Polar integration may discard the cut locus

Dimension zero. If $n=0$ then $S_{p}M=\mathrm{\varnothing }$ and $\mathrm{Cut}(p)=\mathrm{\varnothing }$ by [F3], so $M\setminus \mathrm{Cut}(p)=M$ and both sides of the identity of (a) are literally the same integral, for $[0,\mathrm{\infty }]$-valued and for integrable Borel $f$ alike; and ${\mathrm{vol}}_g$ is counting measure by [F13], which is why the polar identity of (b) is not asserted for $n=0$: there $S_{p}M=\mathrm{\varnothing }$ makes its right-hand side $0$, while ${\int }_{M}fd{\mathrm{vol}}_g$ can be nonzero. Assume $n\ge 1$ from here on.

A linear isometry, the transported sphere measure, and its integral transformation. By [F18] choose an orthonormal basis $(e_1^0,\dots ,e_n^0)$ of $(T_{p}M,g_p)$ and let $\phi :T_{p}M\to {\mathbb{R}}^n$ be its coefficient map $\phi ({\sum }_i{x}_i{e}_i^0)=(x_1,\dots ,x_n)$. This is a single existential instantiation. By [F18], $\phi$ is a linear isometric isomorphism and therefore a homeomorphism with $\phi (S_{p}M)=S^{n-1}$. Define ${\sigma }_p(E):=\sigma (\phi (E))$ for Borel $E\subseteq S_{p}M$, where $\sigma$ is the polar surface measure of [F17]; since $\phi$ is a bijection, this is the image of $\sigma$ under ${\phi }^{-1}$, so ${\sigma }_p$ is a Borel measure with ${\sigma }_p(S_{p}M)=\sigma (S^{n-1})<\mathrm{\infty }$ by [F17], and the defining formula of an image measure gives, for every nonnegative Borel $g:S_{p}M\to [0,\mathrm{\infty }]$, ${\int }_{S_{p}M}gd{\sigma }_p={\int }_{S^{n-1}}(g\circ {\phi }^{-1})d\sigma \mathrm{.}$ Indeed, for $g={\mathbf{1}}_E$ with Borel $E\subseteq S_{p}M$ both sides are $\sigma (\phi (E))$, for nonnegative simple $g$ the identity is finite additivity of $\sigma$, and for general nonnegative Borel $g$ it follows by increasing simple approximation and monotone convergence [F33]. If ${\phi }^{\mathrm{\prime }}:T_{p}M\to {\mathbb{R}}^n$ comes from a second orthonormal basis, then ${\phi }^{\mathrm{\prime }}\circ {\phi }^{-1}$ is orthogonal by [F19], so for every Borel $E\subseteq S_{p}M$ the orthogonal invariance of $\sigma$ recorded in [F19] gives $\sigma ({\phi }^{\mathrm{\prime }}(E))=\sigma (({\phi }^{\mathrm{\prime }}\circ {\phi }^{-1})(\phi (E)))=\sigma (\phi (E))$; thus ${\sigma }_p$ does not depend on the chosen isometry. Consequently, for every nonnegative Borel $G:S^{n-1}\to [0,\mathrm{\infty }]$, ${\int }_{S^{n-1}}Gd\sigma ={\int }_{S_{p}M}(G\circ \phi )d{\sigma }_p,$ the case $g=G\circ \phi$ of the displayed identity.

The normal chart and its metric matrix. Put $\mathrm{\Omega }:=\phi (D_p)\subseteq {\mathbb{R}}^n\setminus \{0\}$ and $U:=M\setminus (\{p\}\cup \mathrm{Cut}(p))$. By [F6] the set $D_p$ is open and ${\exp }_p{\mathrm{\mid }}_{D_p}:D_p\to U$ is a diffeomorphism onto the open submanifold $U$, so $x:=\phi \circ ({\exp }_p{\mathrm{\mid }}_{D_p}{)}^{-1}:U\to \mathrm{\Omega }$ is a diffeomorphism onto the open set $\mathrm{\Omega }$ with inverse $x^{-1}(\xi )={\exp }_p({\phi }^{-1}(\xi ))$ for $\xi \in \mathrm{\Omega }$ [F16, F25]. Write $c_{\phi }(\omega ):=c_p({\phi }^{-1}(\omega ))$ for $\omega \in S^{n-1}$; then $\mathrm{\Omega }=\phi (D_p)=\{t\omega :\omega \in S^{n-1},0<t<c_{\phi }(\omega )\},$ because $\phi$ is linear and $\phi (S_{p}M)=S^{n-1}$ by [F18]. For $\xi \in \mathrm{\Omega }$ with $w:={\phi }^{-1}(\xi )$ let $G(\xi )$ be the metric matrix of the chart $x$ at $\xi$, $G_{ij}(\xi )=g_{x^{-1}(\xi )}(d({\exp }_p{)}_w(e_i^0),d({\exp }_p{)}_w(e_j^0)),1\le i,j\le n;$ this is the matrix of the Riemannian metric in this chart, since by the chain rule [F25] the coordinate frame is ${\mathrm{\partial }}_i{\mathrm{\mid }}_{\xi }=d(x^{-1}{)}_{\xi }(e_i^{\mathrm{s}\mathrm{t}\mathrm{d}})=d({\exp }_p{)}_w(e_i^0)$. The coefficient of the Riemannian volume density in this chart is $\sqrt{\det G(\xi )}$, positive and smooth on the open set $\mathrm{\Omega }$, by [F13].

The Jacobi fields, the frame, and the matrix $a_v$. Fix $v\in S_{p}M$. The Jacobi fields $J_i$ with $J_i(0)=0$, $D_t{J}_i(0)=e_i$ and the parallel sections $E_i$ with $E_i(0)=e_i$ exist, are unique on all of $[0,\mathrm{\infty })$, and are smooth in $t$, by [F9] and [F10]; moreover $E_1(t),\dots ,E_{n-1}(t)$ is an orthonormal frame of ${\dot{\gamma }}_v(t{)}^{\perp }$ by [F8] and [F9]. Define $a_{ij}(t):=g_{{\gamma }_v(t)}(J_i(t),E_j(t))$ for $t\ge 0$. Since $E_1(t),\dots ,E_{n-1}(t)$ is an orthonormal basis of ${\dot{\gamma }}_v(t{)}^{\perp }$ and each $J_i(t)$ lies in that subspace by the Gauss lemma [F12] applied with $w=tv$ and $z=t{e}_i$, the expansion in an orthonormal basis gives $J_i(t)={\sum }_j{a}_{ij}(t)E_j(t)$; hence the Gram matrix $A(t):=(g_{{\gamma }_v(t)}(J_i(t),J_j(t)){)}_{i,j}$ satisfies $A(t)=a(t)a(t{)}^{\mathsf{T}}$ and therefore $\det A(t)=\det (a(t){)}^2$ by [F26]. The entries $a_{ij}$ are smooth in $t$ by [F27], and the metric product rule [F27] evaluated at $t=0$, where $D_t{J}_i(0)=e_i$, $J_i(0)=0$, $E_j(0)=e_j$ and $D_t{E}_j=0$ by [F9] and [F10], gives $a_{ij}^{\mathrm{\prime }}(0)=g_p(e_i,e_j)+g_p(0,0)={\delta }_{ij}\mathrm{.}$ Consequently $a(t)=tI+o(t)$ as $t\to 0^{+}$ by the definition of the derivative [F27], so $t^{-1}a(t)\to I$ entrywise; the determinant is a finite sum of products of the entries [F26], so the limit rules [F28] give $\det (t^{-1}a(t))\to \det I=1$, and therefore $\det a(t)=t^{n-1}\det (t^{-1}a(t))>0$ for every sufficiently small $t>0$ (for $n=1$ the matrix $a(t)$ is empty and $\det a(t)=1>0$ for all $t$ by the empty-determinant convention [F26]). Next, $\det a(t)\mathrm{\ne }0$ for every $0<t<c_p(v)$: if $\det a(t)=0$, then $\det A(t)=0$, so the list $J_1(t),\dots ,J_{n-1}(t)$ is linearly dependent by [F26]; choose a nonzero coefficient vector $(c_i)$ with ${\sum }_i{c}_i{J}_i(t)=0$. The combination $J:={\sum }_i{c}_i{J}_i$ is a Jacobi field along ${\gamma }_v$ because $\mathcal{J}({\gamma }_v)$ is a real vector space [F10], it satisfies $J(0)=0$ and $J(t)=0$, and it is not the zero field because $D_{t}J(0)={\sum }_i{c}_i{e}_i\mathrm{\ne }0$ by the independence of the $e_i$ and the uniqueness in [F10]. Hence $p={\gamma }_v(0)$ and ${\gamma }_v(t)$ are conjugate along ${\gamma }_v{\mathrm{\mid }}_{[0,t]}$ by [F29], and the conjugacy clause of [F29] gives $c_p(v)\le t$, contradicting $t<c_p(v)$. Since $t\mapsto \det a(t)$ is continuous on $(0,\mathrm{\infty })$ (its entries are smooth [F27]) and has no zero on the interval $(0,c_p(v))$, while it is positive near $0$, the intermediate value theorem [F30] shows that $\det a(t)>0$ for every $0<t<c_p(v)$. Finally, $\det a(t)$ does not depend on the choice of the orthonormal basis of $v^{\perp }$: if $e_i^{\mathrm{\prime }}={\sum }_k{O}_{ik}{e}_k$ with $O$ orthogonal, then uniqueness in [F9] and [F10] gives $J_i^{\mathrm{\prime }}={\sum }_k{O}_{ik}{J}_k$ and $E_i^{\mathrm{\prime }}={\sum }_k{O}_{ik}{E}_k$, hence $a^{\mathrm{\prime }}=Oa{O}^{\mathsf{T}}$ and $\det a^{\mathrm{\prime }}=\det a$ by [F26].

The complement of $U$ is ${\mathrm{vol}}_g$-null. The set $\mathrm{Cut}(p)$ is closed by [F5] and ${\mathrm{vol}}_g(\mathrm{Cut}(p))=0$ by [F4]. For the singleton, choose a smooth chart $(V,y)$ of $M$ with $p\in V$, which exists by [F16], and let $r$ be the coefficient of the Riemannian volume density in this chart; then $\{p\}$ is Borel (singletons are closed in the Hausdorff manifold and closed sets are Borel [F5, F16]), and the chart-restriction formula [F14] applied to the measure ${\mathrm{vol}}_g={\mu }_g$ of [F13] gives ${\mathrm{vol}}_g(\{p\})={\int }_{y(\{p\})}rd{\lambda }_n={\int }_{\{y(p)\}}rd{\lambda }_n=0,$ the last equality because $\{y(p)\}$ is a singleton, hence ${\lambda }_n$-null by [F24], and integrals of nonnegative functions over null sets vanish by [F21]. Therefore ${\mathrm{vol}}_g(\{p\}\cup \mathrm{Cut}(p))\le {\mathrm{vol}}_g(\{p\})+{\mathrm{vol}}_g(\mathrm{Cut}(p))=0$ by finite subadditivity [F20], so $M\setminus U=\{p\}\cup \mathrm{Cut}(p)$ is a Borel set of ${\mathrm{vol}}_g$-measure zero.

Part (a) for nonnegative integrands. Let $C:=\mathrm{Cut}(p)$, a Borel null set by [F4, F5]. For Borel $f:M\to [0,\mathrm{\infty }]$, the functions $f{\chi }_C$ and $f{\chi }_{M\setminus C}$ are nonnegative Borel [F21] and $f=f{\chi }_C+f{\chi }_{M\setminus C}$ pointwise, so additivity of the nonnegative integral [F22] gives ${\int }_{M}fd{\mathrm{vol}}_g={\int }_{M}f{\chi }_{C}d{\mathrm{vol}}_g+{\int }_{M}f{\chi }_{M\setminus C}d{\mathrm{vol}}_g\mathrm{.}$ The first term equals ${\int }_{C}fd{\mathrm{vol}}_g=0$ by the definition of the integral over a measurable set and the null-set theorem [F21], and the second is ${\int }_{M\setminus \mathrm{Cut}(p)}fd{\mathrm{vol}}_g$ by the same definition. This is the asserted identity.

Part (a) for integrable integrands. Let $f\in L^1({\mathrm{vol}}_g)$ be real or complex and Borel, and put $C:=\mathrm{Cut}(p)$ and $g:=f{\chi }_{M\setminus C}$. Then $g$ is Borel with $\mathrm{\mid }g\mathrm{\mid }\le \mathrm{\mid }f\mathrm{\mid }$, so $g\in L^1({\mathrm{vol}}_g)$ [F21, F23]; $f$ and $g$ agree outside the null set $C$, so $f=g$ almost everywhere [F21]; and ${\int }_{M}gd{\mathrm{vol}}_g={\int }_{M\setminus \mathrm{Cut}(p)}fd{\mathrm{vol}}_g$ by the definition of the integral over a measurable set [F21]. The almost-everywhere equality theorem [F23], applied with $A=M$, gives ${\int }_{M}fd{\mathrm{vol}}_g={\int }_{M}gd{\mathrm{vol}}_g$, which is the asserted identity.

The chart formula on $U$. By [F16] the open set $U$ carries the structure of a smooth manifold and $x:U\to \mathrm{\Omega }$ is a smooth chart of it; the restricted Riemannian metric makes it a Riemannian manifold whose volume density is the restriction of ${\mu }_g$, because for a Borel set $E\subseteq U$ the chart-restriction formula [F14] applied in $M$ and in the open submanifold $U$ gives in both cases the value ${\int }_{x(E)}\sqrt{\det G}d{\lambda }_n$, the coefficient of the Riemannian density being the same positive function $\sqrt{\det G}$ by [F13]. The one-member family consisting of the chart $x$ and the constant function $1$ is admissible chart-partition gluing data [F15], so the density-integration theorem [F15] applied to the manifold $U$ with this chart partition gives, for every nonnegative Borel $F:U\to [0,\mathrm{\infty }]$, ${\int }_{U}Fd{\mathrm{vol}}_g={\int }_{\mathrm{\Omega }}F(x^{-1}(\xi ))\sqrt{\det G(\xi )}d{\lambda }_n(\xi )\mathrm{.}$

The determinant identity. Fix $\omega \in S^{n-1}$ and $0<t<c_{\phi }(\omega )$, and put $v:={\phi }^{-1}(\omega )$ and $w:=tv$. Then $w\in D_p$ with $x^{-1}(t\omega )={\exp }_p(w)={\gamma }_v(t)$ by 1.3, and $e_n:=v$ together with an orthonormal basis $(e_1,\dots ,e_{n-1})$ of $v^{\perp }$ is an orthonormal basis $(e_1,\dots ,e_n)$ of $T_{p}M$ [F18]. The Gram matrix of the vectors $d({\exp }_p{)}_w(e_1),\dots ,d({\exp }_p{)}_w(e_n)$ differs from $G(t\omega )$ by conjugation with the orthogonal change-of-basis matrix, so it has the same determinant by [F26]. Its entries are computed as follows. For $i,j\le n-1$, linearity of the differential [F25] and the Jacobi identification [F11] give $d({\exp }_p{)}_w(t{e}_i)=d({\exp }_p{)}_{tv}\left(t{e}_i\right)=J_i(t),$ where the last equality uses [F11] for the base vector $tv\in T_{p}M$ (which lies in the exponential domain by [F7]) and the initial vector $t{e}_i$, together with the uniqueness in [F10] which identifies the resulting field with $s\mapsto J_i(st)$; hence $g(d({\exp }_p{)}_w(e_i),d({\exp }_p{)}_w(e_j))=t^{-2}g(J_i(t),J_j(t))=t^{-2}A(t{)}_{ij}$ with $A(t)=a(t)a(t{)}^{\mathsf{T}}$ as in 1.4. For $i\le n-1$ the Gauss lemma [F12] with base vector $tv$ and second vector $t{e}_i$ gives $g(d({\exp }_p{)}_{tv}(tv),d({\exp }_p{)}_{tv}(t{e}_i))=g_p(tv,t{e}_i)=t^2{g}_p(v,e_i)=0$, and $d({\exp }_p{)}_{tv}(tv)=td({\exp }_p{)}_{tv}(v)$ by linearity [F25], so $g(d({\exp }_p{)}_w(v),d({\exp }_p{)}_w(e_i))=0$; taking the pair $(tv,tv)$ gives $g(d({\exp }_p{)}_w(v),d({\exp }_p{)}_w(v))=1$ in the same way. Therefore the Gram matrix in the adapted basis is block diagonal with blocks $1$ and $t^{-2}A(t)$, so $\det G(t\omega )=t^{-2(n-1)}\det A(t)=t^{-2(n-1)}(\det a(t){)}^2$ by [F26]; since $\det a(t)>0$ on $(0,c_p(v))$ by 1.4, taking square roots gives $\sqrt{\det G(t\omega )}{t}^{n-1}=\det a(t)\mathrm{.}$ For $n=1$ the basis $(e_1,\dots ,e_{n-1})$ is empty, $G(t\omega )$ is the $1\times 1$ matrix with entry $1$, and both sides are $1$ by the empty-determinant convention [F26].

Reduction to a Euclidean polar integral. Let $f:M\to [0,\mathrm{\infty }]$ be Borel and let $F:=f{\mathrm{\mid }}_U$. Since $M\setminus U=\{p\}\cup \mathrm{Cut}(p)$ is null by 1.5, additivity [F22] and the null-set theorem [F21] give ${\int }_{M}fd{\mathrm{vol}}_g={\int }_{U}fd{\mathrm{vol}}_g$. Inserting the chart formula of 1.8, the integral becomes ${\int }_{\mathrm{\Omega }}f(x^{-1}(\xi ))\sqrt{\det G(\xi )}d{\lambda }_n(\xi )={\int }_{{\mathbb{R}}^n}hd{\lambda }_n$ for the function $h$ which equals $f(x^{-1}(\xi ))\sqrt{\det G(\xi )}$ on the open set $\mathrm{\Omega }$ and $0$ elsewhere; this $h$ is nonnegative and Borel, because $f$ is Borel, $x^{-1}$ is smooth hence continuous [F16, F25], $\sqrt{\det G}$ is continuous on $\mathrm{\Omega }$ by [F13], and $\mathrm{\Omega }$ is open [F6, F18]. Since by 1.3 the ray through $\omega \in S^{n-1}$ meets $\mathrm{\Omega }$ exactly in $\{t\omega :0<t<c_{\phi }(\omega )\}$, the polar formula [F17] gives ${\int }_{{\mathbb{R}}^n}hd{\lambda }_n={\int }_0^{\mathrm{\infty }}{\int }_{S^{n-1}}{\mathbf{1}}_{\{t<c_{\phi }(\omega )\}}f({\gamma }_{{\phi }^{-1}(\omega )}(t))\sqrt{\det G(t\omega )}{t}^{n-1}d\sigma (\omega )dt,$ and by the determinant identity of 2.1 the two integrands ${\mathbf{1}}_{\{t<c_{\phi }(\omega )\}}f({\gamma }_{{\phi }^{-1}(\omega )}(t))\sqrt{\det G(t\omega )}{t}^{n-1}$ and ${\mathbf{1}}_{\{t<c_{\phi }(\omega )\}}f({\gamma }_{{\phi }^{-1}(\omega )}(t))\det a_{{\phi }^{-1}(\omega )}(t)$ agree at every point of $S^{n-1}\times (0,\mathrm{\infty })$: on $0<t<c_{\phi }(\omega )$ by 2.1, and off that set because both carry the indicator and $\det a_v$ is defined for all $t\ge 0$ by 1.4. Hence ${\int }_{M}fd{\mathrm{vol}}_g={\int }_0^{\mathrm{\infty }}{\int }_{S^{n-1}}{\mathbf{1}}_{\{t<c_{\phi }(\omega )\}}f({\gamma }_{{\phi }^{-1}(\omega )}(t))\det a_{{\phi }^{-1}(\omega )}(t)d\sigma (\omega )dt\mathrm{.}$

Tonelli's theorem, transport back to $S_{p}M$, and the polar identity. Put $K(\omega ,t):={\mathbf{1}}_{\{t<c_{\phi }(\omega )\}}f({\gamma }_{{\phi }^{-1}(\omega )}(t))\det a_{{\phi }^{-1}(\omega )}(t),(\omega ,t)\in S^{n-1}\times (0,\mathrm{\infty })\mathrm{.}$ This is product-measurable: by 2.1 and 1.3 it equals $k(t\omega )$, where $k(\xi ):={\mathbf{1}}_{\mathrm{\Omega }}(\xi )f(x^{-1}(\xi ))\sqrt{\det G(\xi )}\mathrm{\mid }\xi {\mathrm{\mid }}^{n-1}$ is a Borel function on ${\mathbb{R}}^n$ (indicator of the open set $\mathrm{\Omega }$, the Borel function $f\circ x^{-1}$, the continuous function $\sqrt{\det G}$ on $\mathrm{\Omega }$, and the continuous function $\mathrm{\mid }\xi {\mathrm{\mid }}^{n-1}$, with value $0$ off $\mathrm{\Omega }$), and the map $(\omega ,t)\mapsto t\omega$ is continuous, hence Borel for the product Borel structure, which is the trace of $\mathcal{B}({\mathbb{R}}^{n+1})$ by [F32]. The measure $\sigma$ is finite by [F17] and Lebesgue measure on $(0,\mathrm{\infty })$ is sigma-finite, so Tonelli's theorem [F31] applies and gives ${\int }_0^{\mathrm{\infty }}{\int }_{S^{n-1}}Kd\sigma dt={\int }_{S^{n-1}}{\int }_0^{\mathrm{\infty }}Kdtd\sigma ={\int }_{S^{n-1}}{\int }_0^{c_{\phi }(\omega )}f({\gamma }_{{\phi }^{-1}(\omega )}(t))\det a_{{\phi }^{-1}(\omega )}(t)dtd\sigma (\omega ),$ the last equality because $K$ vanishes for $t\ge c_{\phi }(\omega )$; moreover the function $G(\omega ):={\int }_0^{c_{\phi }(\omega )}f({\gamma }_{{\phi }^{-1}(\omega )}(t))\det a_{{\phi }^{-1}(\omega )}(t)dt$ is Borel on $S^{n-1}$ by the measurability clause of [F31]. Since $G$ is nonnegative and Borel, the integral transformation of 1.2 applied to $G\circ \phi$ gives ${\int }_{S^{n-1}}Gd\sigma ={\int }_{S_{p}M}G(\phi (v))d{\sigma }_p(v)\mathrm{.}$ For $v\in S_{p}M$ one has $\phi (v)\in S^{n-1}$, $c_{\phi }(\phi (v))=c_p(v)$ and ${\gamma }_{{\phi }^{-1}(\phi (v))}={\gamma }_v$ by 1.3, so $G(\phi (v))={\int }_0^{c_p(v)}f({\gamma }_v(t))\det a_v(t)dt$. Combining this with 3.1 proves ${\int }_{M}fd{\mathrm{vol}}_g={\int }_{S_{p}M}{\int }_0^{c_p(v)}f({\gamma }_v(t))\det a_v(t)dtd{\sigma }_p(v)\mathrm{.}$