Cut locus of a point has riemannian volume zero

Dimension zero. [F1, F30, F36, given, cases] If $n=0$ then [F1] gives $S_{p}M=\mathrm{\varnothing }$ and hence $\mathrm{Cut}(p)=\mathrm{\varnothing }$, and ${\mathrm{vol}}_g(\mathrm{\varnothing })=0$ because [F30] makes ${\mathrm{vol}}_g$ a measure and measures assign value zero to the empty set [F36]. This is the assertion in dimension zero, so assume $n\ge 1$ from here on.

The square root is smooth on $(0,\mathrm{\infty })$. [F15, F16, F17, F18, algebra] Fix $x_0>0$ and put $t_0:=\sqrt{x_0}>0$, so $t_0^2=x_0$ by [F18]. The function $f(t)=t^2$ has $f^{\mathrm{\prime }}(t_0)=2t_0\mathrm{\ne }0$ by [F15], so the inverse function theorem [F16] supplies open neighbourhoods $V$ of $t_0$ and $W$ of $x_0$ and a $C^1$ inverse $g:W\to V$ with $f(g(y))=y$; since $f$ is smooth, [F17] makes this same local inverse smooth. For $y\in W$ we have $g(y)\in V\subseteq (0,\mathrm{\infty })$ and $g(y{)}^2=y$, so the uniqueness clause of [F18] forces $g(y)=\sqrt{y}$. Hence the square root agrees near the arbitrary point $x_0>0$ with a smooth map and is therefore smooth on $(0,\mathrm{\infty })$.

Coordinates, isometry and transported cut data. [F2, F4, F5, F6, F7, F8, F42, F43, F45, algebra] By [F4] and [F5] choose an ordered orthonormal basis $(e_1,\dots ,e_n)$ of $(T_{p}M,g_p)$; this is a single existential instantiation. Let $\lambda :T_{p}M\to {\mathbb{R}}^n$ be the coefficient map of [F6], $\lambda ({\sum }_i{x}_i{e}_i)=(x_1,\dots ,x_n)$. Expanding with the bilinearity of $g_p$ [F7] and the orthonormality of the basis [F5], $g_p(w,w)={\sum }_{i,j}{x}_i{x}_j{g}_p(e_i,e_j)={\sum }_{i=1}^n{x}_i^2=\mathrm{\mid }\lambda (w){\mathrm{\mid }}_2^2$ for $w={\sum }_i{x}_i{e}_i$, so that $\mathrm{\mid }w{\mathrm{\mid }}_g=\mathrm{\mid }\lambda (w){\mathrm{\mid }}_2$ by [F8]; hence $\lambda$ is a linear isometry and, being bijective, a linear isometric isomorphism [F42], so it is a homeomorphism onto ${\mathbb{R}}^n$ by [F43]. Therefore $\lambda (S_{p}M)=S^{n-1}:=\{y\in {\mathbb{R}}^n:\mathrm{\mid }y{\mathrm{\mid }}_2=1\}$. Define $f:=c\circ (\lambda {\mathrm{\mid }}_{S_{p}M}{)}^{-1}:S^{n-1}\to (0,+\mathrm{\infty }],E:=\{c(v)v:v\in S_{p}M,c(v)<+\mathrm{\infty }\}\mathrm{.}$ The restriction $\lambda {\mathrm{\mid }}_{S_{p}M}:S_{p}M\to S^{n-1}$ is a homeomorphism onto $S^{n-1}$, its inverse being the restriction of the homeomorphism ${\lambda }^{-1}$ [F43]; so $f$ is continuous as a composite of continuous maps [F2, F45]. Finally $\lambda (E)=\{f(y)y:y\in S^{n-1},f(y)<+\mathrm{\infty }\}$: indeed $\lambda (c(v)v)=c(v)\lambda (v)=f(y)y$ with $y:=\lambda (v)$, and conversely every $y\in S^{n-1}$ is $\lambda (v)$ for the unit vector $v:={\lambda }^{-1}(y)$.

Smooth coordinate expressions on Euclidean domains. [F19, F20, F21, step 1.2, algebra] For an open set $U\subseteq {\mathbb{R}}^m$ the coordinate function $x\mapsto x_k$ is smooth on $U$: its iterated partial derivatives are the constants $0$ and $1$, so it is $C^k$ for every $k$ by the definition [F20], that is smooth by [F21]. Consequently, by the closure theorem [F19], all finite sums, products and scalar multiples of coordinate functions are smooth on $U$; moreover if a polynomial expression $q$ is strictly positive on $U$ then $x\mapsto \sqrt{q(x)}$ is smooth on $U$, because $q$ takes its values in the domain $(0,\mathrm{\infty })$ of the smooth square root of step 1.2 and the composite of composable $C^k$ maps is $C^k$ for every $k$ [F19].

Sphere parametrisations. [F12, F14, F18, F27, step 2.1, algebra] Put $U:=\{x\in {\mathbb{R}}^{n-1}:\mathrm{\mid }x{\mathrm{\mid }}_2<1\}$, an open subset of ${\mathbb{R}}^{n-1}$. For $i\in \{1,\dots ,n\}$ and $\epsilon \in \{+1,-1\}$ define ${\psi }_{i,\epsilon }:U\to {\mathbb{R}}^n$ by inserting $\epsilon \sqrt{1-\mathrm{\mid }x{\mathrm{\mid }}_2^2}$ in the $i$-th coordinate and listing the coordinates of $x$ in the remaining positions in increasing order. Each ${\psi }_{i,\epsilon }$ is smooth: its components are either coordinate functions or the square root of the strictly positive polynomial $1-\mathrm{\mid }x{\mathrm{\mid }}_2^2$ on $U$, so step 2.1 applies; since $U$ and ${\mathbb{R}}^n$ are smooth manifolds with their standard structures [F27] and smoothness of a map on an open Euclidean set agrees with the chart-based notion [F14], each ${\psi }_{i,\epsilon }$ is continuous by [F12]. Moreover ${\psi }_{i,\epsilon }(U)=\{y\in S^{n-1}:\epsilon y_i>0\},$ since $\mathrm{\mid }{\psi }_{i,\epsilon }(x){\mathrm{\mid }}_2^2=\mathrm{\mid }x{\mathrm{\mid }}_2^2+(1-\mathrm{\mid }x{\mathrm{\mid }}_2^2)=1$ and $\epsilon {\psi }_{i,\epsilon }(x{)}_i=\sqrt{1-\mathrm{\mid }x{\mathrm{\mid }}_2^2}>0$ for $x\in U$, while for $y\in S^{n-1}$ with $\epsilon y_i>0$ the list $x$ of its remaining coordinates satisfies $\mathrm{\mid }x{\mathrm{\mid }}_2^2=1-y_i^2<1$ and ${\psi }_{i,\epsilon }(x)=y$, because $\sqrt{y_i^2}=\mathrm{\mid }{y}_i\mathrm{\mid }=\epsilon y_i$ by [F18]. Hence $S^{n-1}={\bigcup }_{i,\epsilon }{\psi }_{i,\epsilon }(U)$: a unit vector is nonzero, so some coordinate $y_i$ does not vanish.

Continuity of the parametrised radial functions. [F2, F45, F46, step 1.3, step 3.1] For each pair $(i,\epsilon )$ put $f_{i,\epsilon }:=f\circ {\psi }_{i,\epsilon }:U\to (0,+\mathrm{\infty }]$. This is continuous: ${\psi }_{i,\epsilon }$ is continuous (step 3.1) and $f$ is continuous at every point of $S^{n-1}$ (step 1.3), so the composite is continuous by [F45]. For $k\in {\mathbb{N}}_0$ put $D_{i,\epsilon ,k}:=\{x\in U:f_{i,\epsilon }(x)\in [k,k+1]\}=f_{i,\epsilon }^{-1}([k,k+1])\mathrm{.}$ This set is closed in $U$: the interval $[k,k+1]$ is closed in the order topology of $(0,+\mathrm{\infty }]$ in which $f$ and hence $f_{i,\epsilon }$ is continuous [F2], and the preimage of a closed set under a continuous map is closed [F46].

Compact slabs covering the parametrised data. [F2, F37, F38, F39, F40, F49, F50, step 1.1, step 3.1, step 4.1, algebra, cases] For $m\ge 2$ let $B_m$ be the closed Euclidean ball of radius $1-1\mathrm{/}m$ about the origin in ${\mathbb{R}}^{n-1}$; it is contained in $U$, and it is compact: if $n\ge 2$ this is [F37] applied in ${\mathbb{R}}^{n-1}$, while if $n=1$ then $B_m=U={\mathbb{R}}^0$ is a one-point space, which is compact because the single open set ${\mathbb{R}}^0$ already covers it [F49]. The balls $B_m$ cover $U$: for $x\in U$ the number ${\epsilon }_0:=1-\mathrm{\mid }x{\mathrm{\mid }}_2$ is positive, so by [F39] there is $m\ge 2$ with $1\mathrm{/}m<{\epsilon }_0$, whence $\mathrm{\mid }x{\mathrm{\mid }}_2<1-1\mathrm{/}m$ and $x\in B_m$. For each quadruple $(i,\epsilon ,k,m)$ define $C_{i,\epsilon ,k,m}:=D_{i,\epsilon ,k}\cap B_m\subseteq U\mathrm{.}$ This set is closed in $B_m$: $D_{i,\epsilon ,k}$ is closed in the subspace $U$, so it is the trace $F\cap U$ of a closed set $F\subseteq {\mathbb{R}}^{n-1}$ [F50], and $C_{i,\epsilon ,k,m}=F\cap B_m$ because $B_m\subseteq U$, which is a trace of a closed set on the subspace $B_m$ and hence closed in $B_m$ [F50]. Therefore $C_{i,\epsilon ,k,m}$ is a closed subset of the compact space $B_m$ and is compact by [F38]. On $C_{i,\epsilon ,k,m}$ the function $f_{i,\epsilon }$ takes values in $[k,k+1]\subseteq \mathbb{R}$, so its restriction is a continuous real-valued function [F45]. Finally every element $f(y)y$ of $\lambda (E)$ with $f(y)<+\mathrm{\infty }$ is captured: write $y={\psi }_{i,\epsilon }(x)$ by step 3.1, put $k:=\lfloor f(y)\rfloor$, so that $k\ge 0$ by the positivity clause of [F2] and $f(y)\in [k,k+1]$ by [F40], and choose $m\ge 2$ with $x\in B_m$ as above; then $x\in C_{i,\epsilon ,k,m}$.

The radial graphs are compact and have content zero. [F23, F44, F45, F50, F52, step 4.1, step 5.1] For each quadruple $(i,\epsilon ,k,m)$ the graph ${\mathrm{\Gamma }}_{i,\epsilon ,k,m}:=\{(x,f_{i,\epsilon }(x)):x\in C_{i,\epsilon ,k,m}\}\subseteq {\mathbb{R}}^{n-1}\times \mathbb{R}={\mathbb{R}}^n$ is the image of the compact set $C_{i,\epsilon ,k,m}$ under the map $g(x):=(x,f_{i,\epsilon }(x))$, whose first $n-1$ components are coordinate projections and whose last component is the restriction of the continuous function $f_{i,\epsilon }$ to the subspace $C_{i,\epsilon ,k,m}$, hence continuous [F45]; the componentwise criterion [F52] therefore makes $g$ continuous, and its image is compact by [F44]. On $C_{i,\epsilon ,k,m}$ the function $f_{i,\epsilon }$ takes values in $[k,k+1]$, so its restriction is a continuous real-valued function on the compact set $C_{i,\epsilon ,k,m}$, and [F23] gives that ${\mathrm{\Gamma }}_{i,\epsilon ,k,m}$ has content zero in ${\mathbb{R}}^{(n-1)+1}={\mathbb{R}}^n$. When $C_{i,\epsilon ,k,m}=\mathrm{\varnothing }$ the graph is empty and has content zero as well.

The smooth cone map pushes the graphs into null sets. [F19, F21, F22, F24, step 1.1, step 3.1, step 6.1, algebra] For fixed $(i,\epsilon )$ define the cone map ${\mathrm{\Psi }}_{i,\epsilon }:U\times \mathbb{R}\to {\mathbb{R}}^n,{\mathrm{\Psi }}_{i,\epsilon }(x,t):=t{\psi }_{i,\epsilon }(x),$ on the open set $U\times \mathbb{R}\subseteq {\mathbb{R}}^n$, where $n\ge 1$ by step 1.1. Its components are products of the coordinate $t$ with the components of ${\psi }_{i,\epsilon }$ (step 3.1), so by the closure theorem [F19] the map ${\mathrm{\Psi }}_{i,\epsilon }$ is $C^k$ on $U\times \mathbb{R}$ for every $k$, that is smooth [F21], and in particular $C^1$. The graph ${\mathrm{\Gamma }}_{i,\epsilon ,k,m}$ of step 6.1 is a compact content-zero subset of the open set $U\times \mathbb{R}$, so [F24] applies and shows that ${\mathrm{\Psi }}_{i,\epsilon }\left({\mathrm{\Gamma }}_{i,\epsilon ,k,m}\right)=\{f(y)y:y={\psi }_{i,\epsilon }(x),x\in C_{i,\epsilon ,k,m}\}$ is compact with content zero, hence null by [F22].

The Euclidean radial-graph set is null. [F22, F25, step 1.3, step 3.1, step 5.1, step 7.1, algebra] By step 1.3 every element of $\lambda (E)$ has the form $f(y)y$ with $y\in S^{n-1}$ and $f(y)<+\mathrm{\infty }$, and by steps 3.1 and 5.1 each such element lies in one of the sets ${\mathrm{\Psi }}_{i,\epsilon }({\mathrm{\Gamma }}_{i,\epsilon ,k,m})$; conversely each of those images consists of vectors $f(y)y$ with $y={\psi }_{i,\epsilon }(x)\in S^{n-1}$ (step 3.1) and $f(y)=f_{i,\epsilon }(x)\in [k,k+1]$ finite, so it is contained in $\lambda (E)$. Therefore $\lambda (E)={\bigcup }_{i,\epsilon ,k,m}{\mathrm{\Psi }}_{i,\epsilon }\left({\mathrm{\Gamma }}_{i,\epsilon ,k,m}\right),$ a finite union over $(i,\epsilon )$ of countable unions over $k\ge 0$ and $m\ge 2$ of null sets (step 7.1), which is null in ${\mathbb{R}}^n$ by the closure properties of [F25]; the implication from content zero to nullity is the one recorded in [F22].

The cut locus is a null subset of $M$. [F1, F9, F10, F11, F13, F14, F26, F27, step 1.3, step 8.1] By completeness and Hopf-Rinow [F10] the fibre domain is ${\mathcal{E}}_p=T_{p}M$, and [F11] makes ${\exp }_p:T_{p}M\to M$ smooth. Smoothness on $T_{p}M$ is read in the canonical linear structure of the finite-dimensional real vector space $T_{p}M$ [F9]: the coefficient map $\lambda$ of step 1.3 is a chart of $T_{p}M$, with respect to which the coordinate representative of ${\lambda }^{-1}:{\mathbb{R}}^n\to T_{p}M$ is the identity map of ${\mathbb{R}}^n$, which is smooth, and by the basis-independence recorded in [F9] the map ${\lambda }^{-1}$ is smooth. Hence the composite $F:={\exp }_p\circ {\lambda }^{-1}:{\mathbb{R}}^n⟶M$ is smooth by [F13] and in particular $C^1$ in the chart-based sense of [F14]. The space ${\mathbb{R}}^n$ is a smooth $n$-manifold by [F27] and $\dim M=n$, so $F$ is an equidimensional $C^1$ map between smooth manifolds; since $\lambda (E)$ is null in ${\mathbb{R}}^n$ by step 8.1, [F26] makes $F(\lambda (E))$ a null subset of $M$. Finally $F(\lambda (E))={\exp }_p({\lambda }^{-1}(\lambda (E)))={\exp }_p(E)=\mathrm{Cut}(p)$, the last equality by the definition of $E$ in step 1.3 and of the cut locus in [F1].

Chartwise nullity. [A1, F28, F29, step 1.1, step 9.1, given] By [F28] and the standing ${\mathrm{A}\mathrm{C}}_{\omega }$ of [A1] choose a countable smooth atlas $\{(U_j,{\phi }_j){\}}_{j\in \mathbb{N}}$ of $M$ whose chart domains have compact closures. Since $\mathrm{Cut}(p)$ is null in $M$ by step 9.1 and $n=\dim M\ge 1$ by step 1.1, the atlas criterion [F29] gives that ${\phi }_j(\mathrm{Cut}(p)\cap U_j)$ is a null subset of ${\mathbb{R}}^n$ for every $j$.

The Riemannian volume vanishes. [A1, F3, F12, F25, F27, F28, F30, F31, F32, F33, F34, F35, F36, F41, F47, F48, F50, F51, F53, step 1.1, step 10.1, algebra] Let $\{(U_j,{\phi }_j){\}}_{j\in \mathbb{N}}$ be the atlas produced in step 10.1 and put $G_j:=\mathrm{Cut}(p)\cap U_j$. Each $G_j$ is Borel: $\mathrm{Cut}(p)$ is closed in $M$ [F3] and $U_j$ is open, so both are Borel subsets of $M$, and the Borel sigma-algebra is closed under finite intersections [F41]; moreover ${\bigcup }_j{G}_j=\mathrm{Cut}(p)$. Fix $j$, write $x:={\phi }_j:U_j\to O:=x(U_j)$ for the chart, and let $r$ be the coefficient of the Riemannian density in this chart; by [F31] the density is positive and smooth with coefficient $r$, so $r:O\to (0,+\mathrm{\infty })$ is smooth, hence continuous as a map of smooth manifolds [F12] (the open set $O\subseteq {\mathbb{R}}^n$ carries the standard smooth structure [F27]). Let $\tilde{r}:{\mathbb{R}}^n\to [0,+\mathrm{\infty })$ be $r$ extended by $0$ outside $O$. Then $\tilde{r}$ is Borel measurable: the family $\mathcal{C}:=\{B\subseteq \mathbb{R}:{\tilde{r}}^{-1}(B)\in \mathcal{B}({\mathbb{R}}^n)\}$ is a sigma-algebra, and it contains every open $V\subseteq \mathbb{R}$, because ${\tilde{r}}^{-1}(V)$ is the union of $O\cap r^{-1}(V)$, which is open in ${\mathbb{R}}^n$ since $O$ is open and $r$ is continuous and a set open in the open subspace $O$ is open in ${\mathbb{R}}^n$ [F50], and of $O^{\mathrm{c}}$ when $0\in V$, which is closed [F41]; since the Borel sigma-algebra is generated by the open sets and is the smallest sigma-algebra containing them [F41, F51], $\mathcal{C}$ contains $\mathcal{B}(\mathbb{R})$, so $\tilde{r}$ is Borel measurable [F47]. Every Borel set is Lebesgue measurable [F53], so $\tilde{r}$ is Lebesgue measurable [F47]. By [F30] the measure ${\mathrm{vol}}_g$ is the Borel measure of the density ${\mu }_g$, so the chart formula [F32] gives ${\mathrm{vol}}_g(G_j)={\int }_{x(G_j)}rd{\lambda }_n={\int }_{x(G_j)}\tilde{r}d{\lambda }_n,$ the second equality because $\tilde{r}$ agrees with $r$ on $O\supseteq x(G_j)$ and the integral over a set is the integral of the product with its characteristic function [F48]. The set $x(G_j)$ is contained in the null set $x(\mathrm{Cut}(p)\cap U_j)$ of step 10.1, hence is null [F25]; by [F33] it has Lebesgue outer measure zero, and by [F34] it is Lebesgue measurable with ${\lambda }_n(x(G_j))=0$. Therefore [F35], applied to the measurable nonnegative function $\tilde{r}$ and this measurable null set, gives ${\int }_{x(G_j)}\tilde{r}d{\lambda }_n=0$ and hence ${\mathrm{vol}}_g(G_j)=0$ for every $j$. Disjointifying, $G_0^{\mathrm{\prime }}:=G_0$ and $G_k^{\mathrm{\prime }}:=G_k\setminus (G_0\cup \cdots \cup G_{k-1})$ for $k\ge 1$, the sets $G_k^{\mathrm{\prime }}$ are Borel [F41], pairwise disjoint, satisfy $G_k^{\mathrm{\prime }}\subseteq G_k$, and have union $\mathrm{Cut}(p)$; countable additivity of the measure ${\mathrm{vol}}_g$ [F36] and the nonnegativity of measure values then give ${\mathrm{vol}}_g(\mathrm{Cut}(p))={\sum }_{k=0}^{\mathrm{\infty }}{\mathrm{vol}}_g(G_k^{\mathrm{\prime }})\le {\sum }_{k=0}^{\mathrm{\infty }}{\mathrm{vol}}_g(G_k)=0,$ so that ${\mathrm{vol}}_g(\mathrm{Cut}(p))=0$.

Boundary, endpoint and choice audit. [A1, F1, F2, F10, F11, F22, F23, F24, F25, F26, F28, F29, F30, F33, F35, F36, F38, F41, F43, F48, F49, F50, step 1.1, step 1.3, step 5.1, step 6.1, step 11.1, given] Dimension zero is discharged in step 1.1, where the empty cut locus has volume zero. Dimension one is covered by the same argument: ${\mathbb{R}}^{n-1}={\mathbb{R}}^0$ is a single point, the parametrisations ${\psi }_{1,\pm }$ are the constants $\pm 1$, the graph theorem [F23] covers compact subsets of ${\mathbb{R}}^0$ (its own proof treats the case $m=0$), the cone map ${\mathrm{\Psi }}_{1,\pm }$ is the map $t\mapsto \pm t$ on the open set ${\mathbb{R}}^0\times \mathbb{R}=\mathbb{R}$, and the image lemma [F24] applies there with ambient dimension $m=n=1$; the only step needing care was the compactness of the one-point ball $B_m$, handled in step 5.1 by the definition of compactness [F49]. The empty manifold carries no point $p$ and is excluded by the hypothesis $p\in M$. Degenerate pieces are harmless: a quadruple with $C_{i,\epsilon ,k,m}=\mathrm{\varnothing }$ contributes the empty graph, content zero by [F22], and the ball $B_m$ has radius $1-1\mathrm{/}m>0$ because $m\ge 2$, so no zero-radius ball is used; the closed intervals $[k,k+1]$ of step 6.1 include both endpoints, which costs nothing because [F23] needs only compactness of the parameter set. The zero vector is never passed to the exponential differential and is not part of $S_{p}M$ or $E$ by [F1], and $\lambda$ is defined on all of $T_{p}M$. Assumption [A1] is used through the suppliers that require it and nowhere else: the Hopf-Rinow and exponential smoothness statements [F10, F11], the countable atlas [F28], the nullity criterion [F29], the countable-union closure of null sets [F25], the equidimensional image theorem [F26], the volume measure [F30] and the measurable-integral facts [F33, F35, F36]; the covering argument of step 5.1 uses the Archimedean property and the integer part, neither of which needs choice, and the choice of basis in step 1.3 is a single existential instantiation. The statement asserts one equality and contains no biconditional, so there is no forward or reverse implication of an equivalence to verify; the only equivalence invoked, nullity versus vanishing outer measure [F33], is used in the direction from nullity to vanishing outer measure. Finally the proof never uses compactness of $M$ and never assumes the cut locus to be a submanifold: nullity is produced from the continuity of the cut time alone, and the closing measure computation uses only the density chart formula.