \to 0$. The segment $B_t \cap \widehat\Pi_\tau$ contains this segment. Therefore, its length $\beta_t^2$ also satisfies the relation $\beta_t^2/t \to \infty$ as $t \to 0$; in addition, it is orthogonal to the segment $B_t \cap \widetilde\Pi$ chosen above. Thus, the assertion is proved for $\beta_t=\min \{\beta_t^1,\beta_t^{2}\}$. Lemma 7 is proved. Recall that $S_t$ is the intersection of $\partial C$ and the halfspace $\{r \colon \langle rr_0, e \rangle \geqslant t\}$. For $\varphi \in (0, \pi/2)$, we let $S_{t,\varphi}$ denote the part of the surface $S_t$ at points $r$ in which we have $\langle n_r, e \rangle \leqslant \cos\varphi$. In other words, it is the set of points in $S_t$ such that the normals at them have slope at least $\varphi$ with respect to $e$. Lemma 8. If $e$ is an extreme point of the set $N(r_0)$, then
$$
\begin{equation*}
\frac{S_{t,\varphi}}{B_t} \to 0 \quad \textit{as} \ \ t \to 0
\end{equation*}
\notag
$$
for any $\varphi \in (0, \pi/2)$. Proof. We consider a system of coordinates $(x,z)$, $x=(x_1, x_2)$ in which the $x$plane coincides with $\Pi$ and the $z$axis is directed along the vector $e$. For some (sufficiently small) $t_0>0$ the intersection of $\Pi_t, t \leqslant t_0$, and $C$ is nonempty. For any regular point $r \in S_t, t \leqslant t_0$, and some $\varphi_0 \in (0, \pi/2)$ we have $\langle n_r, e \rangle \geqslant \cos\varphi_0$. Without loss of generality we can take $\varphi<\varphi_0$; then $\langle n_r, e \rangle \leqslant \cos\varphi$ for regular points $r \in S_{t,\varphi}$.
In the above system of coordinates the body $C_t$ lies between the planes $z=0$ and $z=t$. We let $D_t$ denote the image of $C_t$ under the natural projection $(x,z) \mapsto x$. The domain $D_t$ contains $B_t$ and is contained in the $(t\cot \varphi)$neighbourhood of $B_t$; thus, its perimeter is at most $P_t= \partial B_t+2\pi t\cot\varphi$.
Applying Lemma 2 to the body $C=C_t$ and the domain $\mathcal{U}=S_{t,\varphi}$, we deduce the inequality
$$
\begin{equation*}
S_{t,\varphi} \leqslant \frac{tP_t}{\sin\varphi}=t \frac{\partial B_t+2\pi t\cot\varphi}{\sin\varphi}.
\end{equation*}
\notag
$$
Owing to Lemmas 3 and 7, $B_t$ contains a circle of radius $c\beta_t$; hence, by Lemma 1
$$
\begin{equation*}
\partial B_t \leqslant \frac{2}{c\beta_t}B_t.
\end{equation*}
\notag
$$
Therefore,
$$
\begin{equation*}
\frac{S_{t,\varphi}}{B_t} \leqslant t \frac{\frac{2}{c\beta_t}B_t+2\pi t\cot\varphi}{\sin\varphi B_t} \leqslant \frac{2}{c\sin\varphi}\,\frac{t}{\beta_t}+\frac{2\cot\varphi}{c^2 \sin\varphi}\, \frac{t^2}{\beta_t^2} \to 0 \quad \text{as} \ \ t \to 0.
\end{equation*}
\notag
$$
Lemma 8 is proved. Now we complete the proof of Theorem 1. We let $\nu_S$ denote the surface area measure generated by the surface $S$. For any $\varphi \in (0, \pi/2)$ we have
$$
\begin{equation*}
\nu_t=\frac{1}{B_t}\nu_{S_{t,\varphi}}+\frac{1}{B_t}\nu_{S_{t} \setminus S_{t,\varphi}}.
\end{equation*}
\notag
$$
By Lemma 8 the measure $\frac{1}{B_t}\nu_{S_{t,\varphi}}$ tends to 0 as $t \to 0$. On the other hand each partial limit of $\frac{1}{B_t}\nu_{S_{t} \setminus S_{t,\varphi}}$ has support in the ‘disc’ of opening angle $\varphi$ around $e$ on $S^2$; thus, the same is also true for the support of each partial limit of the measure $\nu_t$. Taking account of the fact that $\varphi>0$ can be arbitrarily small and using equality (7), which is valid for each partial limit $\nu_*$, we conclude that the limit of $\nu_t$ exists and is equal to $\delta_e$. Theorem 1 is proved.
§ 3. Proof of Theorem 2 In the proof of Theorem 2 we use the fact that the surface area measure is continuous in the Hausdorff topology in the space of convex bodies. More precisely, a family of convex bodies $C_t$, $t>0$, in $\mathbb R^d$ is said to converge to a convex body $C \subset \mathbb R^d$ as $t \to 0$ in the sense of Hausdorff, denoted by $C_t \xrightarrow[t\to0]{} C$, if for any $\varepsilon>0$ there is $t_0>0$ such that $C_t$ lies in the $\varepsilon$neighbourhood of $C$ and $C$ lies in the $\varepsilon$neighbourhood of $C_t$ for all $t \leqslant t_0$. It is known that if $C_t \xrightarrow[t\to0]{} C$, then $\nu_{\partial C_t} \to \nu_{\partial C}$ as $t \to 0$ (see, for example, the paragraph before (4.2.12) in [13], where the surface area measure of a convex body $K$ is denoted by $S_{n1}(K, \cdot)$; this property is noted as a simple consequence of Theorem 4.2.1 in that book). Recall that $\Pi_t$ denotes the plane $\langle rr_0, e \rangle=t$. The intersection of this plane with the tangent cone is a convex plane figure, which we denote by $K_t$, and $B_t=C \cap \Pi_t$ is a convex figure contained in $K_t$. We choose $\sigma>0$ such that $K_{\sigma}=1$. Recall that the measure $\nu_\star$ is generated by the part of the surface of the tangent cone that is cut off by the plane $\Pi_{\sigma}$. Assume that the origin coincides with $r_0$, that is, $r_0=\vec 0$; then the dilation of a set $\mathcal{A}$ with centre $r_0$ and coefficient $k$ is $k\mathcal{A}$ (Figure 9). Lemma 9. The following limit relation holds:
$$
\begin{equation*}
\frac{\sigma}{t} B_t \xrightarrow[t\to0]{} K_{\sigma}.
\end{equation*}
\notag
$$
Proof. For any positive numbers $t_1$ and $t_2$ we have
$$
\begin{equation*}
\frac{t_1}{t_2}\Pi_{t_2}=\Pi_{t_1}\quad\text{and} \quad \frac{t_1}{t_2}K_{t_2}=K_{t_1}.
\end{equation*}
\notag
$$
In addition, since $K_t$ contains $B_t$,
$$
\begin{equation*}
\frac{\sigma}{t}B_t \subset \frac{\sigma}{t}K_t=K_\sigma.
\end{equation*}
\notag
$$
Let $0<t_1 \leqslant t_2$. Since $\vec 0$ and $B_{t_2}$ lie in $C$, the same is valid for their linear combinations:
$$
\begin{equation*}
\frac{t_1}{t_2}B_{t_2}=\biggl(1\frac{t_1}{t_2}\biggr) \vec 0+\frac{t_1}{t_2}B_{t_2} \subset C.
\end{equation*}
\notag
$$
On the other hand $\frac{t_1}{t_2}B_{t_2} \subset \frac{t_1}{t_2}\Pi_{t_2}=\Pi_{t_1}$. Consequently, $\frac{t_1}{t_2}B_{t_2} \subset C \cap \Pi_{t_1}=B_{t_1}$. We conclude that
$$
\begin{equation*}
\frac{\sigma}{t_2}B_{t_2} \subset \frac{\sigma}{t_1}B_{t_1};
\end{equation*}
\notag
$$
hence the figures $\frac{\sigma}{t} B_t$, $t>0$, form a nested family contained in $K_{\sigma}$.
Assume that the bodies $\frac{\sigma}{t} B_t$ do not converge to $K_{\sigma}$. Then the closure of the union
$$
\begin{equation*}
\overline{\bigcup_{t>0}\frac{\sigma}{t} B_t}=: \widetilde B_{\sigma}
\end{equation*}
\notag
$$
lies in $K_{\sigma}$ but does not coincide with it.
The union
$$
\begin{equation*}
\bigcup_{t>0}\frac{t}{\sigma} \widetilde B_{\sigma}=: L
\end{equation*}
\notag
$$
is a cone with vertex $r_0$; it is contained in the tangent cone but does not coincide with it. On the other hand
$$
\begin{equation*}
C=\bigcup_{t\geqslant0}B_t \subset \bigcup_{t\geqslant0}\frac{t}{\sigma} \widetilde B_{\sigma}=L;
\end{equation*}
\notag
$$
hence $C$ lies in the cone $L$, which is smaller than the tangent cone. This contradiction implies our assertion.
Lemma 9 is proved. It follows from Lemma 9, in particular, that
$$
\begin{equation}
\lim_{t\to0} \biggl\frac{\sigma}{t} B_t\biggr=K_{\sigma}=1;
\end{equation}
\tag{15}
$$
therefore,
$$
\begin{equation}
\nu_{\frac{\sigma}{t} B_t}=\biggl\frac{\sigma}{t} B_t\biggr \delta_{e} \longrightarrow K_{\sigma} \delta_{e}=\nu_{K_{\sigma}}, \qquad t \to 0.
\end{equation}
\tag{16}
$$
We introduce the notation
$$
\begin{equation*}
\Sigma^t_\sigma :=\operatorname{conv}\biggl(\frac{\sigma}{t}B_t \cup r_0\biggr).
\end{equation*}
\notag
$$
In view of the fact that the convex body $\frac{\sigma}{t}C_t$ contains $r_0$ and $\frac{\sigma}{t}B_t$, we conclude that $\Sigma^t_\sigma \subset \frac{\sigma}{t}C_t$. We let $W_\sigma$ denote the part of the tangent cone containing $r_0$ that is cut off by the plane $\Pi_\sigma$ and let $V_\sigma$ denote its lateral surface. We have $W_{\sigma}=\operatorname{conv}(K_{\sigma} \cup r_0)$. According to Lemma 9, $\frac{\sigma}{t} B_t \xrightarrow[t\to0]{} K_{\sigma}$, which implies that
$$
\begin{equation*}
\operatorname{conv}\biggl(\frac{\sigma}{t}B_t \cup r_0\biggr) \xrightarrow[t\to0]{} \operatorname{conv}(K_{\sigma} \cup r_0);
\end{equation*}
\notag
$$
hence
$$
\begin{equation*}
\Sigma^t_\sigma \xrightarrow[t\to0]{} W_{\sigma}.
\end{equation*}
\notag
$$
Using this relation and the double embedding
$$
\begin{equation*}
\Sigma^t_\sigma \subset \frac{\sigma}{t}C_t \subset W_\sigma,
\end{equation*}
\notag
$$
we deduce that $\frac{\sigma}{t}C_t$ converges to $W_\sigma$ in the Hausdorff sense. Consequently, it is true that
$$
\begin{equation*}
\nu_{\frac{\sigma}{t}\partial C_t} \to \nu_{\partial W_\sigma} \quad \text{as} \ \ t \to 0.
\end{equation*}
\notag
$$
Taking account of the fact that $\frac{\sigma}{t}\partial C_t=\frac{\sigma}{t} S_t \cup \frac{\sigma}{t} B_t$ and $\partial W_\sigma=V_\sigma \cup K_\sigma$ and using (16) we obtain
$$
\begin{equation*}
\nu_{\frac{\sigma}{t} S_t} \to \nu_{V_\sigma} \quad \text{as} \ \ t \to 0;
\end{equation*}
\notag
$$
by (15) we arrive at the relation
$$
\begin{equation*}
\lim_{t\to0} \nu_t=\lim_{t\to0} \frac{1}{B_t}\nu_{S_t}= \frac{1}{\lim_{t\to0}\frac{\sigma}{t}B_t} \lim_{t\to0} \nu_{\frac{\sigma}{t}S_t}= \nu_{V_\sigma}=\nu_\star.
\end{equation*}
\notag
$$
Theorem 2 is proved.
§ 4. Proof of Theorem 3 Proceeding to the proof of Theorem 3 we recall that its assumptions can be fulfilled when or First we estimate the area of the plane domain $B_t$. The set $B_0$ is a (nondegenerate) line segment. Let $l>0$ be the length of this segment and let $M$ and $N$ be its endpoints. In the case when the point $r_0$ is conical, one of these endpoints is $r_0$. Then we change the notation by letting $r_0$ denote any interior point of the line segment $MN$. This is a ridge point. We thus find ourselves within the assumptions of part (a), where the corresponding sets $B_t$ and $S_t$ remain the same; therefore, the limiting behaviour of the measure $\nu_t$ does not change. So, without loss of generality, we regard $r_0$ as a ridge point. We let $l_i^t$, $i=1,2$, denote the intersection of the plane $\Pi_i\colon\langle rr_0, e_i \rangle=0$ and the cutting plane $\Pi^t\colon\langle rr_0, e \rangle=t$. In other words, $l_i^t=\Pi_i \cap \Pi^t$. We draw planes orthogonal to $B_0$ through the points $M$ and $N$, and let $M_i=M_{i}^t$ and $N_i=N_{i}^t$, $i=1,2$, denote the points of intersection of these planes and the lines $l_i^t$ (Figure 10, (a)). The triangle $M_1^t M_2^t M$ lies in a plane parallel to the vectors $e_1$ and $e_2$. Its sides $MM_1^t$, $MM_2^t$, and $M_1^t M_2^t$ are orthogonal to the vectors $e_1$, $e_2$, and $e$, respectively; their lengths are $MM_1^t=\lambda_1 c_0 t$, $MM_2^t=\lambda_2 c_0 t$ and $M_1^t M_2^t=c_0 t$ where $c_0$ is a positive constant depending only on $e_1$, $e_2$ and $e$; more precisely,
$$
\begin{equation}
c_0=\frac{\langle e_1, e\rangle}{\sqrt{1\langle e_1, e\rangle^2}}+\frac{\langle e_2, e\rangle}{\sqrt{1\langle e_2, e\rangle^2}}
\end{equation}
\tag{17}
$$
(Figure 10, (b)). The set $B_t$ lies between the lines $l_1^t$ and $l_2^t$ (Figure 11). We are going to prove that the length of the orthogonal projection of $B_t$ onto $l_1^t$ is at most $l+o(1)$ as ${t \to 0}$; hence
$$
\begin{equation}
B_t \leqslant l c_0 t(1+o(1)), \qquad t \to 0.
\end{equation}
\tag{18}
$$
Assume the opposite. Then there exist a sequence $(t_i)_{i \in \mathbb{N}}$ converging to 0 and a sequence of points $P_{t_i} \in B_{t_i}$ such that the distance between $P_{t_i}$ and the rectangle $M_1^{t_i} N_1^{t_i} N_2^{t_i} M_2^{t_i}$ remains above some positive constant. The sequence $(P_{t_i})_{i \in \mathbb{N}} \subset C$ is bounded, and thus it has at least one limit point $P$. This point is in $C$ and lies in the plane $\Pi^0\colon\langle rr_0, e\rangle=0$; consequently, $P \in B_0$. However, on the other hand the distance between $P$ and $B_0=MN$ is larger than some positive constant. We have obtained a contradiction. We take a positive number $\varepsilon<l$ and let $M_\varepsilon$ and $N_\varepsilon$ denote the points on the line segment $B_0= [M, N]$ lying at a distance of $\varepsilon/2$ from $M$ and $N$, respectively. Let $\Pi^\perp_{M\varepsilon}$ and $\Pi^\perp_{N\varepsilon}$ be the planes orthogonal to $MN$ that contain $M_\varepsilon$ and $N_\varepsilon$ respectively. We let $M_{\varepsilon 1}=M_{\varepsilon 1}^t$, $N_{\varepsilon 1}= N_{\varepsilon 1}^t$, $M_{\varepsilon 2}=M_{\varepsilon 2}^t$ and $N_{\varepsilon 2}=N_{\varepsilon 2}^t$ denote the points of intersection of these planes and the lines $l_1^t$ and $l_2^t$. The area of the rectangle $M_{\varepsilon 1} N_{\varepsilon 1} N_{\varepsilon 2} M_{\varepsilon 2}$ is obviously $ \square M_{\varepsilon 1} N_{\varepsilon 1} N_{\varepsilon 2} M_{\varepsilon 2}=c_0 (l\varepsilon) t$. In view of the fact that the tangent cones at $M_\varepsilon$ and $N_\varepsilon$ coincide with the dihedral angle $\langle rr_0, e_1\rangle \leqslant 0,\ \langle rr_0, e_2\rangle \leqslant 0$ formed by the planes $\Pi_1$ and $\Pi_2$, we conclude that the tangent cone to the twodimensional convex figure $C \cap \Pi^\perp_{M\varepsilon}$ in the plane $\Pi^\perp_{M\varepsilon}$ is the angle $M_{\varepsilon 1} M_\varepsilon M_{\varepsilon 2}$, whereas the tangent cone to the twodimensional convex figure $C \cap \Pi^\perp_{N\varepsilon}$ in the plane $\Pi^\perp_{N\varepsilon}$ is the angle $N_{\varepsilon 1} N_\varepsilon N_{\varepsilon 2}$ (Figure 12). Hence the intersection of $C$ and the line $M_{\varepsilon 1} M_{\varepsilon 2}$ (which is nonempty for sufficiently small $t$) is a segment of $M_{\varepsilon 1} M_{\varepsilon 2}$ (we denote it by $\widetilde M_{\varepsilon 1} \widetilde M_{\varepsilon 2}$) such that the distances $M_{\varepsilon 1} \widetilde M_{\varepsilon 1}$ and $M_{\varepsilon 2} \widetilde M_{\varepsilon 2}$ are $o(t)$. In a similar way, for sufficiently small $t$ the intersection of $C$ and the line $N_{\varepsilon 1} N_{\varepsilon 2}$ is a segment $\widetilde N_{\varepsilon 1} \widetilde N_{\varepsilon 2}$ of $N_{\varepsilon 1} N_{\varepsilon 2}$, and the distances $N_{\varepsilon 1} \widetilde N_{\varepsilon 1}$ and $N_{\varepsilon 2} \widetilde N_{\varepsilon 2}$ are $o(t)$. The quadrilateral $\widetilde M_{\varepsilon 1} \widetilde N_{\varepsilon 1} \widetilde N_{\varepsilon 2} \widetilde M_{\varepsilon 2}$ is contained in $B_t$, and its area is $\square \widetilde M_{\varepsilon 1} \widetilde N_{\varepsilon 1} \widetilde N_{\varepsilon 2} \widetilde M_{\varepsilon 2}=c_0 (l\varepsilon) t (1+o(1))$ as $t \to 0$. Therefore,
$$
\begin{equation}
B_t \geqslant c_0 (l\varepsilon) t (1+o(1)), \qquad t \to 0.
\end{equation}
\tag{19}
$$
Taking account of the fact that $\varepsilon>0$ is arbitrary and using relations (18) and (19) we infer that
$$
\begin{equation}
B_t=c_0 l t (1+o(1)), \qquad t \to 0.
\end{equation}
\tag{20}
$$
Now we consider the surface $S_t$. The plane $\Pi_0$ specified by the equation $\langle {r r_0}, e_1e_2\rangle=0$ is the bisector of the dihedral angle; it contains the segment $B_0$ and divides $S_t$ into two parts: $S_t=S_t^1 \cup S_t^2$, where $S_t^1$ lies between $\Pi_1$ and $\Pi_0$, while $S_t^2$ lies between $\Pi_2$ and $\Pi_0$. We prove that: (i) the surface area measure of $S_t^1$ divided by $t$ converges weakly to $\lambda_1 c_0 l\delta_{e_1}$; (ii) the surface area measure of $S_t^2$ divided by $t$ converges weakly to $\lambda_2 c_0 l\delta_{e_1}$. Then we derive the assertion of Theorem 3 from (i), (ii) and formula (20). We prove only (i), since the proof of (ii) is completely similar. In what follows we use the following property: We fix a positive number $\varepsilon<l$; for each $t>0$ we consider several prisms each of which is bounded by the planes $\Pi_1$, $\Pi_2$, $\Pi^t$ and by the two planes orthogonal to $B_0$ at the endpoints. The great prism is the least prism that has these properties and contains $C_t$. The great prism is subdivided into the central prism and two lateral prisms. The central prism is bounded at the end faces by the planes $\Pi^\perp_{M\varepsilon}$ and $\Pi^\perp_{N\varepsilon}$ orthogonal to $B_0$; clearly, it lies in the great prism for sufficiently small $t$. The lateral prisms are obtained by removing the central prism from the great prism. Each of these is bounded by the plane $\Pi^\perp_{M\varepsilon}$ or $\Pi^\perp_{N\varepsilon}$ at one end face. The full surface area measure of the central prism is asymptotically proportional to $t$; more precisely, it is $c_0 (1+ \lambda_1+\lambda_2) (l\varepsilon) t+o(t)$, where the term $o(t)$ corresponds to the lateral surface of the prism. The full surface area measure of each lateral prism is
$$
\begin{equation*}
c_0 (1+\lambda_1+\lambda_2) \, \frac{\varepsilon t}2+o(t), \qquad t \to 0.
\end{equation*}
\notag
$$
The central prism is partitioned by the plane $\Pi_0$ into two parts,
$$
\begin{equation*}
\operatorname{Prism}_{\varepsilon,1}^t=\operatorname{Prism}_{\varepsilon,1}\quad\text{and} \quad \operatorname{Prism}_{\varepsilon,2}^t=\operatorname{Prism}_{\varepsilon,2};
\end{equation*}
\notag
$$
thus, each prism $\operatorname{Prism}_{\varepsilon,i}$, $i=1,2$, is bounded by the planes $\Pi_{i}$, $\Pi_{0}$, $\Pi^t$, $\Pi^\perp_{M\varepsilon}$ and $\Pi^\perp_{N\varepsilon}$. Correspondingly, each of the surfaces $S_t^1$ and $S_t^2$ is also subdivided into two parts by these planes: $S_t^i=S_t^{i,\varepsilon} \cup S_t^{i,\mathrm{lat}}$, where $S_t^{i,\varepsilon}$ lies in $\operatorname{Prism}_{\varepsilon,i}$ and $S_t^{i,\mathrm{lat}}$ lies in the union of the lateral prisms. The last fact means that the area of $S_t^{i,\mathrm{lat}}$ is at most the sum of the surface areas of the lateral prisms:
$$
\begin{equation*}
c_0 (1+\lambda_1+\lambda_2) \varepsilon t+o(t), \qquad t \to 0.
\end{equation*}
\notag
$$
Thus, we have decomposed $\nu_t$ in the form $\nu_t=\nu_{t,\varepsilon}^1+\nu_{t,\varepsilon}^2+ \nu_{t,\varepsilon}^{\mathrm{lat}}$, where
$$
\begin{equation*}
\nu_{t,\varepsilon}^{\mathrm{lat}}(S^2) \leqslant (1+\lambda_1+\lambda_2) \frac{\varepsilon}{l}+o(1), \qquad t \to 0.
\end{equation*}
\notag
$$
Now we estimate the area of $S_t^{i,1}$. First, with this aim in view, we compare the surfaces of the convex bodies $\operatorname{Prism}_{\varepsilon,1}$ and $C_t \cap\operatorname{Prism}_{\varepsilon,1}$. We calculate the surface area measure of the body $\operatorname{Prism}_{\varepsilon,1}$. Its surface is the union of the three rectangles lying in the planes $\Pi_1$, $\Pi^t$ and $\Pi_0$ that are lateral faces of the prism and the two triangles lying in the planes $\Pi^\perp_{M\varepsilon}$ and $\Pi^\perp_{N\varepsilon}$ that are the bases of the prism. The outer normals to the rectangles are $e_1$, $e$ and $\frac{e_2e_1}{e_2e_1}$, respectively. The sides of the rectangles parallel to the straight line $MN$ are of length $l\varepsilon$, whereas the lengths of the sides orthogonal to this line are, respectively,
$$
\begin{equation*}
M_1^t M=c_0 t \lambda_1, \qquad M_1^t D= c_0 t\frac{\lambda_1}{\lambda_1+\lambda_2}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
MD=c_0 t\sqrt{\lambda_1\lambda_2} \sqrt{1(\lambda_1+ \lambda_2)^{2}},
\end{equation*}
\notag
$$
where $MD$ is the bisector of the angle $M$ in the triangle $M_1^tM_2^tM$ (see Figure 10, (b)). Hence the areas of the rectangles are
$$
\begin{equation*}
c_0 t (l\varepsilon) \lambda_1, \qquad c_0 t (l\varepsilon) \frac{\lambda_1}{\lambda_1+\lambda_2}\quad\text{and} \quad c_0 t (l\varepsilon) \sqrt{\lambda_1\lambda_2} \sqrt{1 (\lambda_1+\lambda_2)^{2}},
\end{equation*}
\notag
$$
while the area of each lateral end face of the prism is equal to the area of the triangle $M_1^t MD$ and is thus of order $t^2$. It follows that the surface area measure of the body $\operatorname{Prism}_{\varepsilon,1}$ is
$$
\begin{equation*}
c_0 t (l\varepsilon)\biggl(\lambda_1 \delta_{e_1}+\frac{\lambda_1}{\lambda_1+\lambda_2}\, \delta_{e}+\sqrt{\lambda_1\lambda_2} \sqrt{1(\lambda_1+\lambda_2)^{2}}\, \delta_{\frac{e_2e_1}{e_2e_1}}\biggr)+o(t), \qquad t \to 0.
\end{equation*}
\notag
$$
The surface of the body $C_t \cap\operatorname{Prism}_{\varepsilon,1}$ is a disjoint union of several surfaces. The part of $\partial (C_t \cap\operatorname{Prism}_{\varepsilon,1})$ contained in $\Pi^t$ induced the measure $c_0 t (l\varepsilon) \frac{\lambda_1}{\lambda_1+\lambda_2}\delta_{e}+ o(t)$; for sufficiently small $t$ the part of $\partial (C_t \cap\operatorname{Prism}_{\varepsilon,1})$ contained in $\Pi_0$ coincides with the part of $\partial (\operatorname{Prism}_{\varepsilon,1}$) contained in $\Pi_0$ and therefore induces the measure $c_0 t (l\varepsilon)\sqrt{\lambda_1\lambda_2} \sqrt{1(\lambda_1+\lambda_2)^{2}} \delta_{\frac{e_2e_1}{e_2e_1}}$; the part of the surface lying in the planes $\Pi^\perp_{M\varepsilon}$ and $\Pi^\perp_{N\varepsilon}$ has area $O(t^2)$; the remaining part of $\partial (C_t \cap\operatorname{Prism}_{\varepsilon,1})$ is $S_t^1$, and this is what we are interested in. Using the integral equalities for both convex bodies we obtain
$$
\begin{equation*}
\begin{aligned} \, \int_{S^2} n\, d\nu_{\operatorname{Prism}_{\varepsilon,1}}(n) &=c_0 t (l\varepsilon) \biggl(\lambda_1 e_1 \frac{\lambda_1}{\lambda_1+\lambda_2}\, e \\ &\qquad +\sqrt{\lambda_1\lambda_2} \sqrt{1\frac{1}{(\lambda_1+\lambda_2)^{2}}}\, \frac{e_2e_1}{e_2e_1}\biggr)+o(t)=\vec 0 \end{aligned}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\begin{aligned} \, \int_{S^2} n\, d\nu_{C_t \cap \operatorname{Prism}_{\varepsilon,1}}(n) &=c_0 t (l\varepsilon) \frac{\lambda_1}{\lambda_1+ \lambda_2}\, e+o(t) \\ &\qquad +c_0 t (l\varepsilon) \sqrt{\lambda_1\lambda_2} \sqrt{1\frac{1}{(\lambda_1+\lambda_2)^{2}}}\, \frac{e_2e_1}{e_2e_1} \\ &\qquad +O(t^2)+\int_{S^2} n\, d\nu_{S_t^1}(n)=\vec 0. \end{aligned}
\end{equation*}
\notag
$$
Equating these two expressions we infer that
$$
\begin{equation*}
\int_{S^2} n\, d\nu_{S_t^1}(n)=c_0 t (l\varepsilon)\lambda_1e_1+o(t) .
\end{equation*}
\notag
$$
We introduce the notation
$$
\begin{equation*}
c=\biggl(1\frac{\varepsilon}l\biggr)\lambda_1.
\end{equation*}
\notag
$$
Dividing both sides of the last equality by $B_t$ we arrive at the relation
$$
\begin{equation*}
\int_{S^2} {n}\, d\nu_{t,\varepsilon}^1({n})=c e_1+o(1), \qquad t \to 0.
\end{equation*}
\notag
$$
Then, comparing the surface areas of $\operatorname{Prism}_{\varepsilon,1}$ and $C_t \cap\operatorname{Prism}_{\varepsilon,1}$, we deduce that $S_{t,\varepsilon}^1=(l\varepsilon) c_0 \lambda_1 t+o(t)$; consequently, $\nu_{t,\varepsilon}^1(S^2)=c+o(1), t \to 0$. We will show that $\nu_{t,\varepsilon}^1(\{{n} \in S^2\colon \langle{n}, e_1\rangle>1\delta\}) \to c$ and, accordingly, $\nu_{t,\varepsilon}^1(\{{n\in S^2}\colon \langle{n}, e_1\rangle \leqslant 1\delta\}) \to 0$ as $t \to 0$ for any $\delta>0$; this implies that $\nu_{t,\varepsilon}^1$ converges weakly to $c\delta_{e_1}$. We have
$$
\begin{equation*}
\int_{S^2} (1\langle n, e_1\rangle )\, d\nu_{t,\varepsilon}^1({n})=\nu_{t,\varepsilon}^1(S^2) \biggl\langle \int_{S^2} {n}\, d\nu_{t,\varepsilon}^1({n}), e_1\biggr\rangle=o(1), \qquad t \to 0.
\end{equation*}
\notag
$$
According to Chebyshev’s inequality,
$$
\begin{equation*}
\nu_{t,\varepsilon}^1(\{{n} \in S^2\colon \langle{n}, e_1\rangle \leqslant 1\delta\}) \leqslant \frac{1}{\delta} \int_{S^2} (1\langle n, e_1\rangle)\, d\nu_{t,\varepsilon}^1({n}) \to 0, \qquad t \to 0.
\end{equation*}
\notag
$$
Therefore,
$$
\begin{equation*}
\begin{aligned} \, &\nu_{t,\varepsilon}^1(\{{n} \in S^2\colon \langle{n}, e_1\rangle>1\delta\}) \\ &\qquad = \nu_{t,\varepsilon}^1(S^2)\nu_{t,\varepsilon}^1(\{{n} \in S^2\colon \langle{n}, e_1\rangle \leqslant 1\delta\})=c+ o(1), \qquad t \to 0. \end{aligned}
\end{equation*}
\notag
$$
Thus, we have proved that
$$
\begin{equation*}
\nu_{t,\varepsilon}^1 \xrightarrow[t\to 0]{}\biggl(1\frac{\varepsilon}l\biggr)\lambda_1\delta_{e_1}.
\end{equation*}
\notag
$$
We obtain similarly that
$$
\begin{equation*}
\nu_{t,\varepsilon}^2 \xrightarrow[t\to 0]{}\biggl(1\frac{\varepsilon}l\biggr)\lambda_2\delta_{e_2}.
\end{equation*}
\notag
$$
Consider a continuous function $f$ on $S^2$. Then
$$
\begin{equation*}
\begin{aligned} \, &\biggl\int_{S^2} f(n)\, d\nu_t(n)(\lambda_1 f(e_1)+\lambda_2 f(e_2))\biggr \\ &\qquad \leqslant \biggl\int_{S^2} f(n)\, d\nu_{t,\varepsilon}^1(n)\lambda_1 f(e_1)\biggr +\biggl\int_{S^2} f(n)\, d\nu_{t,\varepsilon}^2(n)\lambda_2 f(e_2)\biggr \\ &\qquad\qquad +\biggl\int_{S^2} f(n)\, d\nu_{t,\varepsilon}^{\mathrm{lat}}(n)\biggr. \end{aligned}
\end{equation*}
\notag
$$
The first term on the righthand side converges to $\frac{\varepsilon}{l}\lambda_1 f(e_1)$, the second term converges to $\frac{\varepsilon}{l}\lambda_2 f(e_2)$, and the upper partial limit of the third term is at most $\frac{\varepsilon}{l}(1+\lambda_1+\lambda_2) \cdot \max f$. Since $\varepsilon>0$ can be arbitrarily small, we conclude that the integral on the lefthand side of this inequality converges to zero. Thus, $\nu_t$ converges weakly to $\lambda_1 \delta_{e_1}+ \lambda_2 \delta_{e_2}$ as $t \to 0$. Theorem 3 is proved.
§ 5. Proof of Theorem 4 We assume here that $B_0$ is a point, that is, $B_0=\{r_0\}$, and $e=\lambda_1 e_1+\lambda_2 e_2$, $\lambda_1>0$, $\lambda_2>0$, where $e_1$ and $e_2$ are extreme points of $N(r_0)$. To prove Theorem 4 we need several lemmas. Lemma 10. The inequality
$$
\begin{equation*}
\limsup_{t\to0} \frac{S_t}{B_t} \leqslant \mathrm{const}
\end{equation*}
\notag
$$
holds. The proof of this lemma is left to the reader. It follows from Lemma 10 that there exists at least one partial limit of the measure $\nu_t$. If the partial limit $\nu_*$ is unique, then $\nu_t$ converges to $\nu_*$. Lemma 11. The support of each partial limit of $\nu_t$ lies on the arc $\Gamma=\Gamma_{e_1,e_2}$. Proof. Note that the intersection of the nested sets $S_t$ is $r_0$: $\bigcap_{t>0} S_t=\{r_0\}$; therefore, for each $\varepsilon>0$ there is $t_0$ such that the set $S_t$ is contained in the $\varepsilon$neighbourhood $B_\varepsilon(r_0)$ of $r_0$ for $t<t_0$. Indeed, otherwise we would have a nested family of nonempty compact sets $S_t \setminus B_\varepsilon(r_0)$ with empty intersection, which is impossible.
Let $K \subset S^2$; we let $\mathcal{O}_K$ denote the set of points ${r} \in \partial C$ such that the outer normal to a supporting plane at ${r}$ lies in $K$. If $K$ is a compact set, then the set $\mathcal{O}_K$ is also compact. Indeed, let ${r}_i$, $i \in \mathbb{N}$, be a sequence of points in $\mathcal{O}_K$ converging to ${r}$. For each $i$ there exists a supporting plane at ${r}_i$ with outer normal $v_i \in K$. Without loss of generality we assume that $v_i$ converges to some vector $v$. Then $v \in K$ and the plane through ${r}$ with outer normal $v$ is supporting. This yields that ${r}$ is in $\mathcal{O}_K$; hence the set $\mathcal{O}_K$ is closed and therefore compact.
We consider an open set $\mathcal{N} \subset S^2$ containing $N(r_0)$. The set $\mathcal{O}_{S^2 \setminus \mathcal{N}} \subset \partial C$ is closed and does not contain $r_0$. Consequently, there is $t_0>0$ such that the set $S_t$ lies in $\partial C \setminus \mathcal{O}_{S^2 \setminus \mathcal{N}}$ for all $t<t_0$. It follows that the outer normals to all supporting planes at points in $S_t$ are contained in $\mathcal{N}$; hence the support of $\nu_t$ lies in the closure of the set $\mathcal{N}$. Since the set $\mathcal{N}$ containing $N(r_0)$ is arbitrary, we conclude that the support of each partial limit of the measure $\nu_t$ as $t \to 0$ lies in $N(r_0)$ .
We fix some partial limit of $\nu_t$ and denote it by $\nu_*$. The above implies that the‘support of $\nu_*$ lies in the closed hemisphere containing $N(r_0)$ and bounded by the great circle containing the arc $\Gamma$. On the other hand, it follows from (7) that the integral $\displaystyle\int_{S^2} n\, d\nu_*(n)$ belongs to this great circle. We derive from this that the support of the measure $\nu_*$ also lies on the great circle and therefore on $\Gamma$.
The lemma is proved. To complete the proof of Theorem 4 it remains to prove the following assertions. Lemma 12. The support of each partial limit of the measure $\nu_t$ contains the points $e_1$ and $e_2$. Proof. It suffices to prove the lemma for $e_1$.
First we consider the case when
(a) $N(r_0)$ has a nonempty interior.
The idea of the proof consists in the following. We fix an arbitrary number $0<\theta<1$ and find, for each $t$, a domain in $S_t$ such that its relative area (that is, the ratio of its area to $B_t$) does not tend to zero as $t \to 0$ and the support of the normalized measure induced by this domain lies at a small distance from $e_1$ (in the sense that the distance tends to zero as $\theta \to 0$).
Recall that the planes $\Pi_1$ and $\Pi_2$ are specified by the equations $\langle rr_0, e_1\rangle=0$ and $\langle rr_0, e_2\rangle=0$ and bound a dihedral angle, the plane $\Pi^{t}$ is specified by the equation $\langle rr_0, e\rangle=t$, and the parallel lines $l_1= l_1^t$ and $l_2=l_2^t$ are the intersections of the plane $\Pi^t$ with $\Pi_1$ and $\Pi_2$. The distance between $l_1^t$ and $l_2^t$ is $c_0 t$, where $c_0$ is the positive constant defined in (17) (also see the caption under Figure 10, (b)).
We let $\widehat B_t$ denote the intersection of the tangent cone and the plane $\Pi_t$. It is a semiinfinite domain bounded by rays of the lines $l_1^t$ and $l_2^t$ and several line segments and containing $B_t$. We choose points $\widehat M_t$ and $\widehat N_t$ in these rays so that the line segment $\widehat M_t \widehat N_t$ is orthogonal to the rays. We let $\widetilde\Pi$ denote the plane containing $r_0$, $\widehat M_t$ and $\widehat N_t$.
For sufficiently small $t$ the intersection of $\widetilde\Pi$ and $B_t$ is a line segment; we denote its endpoints by $M= M_t$ and $N=N_t$. This segment lies on $\widehat M_t \widehat N_t$ and
$$
\begin{equation*}
\frac{M_t N_t}{\widehat M_t \widehat N_t} \to 1\quad\text{as } t \to 0.
\end{equation*}
\notag
$$
We let $\widehat C_t \widehat D_t$ denote the orthogonal projection of $B_t$ onto the line $l_1$. For sufficiently small $t$ the segment $\widehat C_t \widehat D_t$ contains the point $\widehat N_t$, and we have
$$
\begin{equation*}
\lim_{t \to 0}\frac{\widehat C_t \widehat N_t}{t}<\infty\quad\text{and} \quad \lim_{t \to 0} \frac{\widehat D_t \widehat N_t}{t}=\infty.
\end{equation*}
\notag
$$
We introduce the notation $\widehat D_t \widehat N_t=d(t)$; thus, $\lim_{t\to0} (d(t)/t)=\infty$.
We have the inequality
$$
\begin{equation}
B_t \leqslant d(t) \cdot c_0 t(1+o(1)).
\end{equation}
\tag{21}
$$
We fix $0<\theta<1$ and fix a point $\widehat F_t$ in $l_1$ between $\widehat N_t$ and $\widehat D_t$ such that $\widehat N_t \widehat F_t/\widehat N_t \widehat D_t=\theta$. Then $\widehat N_t \widehat F_t=\theta d(t)$ and $\widehat F_t \widehat D_t=(1\theta) d(t)$.
Let $D_t$ be a point in $\partial B_t$ whose projection is $\widehat D_t$; if it is not unique (the set of points of this kind forms a line segment), then we take any of these points, for example, the one closest to $l_1$. We let $\Pi_{[t]}$ denote the plane containing $D_t$ and the edge of the dihedral angle formed by $\Pi_1$ and $\Pi_2$, and we let $H_t$ denote the point of intersection of this plane with $M_t N_t$.
We introduce the notation $H_t \widehat N_t=h(t) \leqslant c_0 t$. It can turn out that the segment $H_t \widehat N_t$ degenerates into a point; in this case $h(t)=0$. We have $H_t N_t=h(t)+o(t)$ as $t \to 0$.
The area of the rectangle $H_t \widehat N_t \widehat F_t E_t$ is
$$
\begin{equation*}
H_t \widehat N_t \widehat F_t E_t=h(t) \cdot \theta d(t).
\end{equation*}
\notag
$$
We let $B_t(\theta)$ denote the part of the domain $B_t$ bounded by the polygonal line $N_t H_t E_t F_t$ and the arc $N_t F_t$ of the boundary $\partial B_t$. In view of the fact that the curve $N_t F_t D_t$ is convex, we conclude that $E_t F_t \geqslant (1\theta) H_t N_t$; hence
$$
\begin{equation*}
\biggl(\biggl(1\frac\theta2\biggr)h(t)+o(t) \biggr)\theta d(t) \leqslant B_t(\theta) \leqslant h(t) \cdot \theta d(t).
\end{equation*}
\notag
$$
It follows that
$$
\begin{equation}
B_t(\theta)=\biggl(\biggl(1\frac{\theta_1}2\biggr)h(t)+o(t) \biggr)\theta d(t)
\end{equation}
\tag{22}
$$
for some $\theta_1=\theta_1(t) \in [0, \theta]$ (Figures 13 and 14).
We draw a plane $\Pi_\theta(t)$ parallel to $\widetilde\Pi$ through the point $E_t$. Let ${r}_{\theta,t}$ be the point of intersection of this plane with this edge. The point of intersection of the line segment $E_t {r}_{\theta,t}$ with the surface $\partial C$ is denoted by $q_{\theta,t}$. We let $v=v(t)$ denote the normal vector to $\Pi_{[t]}$ directed toward $l_1$. We let $C_t(\theta)$ denote the part of the body $C$ bounded by the planes $\Pi_1$, $\Pi^t$, $\Pi_{[t]}$, $\widetilde\Pi$ and $\Pi_\theta(t)$ and compare it with the prism whose bases are the triangles $H_t \widehat N_t r_0$ and $E_t \widehat F_t {r}_{\theta,t}$. We are going to compare the corresponding parts of the surfaces of the body and the prism, namely, compare the part of the surface $\partial C$ bounded by the closed curve $N_t F_t q_{\theta,t} r_0$ with the rectangle $r_0 {r}_{\theta,t} \widehat F_t \widehat N_t$. Let the distance between the lines $r_0 r_{\theta,t}$ and $H_t E_t$ be $c_1 t$, where $c_1= c_1(t)$ is a function depending on $t$ and bounded above and below by positive constants. Then $r_0 H_t E_t {r}_{\theta,t}=c_1t\cdot \theta d(t)$. The area of the curvilinear quadrilateral $r_0 H_t E_t q_{\theta,t}$ satisfies the inequalities
$$
\begin{equation*}
\biggl(1\frac\theta2\biggr) c_1 t\cdot \theta d(t) \leqslant \square r_0 H_t E_t q_{\theta,t} \leqslant \square r_0 H_t E_t {r}_{\theta,t}=c_1 t\cdot \theta d(t).
\end{equation*}
\notag
$$
It follows that
$$
\begin{equation}
\square r_0 H_t E_t q_{\theta,t}=\biggl(1\frac{\theta_2}2\biggr) c_1 t\cdot \theta d(t)
\end{equation}
\tag{23}
$$
for some $\theta_2=\theta_2(t) \in [0, \theta]$. Furthermore, we have
$$
\begin{equation*}
\square r_0 {r}_{\theta,t} \widehat F_t \widehat N_te_1=c_1 t\cdot \theta d(t)v+h(t) \cdot \theta d(t) e;
\end{equation*}
\notag
$$
therefore,
$$
\begin{equation*}
e_1=\frac{v+\frac{h(t)}{c_1 t}\,e}{\biglv+\frac{h(t)}{c_1 t}\,e\bigr}.
\end{equation*}
\notag
$$
We let $S_{t,\theta}$ denote the part of $\partial C$ bounded by the closed curve $N_t F_t q_{\theta,t} r_0$ and consider the corresponding measure $\nu_{S_{t,\theta}}$. Then we have
$$
\begin{equation*}
\begin{aligned} \, &\int_{S^2} n\, d\nu_{S_{t,\theta}}(n) \\ &\qquad=B_t(\theta)e+r_0 H_t E_t q_{\theta,t}v +r_0 H_t N_t\frac{e_1 \times e_2}{e_1 \times e_2}q_{\theta,t} E_t F_t\frac{e_1 \times e_2}{e_1 \times e_2}. \end{aligned}
\end{equation*}
\notag
$$
Using (22) and (23) and taking account of the fact that $OH_t N_t= O(t^2)$ and $q_{\theta,t} E_t F_t=O(t^2)$ as $t \to 0$, we obtain
$$
\begin{equation*}
\begin{aligned} \, &\frac{1}{B_t} \int_{S^2} n\, d\nu_{S_{t,\theta}}(n) \\ &\qquad=\frac{\theta d(t) c_1 t}{B_t} \biggl[ \biggl(\biggl(1\frac{\theta_1}{2}\biggr) \frac{h(t)}{c_1 t}+o(1) \biggr)e +\biggl(1\frac{\theta_2}{2}\biggr)v+\frac{O(t)}{d(t)}\biggr]. \end{aligned}
\end{equation*}
\notag
$$
Note that $\frac{O(t)}{d(t)}=\frac{O(1)}{d(t)/t}=o(1)$. The angle between $e_1$ and $\displaystyle\int_{S^2} n\, d\nu_{S_{t,\theta}(n)}$ is equal to the angle between $v+ \frac{h(t)}{c_1 t}\, e$ and $[v+\frac{h(t)}{c_1 t} \, e][\frac{\theta_2}{2}\, v+ \frac{\theta_1}{2}\,\frac{h(t)}{c_1 t}\, e]+o(1)$. The angle between the lines containing $v$ and $e$ is always greater than the least angle between the lines containing $e$ and $e_i$, $i=1,2$. This can be seen in Figure 15: $e$ lies on the arc of a great circle bounded by $e_1$ and $e_2$, whereas $v$ is on the arc of the same circle between $e_1$ and $e_2$. Therefore, the norm of the vector $v+\frac{h(t)}{c_1 t}\, e$ is greater that some positive number $C_1$ depending only on $e$, $e_1$ and $e_2$. On the other hand, in view of the inequalities $\frac{h(t)}{c_1 t} \leqslant \frac{c_0}{c_1}$ and $0 \leqslant \theta_i \leqslant \theta$, $i= 1,2$, we conclude that the limit of the norm of the vector $\frac{\theta_2}{2}\, v+\frac{\theta_1}{2}\,\frac{h(t)}{c_1 t}\, e+o(1)$ as $t \to 0$, is smaller than the product of $\theta$ and some positive number $C_2$ depending on only $e$, $e_1$ and $e_2$. Furthermore, using (21) we infer that
$$
\begin{equation*}
\frac{\theta d(t)c_1 t}{B_t} \geqslant \frac{\theta c_1}{2c_0} \quad \text{for sufficiently small} \ \ t.
\end{equation*}
\notag
$$
Therefore, the quantity $\frac{1}{B_t}\nu_{S_{t,\theta}}(S^2)$ satisfies the inequality
$$
\begin{equation*}
\frac{1}{B_t}\nu_{S_{t,\theta}}(S^2) \geqslant \frac{1}{B_t} \biggl\int_{S^2} n\,d\nu_{S_{t,\theta}}(n)\biggr \geqslant \frac{\theta c_1}{2c_0}(C_1C_2 \theta+o(1)) \quad \text{as} \ \ t \to 0;
\end{equation*}
\notag
$$
hence for $\theta<C_1/C_2$ each partial limit of $\frac{1}{B_t}\nu_{S_{t,\theta}^{1}}$ is nonzero. Let $\nu_*$ be some partial limit of the measure $\nu_t$; thus, $\nu_*=\lim_{i \to \infty} \nu_{t_i}$ for some sequence $t_i$ converging to zero. The centre of mass of each partial limit of the measure $\frac{1}{B_{t_i}}\nu_{S_{{t_i},\theta}}$ belongs to some open convex cone $\mathcal{U}_\theta(e_1)$ that shrinks to the ray containing $\{e_1\}$ in the limit as $\theta \to 0$. Since $\frac{1}{B_t}\nu_{S_{t,\theta}} \leqslant \nu_t$ and the support of the measure $\nu_*$ lies on the arc $\Gamma$, the support of the partial limit of the measure $\frac{1}{B_{t_i}}\nu_{S_{{t_i},\theta}}$ also lies on $\Gamma$ and, in addition, has a nonempty intersection with $\mathcal{U}_\theta(e_1)$ (otherwise, its centre of mass would not belong to $\mathcal{U}_\theta(e_1)$). It follows that $\operatorname{spt}\nu_*$ also has a nonempty intersection with $\mathcal{U}_\theta(e_1)$. Throughout what follows $\operatorname{spt}\nu$ denotes the support of the measure $\nu$. Since $\theta$ can be arbitrarily small, $\operatorname{spt}\nu_*$ contains points that are arbitrarily close to $e_1$, and thus it contains $e_1$ as well. Now we consider the case when (b) $N(r_0)$ is a dihedral angle. We divide $C$ into two convex bodies $C_1$ and $C_2$ by the plane $\Pi_\perp$ containing $r_0$ and parallel to $e_1$ and $e_2$. The vector $e_\perp=e_1 \times e_2/e_1 \times e_2$ is a normal to this plane. The plane figure $B_t$ and the surface $S_t$ are subdivided by this plane into two figures $B_t^1$ and $B_t^2$ and two surfaces $S_t^1$ and $S_t^2$, respectively. The point $r_0$ is conical for the bodies $C_1$ and $C_2$. Let $\nu_t^{(1)}$ and $\nu_t^{(2)}$ be the normalized measures corresponding to these bodies in a neighbourhood of the same point $r_0$ with the same cutoff planes. These measures are induced by the surfaces $S_t^1$ and $S_t^2$, respectively, and also by plane domains of area $s_t=O(t^2)$ in $\Pi_\perp$. The contribution of these plane domains to the measures $\nu_t^{(1)}$ and $\nu_t^{(2)}$ tends to zero as $t \to 0$. The measure $\nu_t$ is a linear combination of the normalized measures generated by $S_t^1$ and $S_t^2$: more precisely,
$$
\begin{equation*}
\nu_t=\frac{1}{S_t} [S_t^1 \nu_t^{(1)}s_t \delta_{v_\perp}]+\frac{1}{S_t} [S_t^2 \nu_t^{(2)}s_t \delta_{v_\perp}].
\end{equation*}
\notag
$$
Consequently, for each partial limit $\nu_*$ of $\nu_{t}$ and for the corresponding sequence $t_i$, we have
$$
\begin{equation*}
\nu_*=\lim_{i\to\infty} \biggl(\frac{S_{t_i}^1}{S_{t_i}} \nu_{t_i}^{(1)}+ \frac{S_{t_i}^2}{S_{t_i}} \nu_{t_i}^{(2)}\biggr).
\end{equation*}
\notag
$$
Choosing a subsequence if necessary, we can assume that the limits $S_{t_i}^j/S_{t_i}$, $\nu_{t_i}^{(j)}$, $i \to \infty$, $j=1,2$, exist. By what we proved in case (a) the support of each partial limit of the measures $\nu_t^{(1)}$ and $\nu_t^{(2)}$ contains the points $e_1$ and $e_2$; thus, the support of $\nu_*$ also contains these points. Lemma 12 is proved. We have thus completely proved Theorem 4.
§ 6. Proof of Theorem 5 Let $\alpha$ be the angle between $e$ and $e_1$ and let $\beta$ be the angle between $e$ and $e_2$, that is, $\langle e, e_1\rangle=\cos\alpha$ and $\langle e, e_2\rangle=\cos\beta$, $\alpha>0$, $\beta>0$, $\alpha+\beta<\pi$. We take an orthonormal system of coordinates ($x$, $y$, $z$) such that its origin is at $r_0$ and the vectors $e$, $e_1$ and $e_2$ have the form $e=(0,0,1)$, $e_1=(\sin\alpha, 0, \cos\alpha)$ and $e_2=(\sin\beta, 0, \cos\beta)$. Then $\lambda_1=\sin\beta/\sin(\alpha+\beta)$, $\lambda_2= \sin\alpha/\sin(\alpha+\beta)$, $r_0=(0,0,0)$, and the equation for the tangent cone (9) takes the form
$$
\begin{equation}
z\cot\alpha \leqslant x \leqslant z\cot\beta.
\end{equation}
\tag{24}
$$
Note that $\cot\alpha+\cot\beta>0$; therefore, the cone lies in the positive halfspace $z \geqslant 0$. Furthermore, we have $C_t=C \cap \{z \leqslant t\}$, $B_t=C \cap \{z=t\}$ and $S_t=\partial C \cap \{z \leqslant t\}$. The image of the arc $\Gamma$ under the orthogonal projection onto the $xz$plane is the arc in $S^1$ with endpoints $(\sin\alpha,\cos\alpha)$ and $(\sin\beta, \cos\beta)$. We identify the points $(\sin\varphi,\cos\varphi)$ in the circle $S^1$ with $\varphi \in [\pi, \pi]$. In particular, the image of $\Gamma$ is identified with the closed interval $[\alpha,\beta]$. The image of a compact set $K \subset \Gamma$ under this projection is denoted by the same letter $K$. Thus, in this notation we have $\{\alpha,\beta\} \subset K \subset [\alpha,\beta]$. Assume that a convex body $C$ and a singular ridge point $r_0$ on its boundary are such that the tangent cone at this point is specified by (9) and the edge $B_0$ is a nondegenerate segment. By Theorem 3 the corresponding limit measure exists and its support is the twopoint set $\{\alpha,\beta\}$. Therefore, it suffices to limit ourselves in what follows to the case when $K$ is not a twopoint set, that is, $K$ includes some interior points of the interval $[\alpha,\beta]$, in addition to $\alpha$ and $\beta$. We fix a finite measure in $S^1$ with support $K$. For example, we can choose a measuregenerating function $F_\mu(\varphi)=\mu([\alpha, \varphi])$ as follows. The open set $(\alpha,\beta) \setminus K$ is a finite or countable union of pairwise disjoint open intervals: $(\alpha,\beta) \setminus K=\bigcup_i (a_i, b_i)$. We associate each $i$ with some number $\xi_i \in (a_i, b_i)$. We set $F_\mu(\varphi)=\alpha+\xi_i$ for $\varphi \in (a_i, b_i)$, $F_\mu(\varphi)=0$ for $\varphi<\alpha$, and $F_\mu(\varphi)=\alpha+\beta$ for ${\varphi \geqslant \beta}$. Then we extend $F_\mu$ to the closure $\overline{\bigcup_i (a_i, b_i)}$ so that the resulting function is monotonically nondecreasing and right semicontinuous. Finally, we extend the function $F_\mu$ to the whole of $\mathbb R \setminus \overline{\bigcup_i (a_i, b_i)}$ (which is a disconnected union of intervals) in such a way that it is affine and strictly monotonically increasing on the closure of each interval. Now we construct a plane curve generating the measure $\mu$ with endpoints on the rays $x= z\cot\alpha$, $z \geqslant 0$, and $x=z\cot\beta$, $z \geqslant 0$, such that the unbounded domain with boundary formed by this curve and these rays is convex (Figure 16). We set
$$
\begin{equation*}
\int_{S^1} n\, d\mu(n)=c_0 n_0,
\end{equation*}
\notag
$$
where $n_0$ is a unit vector. For any vector $n \in \Gamma$ we have $\langle n, e_1+e_2 \rangle> 0$; hence
$$
\begin{equation*}
\biggl\langle \int_{S^1} n\, d\mu(n), e_1+e_2 \biggr\rangle>0.
\end{equation*}
\notag
$$
It follows that $c_0 \ne 0$. The centre of mass of the auxiliary measure $\overline\mu={\mu+c_0 \delta_{n_0}}$ is at the origin; therefore, according to a theorem due to A. D. Aleksandrov (see [1], § 3), there exists a unique (up to parallel translation) plane convex figure that induces the measure $\widehat\mu$. (In Figure 16, (a), this figure is the quadrilateral $ABCD$; in Figure 16, (b), the figure is bounded by the curve $l$ and the segment connecting its endpoints $A$ and $D$.) The boundary of this figure is the union of the line segment corresponding to the measure $c_0 \delta_{n_0}$ and the curve corresponding to the measure $\mu$. The tangent vectors to this curve at its endpoints are $(\cos\alpha, \sin\alpha)$ and $(\cos\beta, \sin\beta)$. We draw the tangent lines to the curve at its endpoints and assume without loss of generality that they intersect at the origin (otherwise we can shift the curve in an appropriate way). The construction is complete. We let $V$ denote the angle $z\cot\alpha \leqslant x \leqslant z\cot\beta$. Let $\mathcal{B}$ be the unbounded closed convex set bounded below by the above curve and the rays $x=z\cot\alpha, z \geqslant 0$, and $x=z\cot\beta$, $z \geqslant 0$, and let $\overline{\mathcal{B}}=V \setminus \mathcal{B}$ (see Figure 16). We take $y \in \mathbb R$ and consider the curve $y^2 l$ homothetic to $l$ with centre of homothety at the origin and coefficient $y^2$ (if $y=0$, then this curve degenerates into the point $(0,0)$), and let $\mathcal{B}_y$ denote the unbounded convex set bounded by $y^2 l$ and the rays forming the angle $V$. We can equivalently define this set by $\mathcal{B}_y=V \setminus y^2 \overline{\mathcal{B}}$. Thus, $\mathcal{B}_y=y^2\mathcal{B}$ for $y \ne 0$ and $\mathcal{B}_0=V$ for $y=0$. We have the monotonicity relation $\mathcal{B}_{y_1} \subset \mathcal{B}_{y_2}$ for $y_1 \geqslant y_2$. We define $C$ in the following way:
$$
\begin{equation*}
C=\{(x,y,z) \colon (x,z) \in \mathcal{B}_y, \ z \leqslant 1\}.
\end{equation*}
\notag
$$
Informally speaking, the sets $\mathcal{B}_y \cap \{z \leqslant 1\}$ occur as the intersections of $C$ with the planes $y=$ const. Note that
$$
\begin{equation*}
\lambda\mathcal{B}_{y_1}+(1\lambda)\mathcal{B}_{y_2}=\mathcal{B}_{\sqrt{\lambda y_1^2+(1\lambda) y_2^2}} \subset \mathcal{B}_{\lambda y_1+(1\lambda)y_2}
\end{equation*}
\notag
$$
for $0 \leqslant \lambda \leqslant 1$ and any $y_1$ and $y_2$. The inclusion in this formula is a direct consequence of the inequality
$$
\begin{equation*}
\sqrt{\lambda y_1^2+(1\lambda) y_2^2} \geqslant \lambda y_1+(1\lambda)y_2.
\end{equation*}
\notag
$$
If $y_1$ and $y_2$ are nonzero, then the equality in this formula is implied by the equality
$$
\begin{equation*}
\lambda y_1^2 \mathcal{B}+(1\lambda) y_2^2 \mathcal{B}=(\lambda y_1^2+(1\lambda) y_2^2) \mathcal{B},
\end{equation*}
\notag
$$
which holds for the convex set $\mathcal{B}$. If $y_1=y_2=0$, then the equality takes the form $\lambda V+ (1\lambda) V=V$. If $y_1 \ne 0$ and $y_2=0$, then the equality takes the form $\lambda y_1^2 \mathcal{B}+ (1\lambda) V=\lambda y_1^2 \mathcal{B}$. We show that $C$ is convex. We take two points ${r}_1=(x_1, y_1, z_1)$ and ${r}_2=(x_2, y_2, z_2)$ in $C$ and show that $\lambda {r}_1+(1\lambda) {r}_2$ lies in $C$ for all $0 \leqslant \lambda \leqslant 1$. In fact, since $(x_1, z_1) \in \mathcal{B}_{y_1}$ and $(x_2, z_2) \in \mathcal{B}_{y_2}$, we obtain
$$
\begin{equation*}
\lambda (x_1, z_1)+(1\lambda) (x_2, z_2) \in \lambda\mathcal{B}_{y_1}+(1\lambda)\mathcal{B}_{y_2} \subset \mathcal{B}_{\lambda y_1+(1\lambda)y_2}.
\end{equation*}
\notag
$$
In addition, it is clear that $\lambda z_1+(1\lambda) z_2 \leqslant 1$. It follows that $\lambda {r}_1+(1\lambda) {r}_2 \in C$. It is straightforward to see that each point $(x,0,z)$, where $(x,z)$ lies in the interior of $V \cap \{z \leqslant 1\}$, is an interior point of $C$. Furthermore, for sufficiently large $y$, we have
$$
\begin{equation*}
\mathcal{B}_y \cap \{z \leqslant 1\}=\varnothing;
\end{equation*}
\notag
$$
hence $C$ is bounded and has a nonempty interior. Now we verify that $C$ is closed. Assume that a sequence $\{{r}_i=(x_i,y_i, z_i)\} \subset C$ converges to ${r}=(x,y,z)$. If $y \ne 0$, then $\frac{1}{y_i^2} (x_i, z_i) \in \mathcal{B}$; since $\mathcal{B}$ is closed, the limit point $\frac{1}{y^2} (x,z)$ is also in $\mathcal{B}$. It follows that $(x,y,z) \in C$. If $y=0$, then
$$
\begin{equation*}
(x_i, z_i) \in V \quad\Longrightarrow\quad (x,z) \in V;
\end{equation*}
\notag
$$
therefore, $(x,0,z) \in C$. So $C$ is convex and compact and has a nonempty interior, that is, it is a convex body. We fix $\varphi \in (\beta, \pi\alpha)$ and draw a ray in the $xz$plane with vertex $(0,0)$ and direction vector $(\cos\varphi, \sin\varphi)$. This ray intersects $\partial\mathcal{B}$ at some point $\tau(\cos\varphi, \sin\varphi)$, $\tau>0$. Thus, the intersection of $\partial C$ and the plane containing the point $r_0=(0,0,0)$ with the normal $(\sin\varphi, 0, \cos\varphi)$ is the curve
$$
\begin{equation*}
(y^2 \tau\cos\varphi, y, y^2 \tau\sin\varphi), \qquad y \leqslant \frac{1}{\sqrt{\tau\sin\varphi}}.
\end{equation*}
\notag
$$
This implies that the $y$axis is tangent to $C$ and therefore lies in any supporting plane containing $r_0$. Furthermore, $C$ lies in the dihedral angle (24) but does not lie in any smaller dihedral angle with the same edge. Therefore, the tangent cone at $r_0$ is defined by (24). The intersection of $\mathcal{B}$ and the line $z=\tau$, $\tau \geqslant 0$, is either empty or is a line segment. We denote its length by $L(\tau)$; it is a nonnegative continuous function vanishing for sufficiently small $\tau$ and satisfying the inequality $L(\tau) \leqslant (\cot\alpha+ \cot\beta)\tau$. Furthermore, for $0<t \leqslant 1$ the set $B_t$ is
$$
\begin{equation*}
B_t=C \cap \{z=t\}=\{(x,y,t) \colon (x,t) \in \mathcal{B}_y\}.
\end{equation*}
\notag
$$
The intersection of $B_t$ and the line $y=\tau$, $z=t$ is either empty or is a line segment (which possibly degenerates into a point), and its length is $y^2 L(t/y^2)$. Consequently, the area of $B_t$ is
$$
\begin{equation*}
B_t=\int_\mathbb R y^2 L\biggl(\frac{t}{y^2}\biggr)\, dy.
\end{equation*}
\notag
$$
This integral is finite, since the integrand is a bounded function vanishing outside some finite interval. Making the change of variables $\xi=y/\sqrt t$ we obtain
$$
\begin{equation*}
B_t=bt^{3/2}, \quad \text{where} \ \ b=\int_\mathbb R \xi^2 L\biggl(\frac1{\xi^2}\biggr)\, d\xi>0.
\end{equation*}
\notag
$$
For $0<t \leqslant 1$
$$
\begin{equation*}
S_t=\partial C \cap \{z \leqslant t\}=\{(x,y,z) \colon (x,z) \in \partial\mathcal{B}_y, \ z \leqslant t\}.
\end{equation*}
\notag
$$
This is a disconnected union $ S_t^+\cup S_t^ \cup S_t^0$, where
$$
\begin{equation*}
S_t^+=S_t \cap \{x=z\cot\beta\}, \qquad S_t^=S_t \cap \{x=z\cot\alpha\}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
S_t^0=S_t \cap \{z\cot\alpha<x<z\cot\beta\};
\end{equation*}
\notag
$$
that is, $S_t^+$ and $S_t^$ are the intersections of $S_t$ and the faces of the dihedral angle (24), while $S_t^0$ is the part of $S_t$ lying in the interior of this angle. The part of $\partial\mathcal{B}$ below the horizontal line $z=\tau$ is the union of three curves: the first and third curves are segments of the rays $x=z\cot\beta$, $z \geqslant 0$, and $x=z\cot\alpha$, $z \geqslant 0$, respectively, while the second curve lies between these rays (and is contained in the curve $l$ or coincides with it). We let $L_(\tau), L_+(\tau)$ and $ L_0(\tau)$ denote the lengths of these curves. These are nonnegative continuous functions of $\tau$ that vanish for sufficiently small $\tau$. The flat surfaces $S_t^+$ and $S_t^$ (without points of the form $(x,0,z)$) consist of the points $(x,y,z)$ such that $z \leqslant t$ and $(x/y^2, z/y^2)$ lies in the intersection of $\partial\mathcal{B}$ and the rays $x=z\cot\beta$ and $x=z\cot\alpha$, respectively. Their areas are
$$
\begin{equation*}
S_t^\pm=\int_\mathbb R y^2 L_\pm\biggl(\frac{t}{y^2}\biggr)\, dy.
\end{equation*}
\notag
$$
Making the change of variables $\xi=y/\sqrt t$ we obtain
$$
\begin{equation*}
S_t^\pm=s_\pm t^{3/2}, \quad \text{where} \ \ s_\pm=\int_\mathbb R \xi^2 L_\pm\biggl(\frac{1}{\xi^2}\biggr)\, d\xi>0.
\end{equation*}
\notag
$$
The outer normals to the surfaces $S_t^$ and $S_t^+$ are $e_1$ and $e_2$; therefore, the normalized measures generated by $S_t^$ and $S_t^+$ are the atoms $\frac{s_}{b}\delta_{e_1}$ and $\frac{s_+}{b}\delta_{e_2}$. We let $\nu_t^0$ denote the normalized measure generated by the surface $S_t^0$; then
$$
\begin{equation*}
\nu_t=\frac{s_}{b}\delta_{e_1}+\frac{s_+}{b}\delta_{e_2}+\nu_t^0.
\end{equation*}
\notag
$$
It remains to describe $\nu_t^0$. We let $l_0$ denote the part of $l$ lying in the interior $z\cot\alpha<x< z\cot\beta$ of the angle $V$ and consider its natural parametrization $(x(s),z(s))$, $s \in [0, s_0]$, starting from the ray $x=z\cot\alpha$. The length $s_0$ of the curve is finite, and the derivative $(x'(s),z'(s))$ exists for almost all $s$ and satisfies $x'^2+z'^2=1$. The surface $S_t^0$ consists of the points $(x,y,z)$ such that $z \leqslant t$ and $(x/y^2, z/y^2) \in l_0$. We parameterize $S_t^0$ using the mapping
$$
\begin{equation*}
(s,y) \mapsto {r}(s,y)=(y^2 x(s), y, y^2 z(s))
\end{equation*}
\notag
$$
defined on the domain $D_t=\{(s,y) \colon 0 \leqslant s \leqslant s_0,\, y^2 z(s) \leqslant t\}$. This mapping is injective on the fullmeasure subset $D_t \setminus \{y \ne 0\}$. An area element of the surface $S_t^0$ has the form ${r}'_s \times r'_y\, ds\, dy$, and the outer normal to $S_t^0$ is $({r}'_s \times r'_y)/{r}'_s \times r'_y$. It is straightforward to see that
$$
\begin{equation*}
{r}'_s(s,y) \times r'_y(s,y)=y^2 (z'(s), 2y\Delta(s), x'(s)), \quad \text{where} \ \ \Delta(s)= \begin{vmatrix} x(s) & z(s)\\ x'(s) & z'(s) \end{vmatrix},
\end{equation*}
\notag
$$
and ${r}'_s(s,y) \times r'_y(s,y)=y^2 \sqrt{1+4y^2 \Delta(s)^2}$. Note that $(x', z')$ is a unit vector and $(x, z) \leqslant C$, where $C$ denotes the greatest distance between points in $l_0$ and the origin; thus, the function $\Delta(s)$ is bounded: $\Delta(s) \leqslant C$. We see that an area element of $S_t^0$ has the form $y^2 \sqrt{1+4y^2 \Delta(s)^2}\, ds\, dy$ and the outer normal to $S_t^0$ at ${r}(s,y)$ is
$$
\begin{equation*}
n(s,y)=\frac{(z'(s), 2y\Delta(s), x'(s))}{\sqrt{1+4y^2\Delta(s)^2}}.
\end{equation*}
\notag
$$
Now we define an auxiliary normalized measure $\widetilde\nu_t$ induced by the family of vectors $\widetilde n(s,y)=(z'(s), 0, x'(s))$ on the measurable space $D_t$ with measure $d\widetilde\mu(s,y)=y^2 \,ds \,dy$ as follows:
$$
\begin{equation*}
\widetilde\nu_t(\mathcal{A})=\frac{1}{b t^{3/2}}\widetilde\mu\bigl(\{(s,y) \colon \widetilde n(s,y) \in \mathcal{A}\} \bigr)
\end{equation*}
\notag
$$
for any Borel set $\mathcal{A} \subset S^2$. The measure $\widetilde\nu_t$ is easier to study than $\nu_t^0$; we show below that it actually does not depend on $t$. We show that $\widetilde\nu_t$ and $\nu_t^0$ are asymptotically equivalent: either both of them converge to the same limit or both diverge. To do this it suffices to show that
$$
\begin{equation*}
\int_{S^2} f(n)\, d(\nu_t^0\widetilde\nu_t)(n) \to 0 \quad \text{as} \ \ t \to 0
\end{equation*}
\notag
$$
for each continuous function $f$ on $S$. This integral can be written as
$$
\begin{equation*}
\begin{aligned} \, &\frac{1}{b t^{3/2}} \iint_{D_t} f \biggl(\frac{(z'(s), 2y\Delta(s), x'(s))}{\sqrt{1+ 4y^2\Delta(s)^2}}\biggr) y^2 \sqrt{1+4y^2\Delta(s)^2}\,ds\,dy \\ &\qquad\qquad \frac{1}{b t^{3/2}} \iint_{D_t} f(z'(s), 0, x'(s))y^2\, ds\, dy \\ &\qquad =\frac{1}{b t^{3/2}} \iint_{D_t} \biggl[f\biggl(\frac{(z'(s), 2y\Delta(s), x'(s))}{\sqrt{1+4y^2\Delta(s)^2}}\biggr)\sqrt{1+4y^2\Delta(s)^2} \\ &\qquad\qquad\qquad f(z'(s), 0, x'(s))\biggr] y^2\,ds\,dy. \end{aligned}
\end{equation*}
\notag
$$
Since the function $f$ is continuous and therefore uniformly continuous on $S^2$ and the function $\Delta(s)$ is bounded, we conclude that the expression in square brackets in the last integral tends to zero as $y \to 0$ uniformly with respect to $s$; hence its absolute value does not exceed some nonnegative even function $\gamma(y)$ that is monotonically nondecreasing for $y \geqslant 0$ and converges to zero as $y \to 0$. Making the change of variables $\xi=y/\sqrt t$ we obtain that the last integral is at most
$$
\begin{equation*}
\begin{aligned} \, \frac{1}{b t^{3/2}} \iint_{D_t} y^2 \gamma(y)\,ds\,dy &=\frac{1}{b t^{3/2}} \int_0^{s_0} ds \int_{\sqrt{t/z(s)}}^{\sqrt{t/z(s)}} y^2 \gamma(y)\,dy \\ &=\frac{1}{b} \int_0^{s_0} ds \int_{1/\sqrt{z(s)}}^{1/\sqrt{z(s)}} \xi^2 \gamma(\sqrt t\, \xi)\, d\xi \end{aligned}
\end{equation*}
\notag
$$
in absolute value. Since the integrand $\xi^2 \gamma(\sqrt t\, \xi)$ is monotonically nondecreasing in $t$ and tends to zero as $t \to 0$, the integral
$$
\begin{equation*}
\frac{1}{b} \int_0^{s_0} ds \int_{1/\sqrt{z(s)}}^{1/\sqrt{z(s)}} \xi^2 \gamma(\sqrt t\, \xi)\, d\xi
\end{equation*}
\notag
$$
also tends to zero as $t \to 0$. It remains to consider the measure $\widetilde\nu_t$. Its support is $\Gamma$ since all vectors $(z', 0, x')$ lie in $\Gamma$. Recall that we identify $\varphi$ with the vector $(\sin\varphi, 0, \cos\varphi)$. We fix an interval $\mathcal{U}=(\varphi_1, \varphi_2) \subset (\alpha,\beta)$. Then we have
$$
\begin{equation*}
\widetilde\nu_t(\mathcal{U})=\frac{1}{b t^{3/2}} \int_{F_\mu(\varphi_1)}^{F_\mu(\varphi_20)}\, ds \int_{\sqrt{t/z(s)}}^{\sqrt{t/z(s)}} y^2\, dy =\frac{1}{b} \int_{F_\mu(\varphi_1)}^{F_\mu(\varphi_20)} \frac{2}{3}\,\frac{1}{z(s)^{3/2}}\,ds.
\end{equation*}
\notag
$$
Consequently, $\widetilde\nu_t$ is independent of $t$: $\widetilde\nu_t=\widetilde\nu$; therefore, $\nu_t^0 \to \widetilde\nu$ as $t \to 0$. If ${\mathcal{U} \cap K=\varnothing}$, then $F_\mu(\varphi_1)= F_\mu(\varphi_20)$ and, according to the above formula, $\widetilde\nu(\mathcal{U})= 0$. If the intersection $\mathcal{U} \cap K$ is nonempty, then $F_\mu(\varphi_1)<F_\mu(\varphi_20)$, and therefore $\widetilde\nu(\mathcal{U})>0$. Thus, $\operatorname{spt}\widetilde\nu \setminus \{\alpha,\beta\}$ coincides with $K \setminus \{ \alpha,\beta\}$. It remains to note that $\nu_t$ converges to $\nu_*=\frac{s_}{b}\, \delta_{e_1}+ \frac{s_+}{b}\, \delta_{e_2}+\widetilde\nu$ and $\operatorname{spt}\nu_*=K$. Theorem 5 is proved. Acknowledgments The author is grateful to G. Wachsmuth and V. A. Alexandrov for useful discussions. The author is also grateful to an anonymous referee for reading the paper carefully and making numerous helpful remarks.



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Citation:
A. Yu. Plakhov, “Local structure of convex surfaces”, Mat. Sb., 215:3 (2024), 119–158; Sb. Math., 215:3 (2024), 401–437
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