In our final lemma preceding the proof of Theorem 4.4, we treat a general
measure-theoretic statement which gives sufficient conditions concerning two
measure spaces to ensure that their completions are isomorphic. Though we
suspect that this result is well-known, we present its proof for completeness.

######  Lemma 4.9.

Let $(X_{1},\Sigma_{1},\nu_{1})$ and $(X_{2},\Sigma_{2},\nu_{2})$ be measure
spaces, let $\varphi:X_{1}\to X_{2}$ be a bijection and denote by
$(X_{i},\Sigma_{i}^{\prime},\nu_{i}^{\prime})$ the completion of the measure
space $(X_{i},\Sigma_{i},\nu_{i})$ for $i=1,2$. Assume that the following two
properties are satisfied:

  1. 1.

For each $A_{1}\in\Sigma_{1}$, $\varphi(A_{1})\in\Sigma_{2}^{\prime}$ and
$\nu_{2}^{\prime}(\varphi(A_{1}))=\nu_{1}(A_{1}).$

  2. 2.

For each $A_{2}\in\Sigma_{2}$, $\varphi^{-1}(A_{2})\in\Sigma_{1}^{\prime}$ and
$\nu_{1}^{\prime}(\varphi^{-1}(A_{2}))=\nu_{2}(A_{2})$.

Then the measure spaces $(X_{1},\Sigma_{1}^{\prime},\nu_{1}^{\prime})$ and
$(X_{2},\Sigma_{2}^{\prime},\nu_{2}^{\prime})$ are isomorphic with point
isomorphism $\varphi$. Specifically,

$\Sigma_{2}^{\prime}=\\{A_{2}\subseteq
X_{2}:\varphi^{-1}(A_{2})\in\Sigma_{1}^{\prime}\\}$ (34)

and

$\nu_{2}^{\prime}(A_{2})=\nu_{1}^{\prime}(\varphi^{-1}(A_{2}))$ (35)

for all $A_{2}\in\Sigma_{2}^{\prime}$.

###### Proof.

Let us first assume that $A_{2}\in\Sigma_{2}^{\prime}$. By definition,
$A_{2}=G_{2}\cup H_{2}$ where $G_{2}\in\Sigma_{2}$ and $H_{2}\subseteq
G_{2,0}\in\Sigma_{2}$ with $\nu_{2}^{\prime}(A_{2})=\nu_{2}(G_{2})$ and
$\nu_{2}^{\prime}(H_{2})=\nu_{2}(G_{2,0})=0$. Consequently,
$\varphi^{-1}(A_{2})=\varphi^{-1}(G_{2})\cup\varphi^{-1}(H_{2})$ and
$\varphi^{-1}(H_{2})\subseteq\varphi^{-1}(G_{2,0})$. In view of Property 2,
$\varphi^{-1}(G_{2}),\varphi^{-1}(G_{2,0})\in\Sigma_{1}^{\prime}$ and we have

$\nu_{1}^{\prime}(\varphi^{-1}(G_{2}))=\nu_{2}(G_{2})=\nu_{2}^{\prime}(A_{2})\hskip
28.45274pt\mbox{and}\hskip
28.45274pt\nu_{1}^{\prime}(\varphi^{-1}(G_{2,0}))=\nu_{2}(G_{2,0})=0.$

In view of the fact that
$(X_{1}^{\prime},\Sigma_{1}^{\prime},\nu_{1}^{\prime})$ is complete,
$\varphi^{-1}(H_{2})\in\Sigma_{1}^{\prime}$ and
$\nu_{1}^{\prime}(\varphi^{-1}(H_{2}))=0$. Consequently, we obtain
$\varphi^{-1}(A_{2})=\varphi^{-1}(G_{2})\cup\varphi^{-1}(H_{2})\in\Sigma_{1}^{\prime}$
and

$\nu_{2}^{\prime}(A_{2})=\nu_{1}^{\prime}(\varphi^{-1}(G_{2}))\leq\nu_{1}^{\prime}(\varphi^{-1}(A_{2}))\leq\nu_{1}^{\prime}(\varphi^{-1}(G_{2}))+\nu_{1}^{\prime}(\varphi^{-1}(H_{2}))=\nu_{2}(G_{2})+0=\nu_{2}^{\prime}(A_{2}).$

From this we obtain that $\Sigma_{2}^{\prime}\subseteq\\{A_{2}\subseteq
X_{2}:\varphi^{-1}(A_{2})\in\Sigma_{1}^{\prime}\\}$ and, for each
$A_{2}\in\Sigma_{2}^{\prime}$,
$\nu_{2}^{\prime}(A_{2})=\nu_{1}^{\prime}(\varphi^{-1}(A_{2}))$. It remains to
prove that

$\\{A_{2}\subseteq
X_{2}:\varphi^{-1}(A_{2})\in\Sigma_{1}^{\prime}\\}\subseteq\Sigma_{2}^{\prime}.$

To this end, let $A_{2}$ be a subset of $X_{2}$ for which
$\varphi^{-1}(A_{2})\in\Sigma_{1}^{\prime}$. By the definition of
$\Sigma_{1}^{\prime}$, we have $\varphi^{-1}(A_{2})=G_{1}\cup H_{1}$ where
$G_{1}\in\Sigma_{1}$, $H_{1}\subseteq G_{1,0}\in\Sigma_{1}$ and
$\nu_{1}^{\prime}(H_{1})=\nu_{1}(G_{1,0})=0$. In view of Property 1,
$\varphi(G_{1})\in\Sigma_{2}^{\prime}$,
$\varphi(H_{1})\subseteq\varphi(G_{1,0})\in\Sigma_{2}^{\prime}$ and
$\nu_{2}^{\prime}(\varphi(G_{1,0}))=\nu_{1}(G_{1,0})=0$. Because
$(X_{2}^{\prime},\Sigma_{2}^{\prime},\nu_{2}^{\prime})$ is complete, we have
$\varphi(H_{1})\in\Sigma_{2}^{\prime}$ and so

$A_{1}=\varphi(\varphi^{-1}(A_{2}))=\varphi(G_{1})\cup\varphi(H_{1})\in\Sigma_{2}^{\prime},$

as desired. ∎

We are finally in a position to prove Theorem 4.4.

###### Proof of Theorem 4.4.

Denote by $\mathcal{C}$ the collection of sets $G\subseteq(0,\infty)\times S$
for which $\psi_{E}(G)\in\mathcal{M}_{d}$ and
$m(\psi_{E}(G))=(\lambda_{P}\times\sigma_{P,E})(G).$ By virtue of Lemma 4.7,
it follows that $\mathcal{C}$ contains all elementary sets, i.e., finite
unions of disjoint measurable rectangles. Using the continuity of measure
(applied to the measures $m$ and $\lambda_{P}\times\sigma_{P,E}$) and the fact
that $\psi_{E}$ is a bijection, it is straightforward to verify that
$\mathcal{C}$ is a monotone class. By the monotone class lemma (Theorem 8.3 of
[23]), it immediately follows that
$\mathcal{L}\times\Sigma_{P,E}\subseteq\mathcal{C}$. In other words, for each
$G\in\mathcal{L}\times\Sigma_{P,E}$,

$\psi_{E}(G)\in\mathcal{M}_{d}\hskip 28.45274pt\mbox{and}\hskip
28.45274ptm(\psi_{E}(G))=(\lambda_{P}\times\sigma_{P,E})(G).$ (36)

We claim that, for each Borel subset $A$ of $\mathbb{R}^{d}\setminus\\{0\\}$,
$\psi_{E}^{-1}(A)\subseteq\mathcal{L}\times\Sigma_{P,E}$. To this end, we
write

$\psi_{E}(\mathcal{L}\times\Sigma_{P,E})=\\{\psi_{E}(G):G\in\mathcal{L}\times\Sigma_{P,E}\\}$

for the $\sigma$-algebra on $\mathbb{R}^{d}\setminus\\{0\\}$ induced by
$\psi_{E}$. In view of Lemma 4.8, $\psi_{E}(\mathcal{L}\times\Sigma_{P,E})$
contains every open subset of $\mathbb{R}^{d}\setminus\\{0\\}$ and therefore

$\mathcal{B}(\mathbb{R}^{d}\setminus\\{0\\})\subseteq\psi_{E}(\mathcal{L}\times\Sigma_{P,E}).$

where $\mathcal{B}(\mathbb{R}^{d}\setminus\\{0\\})$ denotes the
$\sigma$-algebra of Borel subsets of $\mathbb{R}^{d}\setminus\\{0\\}$ thus
proving our claim.

Together, the results of the two preceding paragraphs show that, for each
$A\in\mathcal{B}(\mathbb{R}^{d}\setminus\\{0\\})$,
$\psi_{E}^{-1}(A)\subseteq\mathcal{L}\times\Sigma_{P,E}$ and
$m(A)=(\lambda_{P}\times\sigma_{P,E})(\psi_{E}^{-1}(A))$. Upon noting that
$\mathcal{L}\times\Sigma_{P,E}\subseteq(\mathcal{L}\times\Sigma_{P,E})^{\prime}$,
we immediately obtain the following statement: For each
$A\in\mathcal{B}(\mathbb{R}^{d}\setminus\\{0\\})$,

$\psi_{E}^{-1}(A)\in(\mathcal{L}\times\Sigma_{P,E})^{\prime}\hskip
28.45274pt\mbox{and}\hskip
28.45274ptm(A)=(\lambda_{P}\times\sigma_{P,E})(\psi_{E}^{-1}(A)).$ (37)

In comparing (36) and (37) with Properties 1 and 2 of Lemma 4.9 and, upon
noting that $((0,\infty)\times
S,(\mathcal{L}\times\Sigma_{P,E})^{\prime},\lambda_{P}\times\sigma_{P,E})$ is
the completion of $((0,\infty)\times
S,\mathcal{L}\times\Sigma_{P,E},\lambda_{P}\times\sigma_{P,E})$ and
$(\mathbb{R}^{d}\setminus\\{0\\},\mathcal{M}_{d},m)$ is the completion of
$(\mathbb{R}^{d}\setminus\\{0\\},\mathcal{B}(\mathbb{R}^{d}\setminus\\{0\\}),m)$,
Property 1 of Theorem 4.4 follows immediately from Lemma 4.9.

It remains to prove Property 2. To this end, let
$f:\mathbb{R}^{d}\to\mathbb{C}$ be Lebesgue measurable. Because
$\mathcal{M}_{d}=\\{A\subseteq\mathbb{R}^{d}\setminus\\{0\\}:\psi_{E}^{-1}(A)\in(\mathcal{L}\times\Sigma_{P,E})^{\prime}\\}$,
it follows that $f\circ\psi_{E}$ is
$(\mathcal{L}\times\Sigma_{P,E})^{\prime}$-measurable. In the case that $f\geq
0$, we may approximate $f$ monotonically by simple functions and, by invoking
Property 1 and the monotone convergence theorem, we find that

$\int_{\mathbb{R}^{d}}f(x)\,dx=\int_{\mathbb{R}^{d}\setminus\\{0\\}}f(x)\,dx=\int_{(0,\infty)\times
S}f\circ\psi_{E}\,d(\lambda_{P}\times\sigma_{P,E}).$ (38)

From this, (29) follows from Fubini’s theorem (see, e.g., Part (a) of Theorem
8.8 and Theorem 8.12 of [23]). Finally, by applying the above result to
$|f|\geq 0$, we obtain $f\in L^{1}(\mathbb{R}^{d})$ if and only if
$f\circ\psi_{E}\in L^{1}((0,\infty)\times
S,(\mathcal{L}\times\Sigma_{P,E})^{\prime},\lambda_{P}\times\sigma_{P,E})$. In
this case, by applying (38) to
$\operatorname{Re}(f)_{+},\operatorname{Re}(f)_{-},\operatorname{Im}(f)_{+}$
and $\operatorname{Im}(f)_{-}$, we find that (38) holds for our integrable $f$
and, by virtue Fubini’s theorem (see, e.g., Part (b) of Theorem 8.8 and
Theorem 8.12 of [23]), the desired result follows. ∎

Our next result, Proposition 4.10, guarantees that, in particular,
$\sigma_{P,E}$ is a Radon measure.

######  Proposition 4.10.

We have:

  1. 1.

$(S,\Sigma_{P,E},\sigma_{P,E})$ is the completion of the measure space
$(S,\mathcal{B}(S),\sigma_{P,E})$. In particular, the measure space
$(S,\Sigma_{P,E},\sigma_{P,E})$ is complete and every $F\in\Sigma_{P,E}$ is of
the form $F=G\cup H$ where $G$ is a Borel set and $H$ is a subset of a Borel
set $Z$ with $\sigma_{P,E}(Z)=0$.

  2. 2.

For each $F\in\Sigma_{P,E}$,

$\sigma_{P,E}(F)=\inf\\{\sigma_{P,E}(\mathcal{O}):F\subseteq\mathcal{O}\subseteq
S\mbox{ and $\mathcal{O}$ is open}\\}$ (39)

and

$\sigma_{P,E}(F)=\sup\\{\sigma_{P,E}(K):K\subseteq F\subseteq S\mbox{ and $K$
is compact}\\}.$ (40)

######  Remark 2.

This proposition can be seen as an application of Proposition 4.2 and Theorem
2.18 of [23]. The proof we give here is distinct and, we believe, nicely
illustrates the utility of (29) of Theorem 4.4.

###### Proof.

Throughout the proof, we shall write $\sigma=\sigma_{P,E}$,
$\Sigma=\Sigma_{P,E}$ and, for each $F\subseteq S$,
$\widetilde{F}=\widetilde{F_{E}}$. We remark that, by standard arguments using
$G_{\delta}$ and $F_{\sigma}$ sets, Item 1 follows immediately from Item 2.
Also, given that $S$ is compact and $\sigma$ is finite, it suffices to prove
(39), i.e., it suffices to prove the statement: For each $F\in\Sigma$ and
$\epsilon>0$, there is an open subset $\mathcal{O}$ of $S$ containing $F$ for
which

$\sigma(\mathcal{O}\setminus F)<\epsilon.$

To this end, let $F\in\Sigma$ and $\epsilon>0$. Given that
$\widetilde{F}\in\mathcal{M}_{d}$ and $m$ is outer regular, there exists an
open set $U\subseteq\mathbb{R}^{d}\setminus\\{0\\}$ for which
$\widetilde{F}\subseteq U$ and
$m(U\setminus\widetilde{F})<\epsilon/(2\mu_{P})$. Since $\widetilde{F}$ is a
subset of the open set $B\setminus\\{0\\}$, we may assume without loss of
generality that $U\subseteq B\setminus\\{0\\}$ and so $m(\widetilde{F})\leq
m(U)<\infty$ and

$m(U\setminus\widetilde{F})=m(U)-m(\widetilde{F})<\epsilon/(2\mu_{P}).$ (41)

For each $0<r<1$, consider the open set

$\mathcal{O}_{r}=S\cap\left(r^{-E}U\right)$

in $S$. Observe that, for each $x\in F$, $r^{E}x\in\widetilde{F}\subseteq U$
and therefore $x\in\mathcal{O}_{r}$. Hence, for each $0<r<1$,
$\mathcal{O}_{r}$ is an open subset of $S$ containing $F$.

We claim that there is at least one $r_{0}\in(0,1)$ for which

$m(\widetilde{\mathcal{O}_{r_{0}}})<m(U)+\epsilon/(2\mu_{P}).$ (42)

To prove the claim, we shall assume, to reach a contradiction, that

$m(\widetilde{\mathcal{O}_{r}})\geq m(U)+\epsilon/(2\mu_{P})$

for all $0<r<1$. By virtue of (29) of Theorem 4.4,

$m(U)=\int_{0}^{\infty}\left(\int_{S}\chi_{U}(r^{E}\eta)\,\sigma(d\eta)\right)r^{\mu_{P}-1}\,dr.$

Upon noting that $U\subseteq B\setminus\\{0\\}$, it is easy to see that

$U=\bigcup_{0<s<1}s^{E}\mathcal{O}_{s}\hskip 28.45274pt\mbox{and}\hskip
28.45274ptr^{E}\eta\in\bigcup_{0<s<1}s^{E}\mathcal{O}_{s}$

if and only if $0<r<1$ and $\eta\in\mathcal{O}_{r}$. Consequently,

$m(U)=\int_{0}^{1}\left(\int_{S}\chi_{\mathcal{O}_{r}}(\eta)\,\sigma(d\eta)\right)\,r^{\mu_{P}-1}\,dr=\int_{0}^{1}\sigma(\mathcal{O}_{r})r^{\mu_{P}-1}\,dr=\int_{0}^{1}\mu_{P}\cdot
m(\widetilde{\mathcal{O}_{r}})\,r^{\mu_{P}-1}\,dr.$

Upon making use of our supposition, we have

$\int_{0}^{1}\mu_{P}\cdot
m(\widetilde{\mathcal{O}_{r}})r^{\mu_{P}-1}\,dr\geq\int_{0}^{1}\mu_{P}\cdot(m(U)+\epsilon/(2\mu_{P}))r^{\mu_{P}-1}\,dr=m(U)+\epsilon/(2\mu_{P})$

and so

$m(U)\geq m(U)+\epsilon/(2\mu_{P}),$

which is impossible. Thus, the stated claim is true.

Given any such $r_{0}$ for which (42) holds, set
$\mathcal{O}=\mathcal{O}_{r_{0}}$. As previously noted, $\mathcal{O}$ is an
open subset of $S$ which contains $F$. In view of (41) and (42), we have

$m(\widetilde{\mathcal{O}})-m(\widetilde{F})<m(U)-m(\widetilde{F})+\epsilon/(2\mu_{P})<\epsilon/(2\mu_{P})+\epsilon/(2\mu_{P})=\epsilon/\mu_{P}$

and therefore

$\sigma(\mathcal{O}\setminus
F)=\sigma(\mathcal{O})-\sigma(F)=\mu_{P}(m(\widetilde{\mathcal{O}})-m(\widetilde{F}))<\epsilon,$

as desired. ∎

###  4.3 The construction is independent of $E\in\operatorname{Exp}(P)$.

In this subsection, we show that the Radon measure $\sigma_{P,E}$ is
independent of the choice of $E\in\operatorname{Exp}(P)$ and complete the
proof of Theorem 1.3. To set the stage for our first result, let
$E_{1},E_{2}\in\operatorname{Exp}(P)$ and consider the associated (respective)
measure spaces $(S,\Sigma_{P,E_{1}},\sigma_{P,E_{1}})$ and
$(S,\Sigma_{P,E_{2}},\sigma_{P,E_{2}})$ produced via the construction in
Subsection 4.1.

######  Proposition 4.11.

These measure spaces are the same, i.e., $\Sigma_{P,E_{1}}=\Sigma_{P,E_{2}}$
and $\sigma_{P,E_{1}}=\sigma_{P,E_{2}}$.

###### Proof.

Throughout the proof, we will write $\Sigma_{i}=\Sigma_{P,E_{i}}$ and
$\sigma_{i}=\sigma_{P,E_{i}}$ for $i=1,2$. In view of the Proposition 4.10, it
suffices to show that

$\sigma_{1}(F)=\sigma_{2}(F)$

for all $F\in\mathcal{B}(S)\subseteq\Sigma_{1}\cap\Sigma_{2}$. To this end, we
let $F\in\mathcal{B}(S)$ be arbitrary but fixed.

Given $n\in\mathbb{N}_{+}$, using the regularity of the measures $\sigma_{1}$
and $\sigma_{2}$, select open sets $\mathcal{O}_{n,1},\mathcal{O}_{n,2}$ and
compact sets $K_{n,1},K_{n,2}$ for which

$K_{n,j}\subseteq F\subseteq\mathcal{O}_{n,j}\hskip 28.45274pt\mbox{and}\hskip
28.45274pt\sigma_{j}(\mathcal{O}_{n,j}\setminus K_{n,j})<1/n$

for $j=1,2$. Observe that $K_{n}=K_{n,1}\cup K_{n,2}$ is a compact set,
$\mathcal{O}_{n}=\mathcal{O}_{n,1}\cap\mathcal{O}_{n,2}$ is an open set and
$K_{n}\subseteq F\subseteq\mathcal{O}_{n}$. Furthermore,

$\sigma_{j}(\mathcal{O}_{n}\setminus
K_{n})\leq\sigma_{j}(\mathcal{O}_{n,j}\setminus K_{n,j})<1/n$

for $j=1,2$. Given that $\mathcal{O}_{n}$ is open in $S$,
$\mathcal{O}_{n}=S\cup U_{n}$ where $U_{n}$ is an open subset of
$\mathbb{R}^{d}$ and, because that $S$ is compact, $K_{n}=K_{n}\cap S$ is a
compact subset of $\mathbb{R}^{d}$. By virtue of Urysohn’s lemma, let
$\phi_{n}:\mathbb{R}^{d}\to[0,1]$ be a continuous function which is compactly
supported in $U_{n}$ and for which $\phi_{n}(x)=1$ for all $x\in K_{n}$. Using
this sequence of functions $\\{\phi_{n}\\}$, we establish the following useful
lemma.

######  Lemma 4.12.

For $j=1,2$ and $n\in\mathbb{N}$, define $g_{n,j}:(0,\infty)\to\mathbb{R}$ by

$g_{n,j}(r)=\int_{S}\phi_{n}(r^{E_{j}}\eta)\,\sigma_{j}(d\eta).$

for $r>0$. Then $g_{n,j}$ is continuous for each $n\in\mathbb{N}$ and $j=1,2$
and

$\sigma_{j}(F)=\lim_{n\to\infty}g_{n,j}(1)$

for $j=1,2$.

###### Proof.

First, we note that, for each $r\in(0,\infty)$, the above integral makes sense
because $\eta\mapsto\phi_{n}(r^{E_{j}}\eta)$ is Borel measurable (because it’s
continuous on $S$) and non-negative. Let $\epsilon>0$ and $r_{0}\in(0,\infty)$
be arbitrary but fixed. It is clear that the function $(0,\infty)\times
S\ni(r,\eta)\mapsto\phi_{n}(r^{E_{j}}\eta)$ is continuous on its domain and
therefore, in view of the compactness of $S$, we can find a $\delta>0$ for
which

$|\phi_{n}(r^{E_{j}}\eta)-\phi_{n}(r_{0}^{E_{j}}\eta)|\leq\frac{\epsilon}{2\sigma_{j}(S)}\hskip
28.45274pt\mbox{whenever}\hskip 28.45274pt|r-r_{0}|<\delta$

for all $\eta\in S$. The triangle inequality guarantees that

$|g_{n,j}(r)-g_{n,j}(r_{0})|\leq\int_{S}|\phi_{n}(r^{E_{j}}\eta)-\phi_{n}(r_{0}^{E_{j}}\eta)|\,\sigma_{j}(d\eta)\leq\epsilon/2<\epsilon$

whenever $|r-r_{0}|<\delta$. Thus, $g_{n,j}$ is continuous.

We observe that

$g_{n,j}(1)=\int_{S}\phi_{n}(\eta)\,\sigma_{j}(d\eta)$

because $1^{E_{j}}=I$. By construction, we have
$\chi_{K_{n}}(\eta)\leq\phi_{n}(\eta)\leq\chi_{\mathcal{O}_{n}}(\eta)$ for all
$\eta\in S$ and $n\in\mathbb{N}_{+}$ and therefore

$\sigma_{j}(K_{n})\leq g_{n,j}(1)\leq\sigma_{j}(\mathcal{O}_{n})$

by the monotonicity of the integral. Since

$\sigma_{j}(F)=\lim_{n\to\infty}\sigma_{j}(K_{n})=\lim_{n\to\infty}\sigma_{j}(\mathcal{O}_{n})$

in view of our choice of $\mathcal{O}_{n}$ and $K_{n}$, the remaining result
follows immediately from the preceding inequality (and the squeeze theorem). ∎

Let us now complete the proof of Proposition 4.11. Given any $0<s<1<t$ and
$n\in\mathbb{N}$, consider the function $f=f_{n,s,t}:\mathbb{R}^{d}\to[0,1]$
given by

$f(x)=\phi_{n}(x)\chi_{[s,t]}(P(x))$

for $x\in\mathbb{R}^{d}$. It is clear that $f$ is Lebesgue measurable on
$\mathbb{R}^{d}$ and non-negative. By virtue of Theorem 4.4 (applied to the
two measures $\sigma_{1}$ and $\sigma_{2}$), we have

$\int_{0}^{\infty}\int_{S}f(r^{E_{1}}\eta)\,\sigma_{1}(d\eta)r^{\mu_{P}-1}\,dr=\int_{\mathbb{R}^{d}}f(x)\,dx=\int_{0}^{\infty}\int_{S}f(r^{E_{2}}\eta)\,\sigma_{2}(d\eta)r^{\mu_{P}-1}\,dr$
(43)

Upon noting that

$f(r^{E_{j}}\eta)=\phi_{n}(r^{E_{j}}\eta)\chi_{[s,t]}\left((P(r^{E_{j}}\eta)\right)=\phi_{n}(r^{E_{j}}\eta)\chi_{[s,t]}(rP(\eta))=\chi_{[s,t]}(r)\phi_{n}(r^{E_{j}}\eta)$

for $r\in(0,\infty)$, $\eta\in S$, and $j=1,2$, we have

$\int_{0}^{\infty}\int_{S}f(r^{E_{j}}\eta)\,\sigma_{j}(d\eta)r^{\mu_{P}-1}\,dr=\int_{[s,t]}\int_{S}\phi_{n}(r^{E_{j}}\eta)\,\sigma_{j}(d\eta)r^{\mu_{P}-1}\,dr=\int_{[s,t]}g_{n,j}(r)r^{\mu_{P}-1}\,dr$

for $j=1,2$. By virtue of the Lemma 4.12, $r\mapsto g_{n,j}(r)r^{\mu_{P}-1}$
is continuous and necessarily bounded on $[s,t]$ and so the final integral
above can be interpreted as a Riemann integral. In this interpretation, we
have

$\int_{s}^{t}g_{n,j}(r)r^{\mu_{P}-1}\,dr=\int_{0}^{\infty}\int_{S}f(r^{E_{j}}\eta)\,\sigma_{j}(d\eta)r^{\mu_{P}-1}\,dr$
(44)

for $j=1,2$ and $0<s<1<t$. In view of (43) and (44), we conclude that

$\int_{s}^{t}g_{n,1}(r)r^{\mu_{P}-1}\,dr=\int_{s}^{t}g_{n,2}(r)r^{\mu_{P}-1}\,dr$

for all $0<s<1<t$. In view of continuity of the integrands, an application of
the fundamental theorem of calculus now guarantees that
$g_{n,1}(1)=g_{n,2}(1)$ for each $n\in\mathbb{N}$. Therefore

$\sigma_{1}(F)=\lim_{n\to\infty}g_{n,1}(1)=\lim_{n\to\infty}g_{n,2}(1)=\sigma_{2}(F)$

by virtue of Lemma 4.12. ∎

In view of Proposition 4.11, we will denote by $\Sigma_{P}$ and $\sigma_{P}$
the unique $\sigma$-algebra and measure on $S$ which, respectively, satisfy

$\Sigma_{P}=\Sigma_{P,E}\hskip 28.45274pt\mbox{and}\hskip
28.45274pt\sigma_{P}=\sigma_{P,E}$

for all $E\in\operatorname{Exp}(P)$. We will henceforth assume this notation.

######  Proposition 4.13.

For any $O\in\operatorname{Sym}(P)$ and $F\in\Sigma_{P}$, $OF\in\Sigma_{P}$
and

$\sigma_{P}(OF)=\sigma_{P}(F).$

That is, the measure $\sigma_{P}$ is invariant under the action by
$\operatorname{Sym}(P)$.

###### Proof.

Let $O\in\operatorname{Sym}(P)$, $F\in\Sigma_{P}$ and, for
$E\in\operatorname{Exp}(P)$, define $E^{\prime}=O^{*}EO$. In view of
Proposition 2.4, we note that $E^{\prime}\in\operatorname{Exp}(P)$. Observe
that

$\widetilde{(OF)_{E}}=\bigcup_{0<r<1}r^{E}(OF)=\bigcup_{0<r<1}O\left(O^{*}r^{E}OF\right)=O\left(\bigcup_{0<r<1}r^{E^{\prime}}F\right)=O\widetilde{F_{E^{\prime}}}$
(45)

thanks to Proposition A.1. In view of Proposition 4.11, we have
$O\widetilde{F_{E^{\prime}}}\in\mathcal{M}_{d}$ because
$F\in\Sigma_{P}=\Sigma_{P,E^{\prime}}$ and $O$ is linear. Using (45), we find
that $\widetilde{(OF)_{E}}\in\mathcal{M}_{d}$ and therefore
$OF\in\Sigma_{P,E}=\Sigma_{P}$ by virtue of Proposition 4.11. In view of the
fact that $O$ is orthogonal,

$\sigma_{P,E}(OF)=\mu_{P}\cdot m\left(\widetilde{(OF)_{E}}\right)=\mu_{P}\cdot
m\left(O\widetilde{F_{E^{\prime}}}\right)=\mu_{P}\cdot
m\left(\widetilde{F}_{E^{\prime}}\right)=\sigma_{P,E^{\prime}}(F)$

and therefore, a final appeal to Proposition 4.11 guarantees that

$\sigma_{P}(OF)=\sigma_{P,E}(OF)=\sigma_{P,E^{\prime}}(F)=\sigma_{P}(F),$

as desired. ∎

###### Proof of Theorem 1.3.

Together, the results of Propositions 4.10, 4.11 and 4.13, guarantee that
$\sigma_{P}$ is a Radon measure satisfying Properties 1 and 2. Property 3
follows directly from Proposition 4.11 and the definition of $\sigma_{P}$ in
terms of $\sigma_{P,E}$ for any $E\in\operatorname{Exp}(P)$. Similarly,
Properties 1 and 2 follow from Theorem 4.4 by virtue of Proposition 4.11. ∎

##  5 Using a smooth structure on $S$ to compute $\sigma$.

In this section, we shall study the special case in which a positive
homogeneous function $P$ on $\mathbb{R}^{d}$ is smooth555Many of the results
in this section remain valid (with appropriate modification) under the weaker
assumption that $P\in C^{k}(\mathbb{R}^{d})$ for $k=1,2,\dots$. In this
setting, $S$ is easily seen to be a $C^{k}$ manifold. Because working in the
smooth category is sufficient for our purposes, we shall not pursue the
greater level of generality but invite the reader to do
so.${}^{\mbox{\tiny{,}}}$666To avoid trivialities, we assume that $d>1$
throughout.. Under this additional assumption, we shall find that $\gradient
P$ is everywhere non-vanishing on $S$ and so $S$ is a smooth compact embedded
hypersurface in $\mathbb{R}^{d}$.

We first set up some notation: For a smooth manifold $M$, we shall denote by
$\mathcal{A}(M)$ its unique maximal atlas. Also on $M$, the collection of
smooth vector fields is denoted by $\mathfrak{X}(M)$ and, for each
$k=1,2,\dots$, the set of (smooth) differential $k$-forms on $M$ will be
denoted by $\Omega^{k}(M)$. In this section, we integrate non-smooth
differential forms and, for the generality needed here, we shall refer the
reader to [19] for background (Another perspective is given in [1]). To this
end, let us denote the Lebesgue $\sigma$-algebra of measurable sets on $M$ by
$\mathcal{L}(M)$. We note that $F\in\mathcal{L}(M)$ if and only if,

$\varphi(F\cap\mathcal{U})\in\mathcal{M}_{d}$

for every chart $(\varphi,\mathcal{U})\in\mathcal{A}(M)$. An $n$-form $\omega$
on $M$, with $n=\dim(M)$, is said to be (Lebesgue) measurable, if in each
coordinate system $(\varphi,\mathcal{U})$, the local representation

$\omega=h_{\varphi}(x)dx^{1}\wedge dx^{2}\wedge\cdots\wedge dx^{n}$

in the coordinates $\varphi=(x^{1},x^{2},\dots,x^{n})$ has $h_{\varphi}(x)$ a
Lebesgue measurable function on
$U=\varphi(\mathcal{U})\subseteq\mathbb{R}^{n}$. The collection of measurable
$n$-forms on $M$ is denoted by $\mathcal{L}(\Lambda^{d}(M))$ and, naturally,
$\Omega^{d}(M)\subseteq\mathcal{L}(\Lambda^{d}(M))$. As standard, we shall use
Einstein’s summation convention without explicit mention.

We view $\mathbb{R}^{d}$ as smooth oriented Riemannian manifold with its
standard Euclidean metric $\overline{g}$, oriented smooth atlas
$\mathcal{A}_{+}(\mathbb{R}^{d})$, and Riemannian volume form
$d\operatorname{Vol}_{\mathbb{R}^{d}}$. Given any $E\in\operatorname{Exp}(P)$,
consider $\mathcal{E}_{E}\in\mathfrak{X}(\mathbb{R}^{d})$ defined, at each
$x\in\mathbb{R}^{d}$, by

$(\mathcal{E}_{E})_{x}(f)=\frac{d}{dt}f(x+t(Ex))\big{|}_{t=0}\hskip
28.45274pt$

for $f\in C^{\infty}(\mathbb{R}^{d})$. In the standard (global) chart with
coordinates $x=(x^{1},x^{2},\dots,x^{d})$, $(\mathcal{E}_{E})_{x}\in
T_{x}(\mathbb{R}^{d})$ is given by

$(\mathcal{E}_{E})_{x}=(Ex)^{\alpha}\partial_{x^{\alpha}}=E^{\alpha}_{\beta}x^{\beta}\partial_{x^{\alpha}}$

where $(E_{\alpha}^{\beta})$ is the standard matrix representation for $E$ and
$\partial_{x^{\alpha}}=\partial/\partial x^{\alpha}$. By an abuse of notation,
we shall write $\gradient P$ to denote both the function

$\mathbb{R}^{d}\ni x\mapsto\gradient P(x)=\left(\frac{\partial P}{\partial
x^{1}},\frac{\partial P}{\partial x^{2}},\dots,\frac{\partial P}{\partial
x^{d}}\right)\in\mathbb{R}^{d},$

where $\frac{\partial P}{\partial x^{\alpha}}=\frac{\partial P}{\partial
x^{\alpha}}|_{x}$ for $\alpha=1,2\dots,d$, and its canonical identification
$\gradient P\in\mathfrak{X}(\mathbb{R}^{d})$ given by

$\gradient P_{x}=\overline{g}^{\alpha\beta}\frac{\partial P}{\partial
x^{\alpha}}\partial_{x^{\beta}}=\delta^{\alpha\beta}\frac{\partial P}{\partial
x^{\alpha}}\partial_{x^{\beta}}=\sum_{\alpha=1}^{d}\frac{\partial P}{\partial
x^{\alpha}}\partial_{x^{\alpha}}$

in standard Euclidean coordinates $x=(x^{\alpha})$. Of course, for each
$x\in\mathbb{R}^{d}$, the Riemannian norm $|\gradient P_{x}|_{\overline{g}}$
of $\gradient P_{x}\in T_{x}(\mathbb{R}^{d})$ coincides with the Euclidean
norm $|\gradient P(x)|$ of $\gradient P(x)\in\mathbb{R}^{d}$. These equivalent
quantities (functions) will be henceforth denoted by $|\gradient P|$.

######  Proposition 5.1.

For each $\eta\in S$,

$\overline{g}(\gradient P,\mathcal{E})_{\eta}=\gradient
P(\eta)\cdot(E\eta)=1.$

In particular, $\gradient P$ (and $dP$) never vanishes on $S$ and so $S$ is a
smooth compact embedded hypersurface in $\mathbb{R}^{d}$.

###### Proof.

Given that $E\in\operatorname{Exp}(P)$ and $P\in C^{\infty}(\mathbb{R}^{d})$,
we differentiate the identity $rP(x)=P(r^{E}x)$ to find that

$P(x)=\frac{d}{dr}P(r^{E}x)=\gradient
P(r^{E}x)\cdot\left(\left(r^{E-I}E\right)x\right)$

for $r>0$ and $x\in\mathbb{R}^{d}$. In particular, when $r=1$ and $x=\eta\in
S$, we have

$1=\gradient P(\eta)\cdot\left(E\eta\right)=\frac{\partial P}{\partial
x^{\alpha}}(E\eta)^{\alpha}=\overline{g}_{\alpha\beta}\left(\gradient
P_{\eta}\right)^{\alpha}(\mathcal{E}_{\eta})^{\beta}=\overline{g}(\gradient
P,\mathcal{E})_{\eta}.$

Thus, our (necessarily) compact level set $S$ is a smooth embedded
hypersurface of $\mathbb{R}^{d}$ in view of the regular level set theorem. ∎

We shall denote by $\iota:S\hookrightarrow\mathbb{R}^{d}$ the canonical
inclusion map and set $d^{\prime}=d-1$. As an embedded submanifold of
$\mathbb{R}^{d}$, $S$ is a Riemannian submanifold of $\mathbb{R}^{d}$ with
metric $g^{S}$ given by

$g^{S}(X,Y)=\overline{g}(\iota_{*}(X),\iota_{*}(Y))$

for $X,Y\in\mathfrak{X}(S)$; here, for each $\eta\in S$,
$\iota_{*}:T_{\eta}(S)\to T_{\eta}(\mathbb{R}^{d})$ is the pushforward of
$\iota$. In view of the preceding proposition, $N:=\gradient P/|\gradient
P|\in\mathfrak{X}(\mathbb{R}^{d})$ is a smooth unit normal vector field along
$S$ and it determines an orientation on the Riemannian manifold $S$. Equipped
with this orientation, $(S,g^{S})$ is an oriented Riemannian manifold and we
shall denote by $d\operatorname{Vol}_{S}$ the Riemannian volume form and by
$\mathcal{A}_{+}(S)$ its corresponding (maximal) oriented atlas. By virtue of
Proposition 15.21 of [18],
$d\operatorname{Vol}_{S}=(N\mathbin{\lrcorner}d\operatorname{Vol}_{\mathbb{R}^{d}})|_{S}$,
i.e.,

$\displaystyle d\operatorname{Vol}_{S}(X_{1},X_{2},\dots,X_{d^{\prime}})$
$\displaystyle=$ $\displaystyle
d\operatorname{Vol}_{\mathbb{R}^{d}}(N,\iota_{*}(X_{1}),\iota_{*}(X_{2}),\dots,\iota_{*}(X_{d^{\prime}}))$
$\displaystyle=$ $\displaystyle\frac{1}{|\gradient
P|}d\operatorname{Vol}_{\mathbb{R}^{d}}(\gradient
P,\iota_{*}(X_{1}),\iota_{*}(X_{2}),\dots,\iota_{*}(X_{d^{\prime}}))$

for any collection $\\{X_{1},X_{2},\dots,X_{d^{\prime}}\\}\in\mathfrak{X}(S)$.
Beyond $d\operatorname{Vol}_{S}\in\Omega^{d^{\prime}}(S)$, we consider the
following smooth $d^{\prime}$-form(s): Given $E\in\operatorname{Exp}(P)$,
define $d\sigma_{P,E}\in\Omega^{d^{\prime}}(S)$ by

$d\sigma_{P,E}(X_{1},X_{2},\dots,X_{d^{\prime}})=d\operatorname{Vol}_{\mathbb{R}^{d}}(\mathcal{E}_{E},\iota_{*}(X_{1}),\iota_{*}(X_{2}),\dots,\iota_{*}(X_{d^{\prime}}))$

for $X_{1},X_{2},\dots,X_{d^{\prime}}\in\mathfrak{X}(S)$. We have

######  Proposition 5.2.

For any $E\in\operatorname{Exp}(P)$,

$d\sigma_{P,E}=\frac{1}{|\gradient P|}d\operatorname{Vol}_{S}.$

In particular, $d\sigma_{P,E}$ is positively oriented and is independent of
$E\in\operatorname{Exp}(P)$.

Before proving the proposition, we first treat a lemma of a purely linear
algebraic nature.

######  Lemma 5.3.

Let $v_{1},v_{2},\dots,v_{d^{\prime}}$ be linearly independent vectors in
$\mathbb{R}^{d}$ and suppose that $w\in\mathbb{R}^{d}\setminus\\{0\\}$ is such
that $w\perp v_{i}$ for all $i$. Then, for any $z\in\mathbb{R}^{d}$ for which
$z\cdot w=1$,

$\operatorname{det}(z,v_{1},v_{2},\dots,v_{d^{\prime}})=\frac{1}{|w|}\operatorname{det}(n,v_{1},v_{2},\dots,v_{d^{\prime}})=\frac{1}{|w|^{2}}\operatorname{det}(w,v_{1},v_{2},\dots,v_{d^{\prime}}).$

where $n:=w/|w|$.

###### Proof.

Given $z\in\mathbb{R}^{d}$ such that $z\cdot w=1$, it follows that

$z=\frac{1}{|w|}n+a_{1}v_{1}+a_{2}v_{2}+\cdots a_{d^{\prime}}v_{d^{\prime}}.$

By the multilinearity of the determinant map, we have

$\displaystyle\operatorname{det}(z,v_{1},v_{2},\dots,v_{d^{\prime}})$
$\displaystyle=$
$\displaystyle\operatorname{det}\left(\frac{1}{|w|}n+a_{1}v_{2}+a_{2}v_{2}+\cdots
a_{d^{\prime}}v_{d^{\prime}},v_{1},v_{2},\dots,v_{d^{\prime}}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{|w|}\operatorname{det}(n,v_{1},v_{2},\dots,v_{d^{\prime}})+\operatorname{det}(a_{1}v_{1}+\cdots+a_{d^{\prime}}v_{d^{\prime}},v_{1},v_{2},\dots,v_{d^{\prime}})$
$\displaystyle=$
$\displaystyle\frac{1}{|w|}\operatorname{det}(n,v_{1},v_{2},\dots,v_{d^{\prime}})+0$

where we have used the fact that the columns of the matrix
$(a_{1}v_{1}+\cdots+a_{d^{\prime}}v_{d^{\prime}},v_{1},v_{2},\dots,v_{d^{\prime}})$
are linearly dependent to conclude that the final determinant is zero. ∎

###### Proof of Proposition 5.2.

We fix $E\in\operatorname{Exp}(P)$ and note that the assertion at hand is a
local one. Thus, it suffices to verify that, for any $\eta\in S$ and
$X_{1},X_{2},\dots,X_{d^{\prime}}\in T_{\eta}(S)$,

$d\sigma_{P,E}(X_{1},X_{2},\dots,X_{d^{\prime}})=\frac{1}{|\gradient
P_{\eta}|}d\operatorname{Vol}_{\mathbb{R}^{d}}(N_{\eta},\iota_{*}(X_{1}),\iota_{*}(X_{2}),\dots,\iota_{*}(X_{d^{\prime}})).$

Fix $\eta\in S$ and let $(\mathcal{O},\varphi)$ be a coordinate chart centered
at $\eta$ with local coordinates $u=(u^{\alpha})$. As usual, denote by
$x=(x^{\alpha})$ the Euclidean coordinates on $\mathbb{R}^{d}$. For each
$i=1,2,\dots,{d^{\prime}}$,

$X_{i}=X_{i}^{\alpha}\partial_{u^{\alpha}}\hskip 28.45274pt\mbox{and}\hskip
28.45274pt\iota_{*}(X_{i})=v_{i}^{\beta}\partial_{x^{\beta}}$

where

$v_{i}^{\beta}=X_{i}^{\alpha}\frac{\partial x^{\beta}}{\partial u^{\alpha}}.$

For each $i=1,2,\dots,d^{\prime}$, we set
$v_{i}=(v_{i}^{1},v_{i}^{2},\dots,v_{i}^{d^{\prime}})\in\mathbb{R}^{d}$. Also,
let $w=\gradient P(\eta)\in\mathbb{R}^{d}$ with $|w|=|\gradient
P(\eta)|=|\gradient P_{\eta}|$, set $n=w/|w|\in\mathbb{R}^{d}$ and note that

$N_{\eta}=\frac{1}{|\gradient P_{\eta}|}\sum_{k=1}^{d}\frac{\partial
P}{\partial x^{k}}\partial_{x^{k}}=n^{\mu}\partial_{x^{\mu}}.$

Given that $\gradient P$ is normal to $S$, we have

$v_{i}\cdot
w=\overline{g}_{\mu,\nu}v_{i}^{\mu}w^{\nu}=\overline{g}(\iota_{*}(X_{i}),\gradient
P)_{\eta}=0$

and therefore $w\perp v_{i}$ for each $i=1,2,\dots,{d^{\prime}}$. Upon
recalling that
$(\mathcal{E}_{E})_{\eta}=(E\eta)^{\alpha}\partial_{x^{\alpha}}$, set
$z=((E\eta)^{1},(E\eta)^{2},\dots,(E\eta)^{d})\in\mathbb{R}^{d}$ and observe
that $z\cdot w=1$ by virtue of Proposition 5.1. An appeal to the lemma
guarantees that

$\displaystyle d\sigma_{P,E}(X_{1},X_{2},\dots,X_{d^{\prime}})$
$\displaystyle=$ $\displaystyle
d\operatorname{Vol}_{\mathbb{R}^{d}}(\mathcal{E}_{E},\iota_{*}(X_{1}),\iota_{*}(X_{2}),\dots,\iota_{*}(X_{d^{\prime}}))$
$\displaystyle=$
$\displaystyle\operatorname{det}(z,v_{1},v_{2},\dots,v_{d^{\prime}})$
$\displaystyle=$
$\displaystyle\frac{1}{|w|}\operatorname{det}(n,v_{1},v_{2},\dots,v_{d^{\prime}})$
$\displaystyle=$ $\displaystyle\frac{1}{|\gradient
P_{\eta}|}d\operatorname{Vol}_{\mathbb{R}^{d}}(N_{\eta},\iota_{*}(X_{1}),\iota_{*}(X_{2}),\dots,\iota_{*}(X_{d^{\prime}}))$
$\displaystyle=$ $\displaystyle\frac{1}{|\gradient
P|}d\operatorname{Vol}_{S}(X_{1},X_{2},\dots,X_{d^{\prime}}).$

∎

By virtue of the preceding proposition, we shall denote by $d\sigma_{P}$ the
unique smooth $d^{\prime}$-form on $S$ which satisfies

$d\sigma_{P}=d\sigma_{P,E}=\frac{1}{|\gradient P|}d\operatorname{Vol}_{S}$
(46)

for all $E\in\operatorname{Exp}(P)$. In this notation, we have this section’s
central result.

######  Theorem 5.4.

Let $P$ be a smooth positive homogeneous function and let
$S=\\{\eta\in\mathbb{R}^{d}:P(\eta)=1\\}$. Then $S$ is a compact smooth
embedded hypersurface of $\mathbb{R}^{d}$. Viewing $\mathbb{R}^{d}$ as an
oriented Riemannian manifold with its usual orientation and metric
$\overline{g}$, $N=\gradient P/|\gradient P|$ is a smooth unit normal vector
field along $S$. As a submanifold of $\mathbb{R}^{d}$, $S$ is a oriented
Riemannian manifold of dimension $d^{\prime}=d-1$ with its induced Riemannian
metric $g^{S}$, volume form $d\operatorname{Vol}_{S}\in\Omega^{d^{\prime}}(S)$
and orientation determined by $N$. The $\sigma$-algebras $\Sigma_{P}$ and
$\mathcal{L}(S)$ on $S$ coincide and the smooth $d^{\prime}$-form
$d\sigma_{P}\in\Omega^{d^{\prime}}(S)$, defined by (46), coincides with the
measure $\sigma_{P}$ in the sense that

$\int_{S}g(\eta)\,\sigma_{P}(d\eta)=\int_{S}g\,d\sigma_{P}$ (47)

for all $g\in L^{1}(S,\Sigma_{P},\sigma_{P})$; here, the left hand side
represents the Lebesgue integral of $g$ with respect to $\sigma_{P}$ and the
right hand side is the integral of the measurable $d^{\prime}$-form
$g\,d\sigma_{P}$. Furthermore, the measure $\sigma_{P}$ and the canonical
Riemannian volume measure $\operatorname{Vol}_{S}$ on $S$ are mutually
absolutely continuous.

Before we are able to prove the above theorem, we first treat two lemmas which
we will find useful in computation.

######  Lemma 5.5.

Let $(\mathcal{U},\varphi)\in\mathcal{A}(S)$ and $E\in\operatorname{Exp}(P)$.
Set $U=\varphi(\mathcal{U})\subseteq\mathbb{R}^{d^{\prime}}$, $V=(0,1)\times
U$ and define $\rho_{E,\varphi}:V\to\mathbb{R}^{d}$ and
$h_{E,\varphi}:U\to\mathbb{R}$, respectively, by

$\rho_{E,\varphi}(y)=\psi_{E}(r,\varphi^{-1})=r^{E}\varphi^{-1}(u)$

for $y=(r,u)\in V$ and

$h_{E,\varphi}(u)=\operatorname{det}\left.\left(E\varphi^{-1}(u)\right|D_{u}\varphi^{-1}(u)\right)$

for $u\in U$; here, the vertical bar separates the first column of the
(necessarily) $d\times d$ matrix from the rightmost $d\times d^{\prime}$
submatrix and $D_{u}$ denotes the Jacobian in the coordinates
$u=(u^{1},u^{2},\dots,u^{d^{\prime}})\in U$. Then, $\rho_{E,\varphi}$ is a
diffeomorphism onto its image
$\rho_{E,\varphi}(V)=\widetilde{\mathcal{U}_{E}}$ and its Jacobian matrix
$D\rho_{E,\varphi}$ has

$\operatorname{det}(D\rho_{E,\varphi}(y))=r^{\mu_{P}-1}h_{E,\varphi}(u)$ (48)

for all $y=(r,u)\in V$. Furthermore, $h_{E,\varphi}$ is everywhere non-zero,
smooth and

$d\sigma_{P}=h_{E,\varphi}(u)\,du^{1}\wedge du^{2}\wedge\cdots\wedge
du^{d^{\prime}}$ (49)

in the coordinates $u=(u^{1},u^{2},\dots,u^{d^{\prime}})\in U$.

###### Proof.

The map $\rho_{E,\varphi}$ is smooth because $\varphi$ is smooth. By virtually
the same argument made in the proof of Proposition 4.1, which here uses the
fact that $P\in C^{\infty}(\mathbb{R}^{d})$, we conclude that
$\rho_{E,\varphi}$ is a diffeomorphism onto

$\rho_{E,\varphi}(V)=\bigcup_{0<r<1}r^{E}\mathcal{U}=\widetilde{\mathcal{U}_{E}}$

with inverse $\rho_{E,\varphi}^{-1}(x)=(P(x),\varphi((P(x))^{-E}x))$. For
$y=(r,u)\in V$, observe that

$\displaystyle D\rho_{E,\varphi}(y)$ $\displaystyle=$
$\displaystyle\left.\left(\frac{d}{dr}(r^{E}\varphi^{-1}(u))\right|D_{u}\left[r^{E}\varphi^{-1}(u)\right]\right)$
$\displaystyle=$
$\displaystyle\left.\left(r^{E-I}E\varphi^{-1}(u)\right|r^{E}D_{u}\varphi^{-1}(u)\right)$
$\displaystyle=$ $\displaystyle
r^{E}\left.\left(\frac{1}{r}E\varphi^{-1}(u)\right|D_{u}\varphi^{-1}(u)\right).$

Using properties of the determinant, we have

$\displaystyle\operatorname{det}(D\rho_{E}(y))$ $\displaystyle=$
$\displaystyle\operatorname{det}(r^{E})\operatorname{det}\left.\left(\frac{1}{r}E\varphi^{-1}(u)\right|D_{u}\varphi^{-1}(u)\right)$
$\displaystyle=$ $\displaystyle r^{\tr
E}r^{-1}\operatorname{det}\left.\left(E\varphi^{-1}(u)\right|D_{u}\varphi^{-1}(u)\right)$
$\displaystyle=$ $\displaystyle r^{\mu_{P}-1}h_{E,\varphi}(u)$

for all $y=(r,u)\in V$ thus proving (48). It is clear that $h_{E,\varphi}$ is
smooth and, by virtue of the fact that $\rho_{E,\varphi}:(0,1)\times
U\to\widetilde{\mathcal{U}_{E}}$ is a diffeomorphism, (48) guarantees that
$h_{E,\varphi}$ is everywhere non-vanishing. Finally, in view of Proposition
5.2 (and (46)), we find that

$\displaystyle h_{E,\varphi}(u)$ $\displaystyle=$
$\displaystyle\operatorname{det}\left.\left(E\varphi^{-1}(u)\right|D_{u}\varphi^{-1}(u)\right)$
$\displaystyle=$
$\displaystyle(d\operatorname{Vol}_{\mathbb{R}^{d}})_{\varphi^{-1}(u)}(\mathcal{E}_{E},\iota_{*}(\partial_{u^{1}}),\iota_{*}(\partial_{u^{2}}),\dots,\iota_{*}(\partial_{u^{d^{\prime}}}))$
$\displaystyle=$
$\displaystyle\left(d\sigma_{P}\right)_{\varphi^{-1}(u)}(\partial_{u^{1}},\partial_{u^{2}},\dots,\partial_{u^{d^{\prime}}})$

for $u\in U$ and so (49) is satisfied. ∎

######  Lemma 5.6.

Let $g\in L^{1}(S,\Sigma_{P},\sigma_{P})$ be supported on the domain of some
chart on $S$. Then, for any $(\mathcal{U},\varphi)\in\mathcal{A}^{+}(S)$ such
that $\operatorname{Supp}(g)\subseteq\mathcal{U}$, the pushforward
$(\varphi^{-1})^{*}(g)=g\circ\varphi^{-1}$ is Lebesgue measurable on
$U=\varphi(\mathcal{U})\subseteq\mathbb{R}^{d^{\prime}}$ and

$\int_{S}g(\eta)\sigma_{P}(d\eta)=\int_{U}(\psi^{-1})^{*}(g\,d\sigma_{P}).$

###### Proof.

Let $(\mathcal{U},\varphi)$ be a chart on $S$ for which
$\operatorname{Supp}(g)\subseteq\mathcal{U}$ and assume the notation of Lemma
5.5. Given $E\in\operatorname{Exp}(P)$, observe that

$f(x)=\mu_{P}\,\chi_{(0,1)}(P(x))g(P(x)^{-E}x),$

defined for $x\in\mathbb{R}^{d}\setminus\\{0\\}$, is supported on
$\widetilde{\mathcal{U}_{E}}=\rho_{E,\varphi}(V)$. An appeal to Corollary 1.5
guarantees that $f$ is absolutely integrable on $\rho_{E,\varphi}(V)$ and

$\int_{S}g(\eta)\,\sigma_{P}(d\eta)=\int_{\rho_{E,\varphi}(V)}f(x)\,dx.$ (50)

Given that $\rho_{E,\varphi}$ is a diffeomorphism, an appeal to Theorem 15.11
of [2] guarantees that

$\int_{\rho_{E,\varphi}(V)}f(x)\,dx=\int_{V}f(\rho_{E,\varphi}(y))|\operatorname{det}(D\rho_{E,\varphi}(y))|\,dy;$
(51)

in particular, $V\ni y\mapsto
f(\rho_{E,\varphi}(y))|\operatorname{det}(D\rho_{E,\varphi}(y))$ is Lebesgue
measurable on $\mathbb{R}^{d}$. Let us now view the Lebesgue measure $dy$ on
$\mathbb{R}^{d}$ as the completion of the product measure $dr\times du$ on the
product space $\mathbb{R}\times\mathbb{R}^{d^{\prime}}$. By virtue of Lemma
5.5, we have

$\displaystyle
f(\rho_{E,\varphi}(y))|\operatorname{det}(D\rho_{E,\varphi}(y))|$
$\displaystyle=$ $\displaystyle
f(r^{E}\varphi^{-1}(u))\,r^{\mu_{P}-1}|h_{E,\varphi}(u)|$ (52)
$\displaystyle=$
$\displaystyle\mu_{P}\,\chi_{(0,1)}(r)g(\varphi^{-1}(u))\,r^{\mu_{P}-1}|h_{E,\varphi}(u)|$
$\displaystyle=$
$\displaystyle\mu_{P}\,r^{\mu_{P}-1}g(\varphi^{-1}(u))|h_{E,\varphi}(u)|$

for $y=(r,u)\in V$. By virtue of Fubini’s theorem, $dr$-almost every
$r\in(0,1)$, the $r$-section

$U\ni u\mapsto\mu_{P}r^{\mu_{P}-1}g(\varphi^{-1}(u))|h_{E,\varphi}(u)|$

is Lebesgue measurable and, upon recalling that $h_{E,\varphi}$ is smooth and
everywhere nonzero, we conclude that
$(\varphi^{-1})^{*}(g)=g\circ\varphi^{-1}$ is Lebesgue measurable on $U$. In
view of (52), Fubini’s theorem also guarantees that

$\displaystyle\int_{V}f(\rho_{E,\varphi}(y))|\operatorname{det}(D\rho_{E,\varphi}(y))|\,dy$
$\displaystyle=$
$\displaystyle\int_{(0,1)}\int_{U}\mu_{P}\,r^{\mu_{P}-1}g(\varphi^{-1}(u))|h_{E,\varphi}(u)|\,du\,dr$
(53) $\displaystyle=$
$\displaystyle\int_{0}^{1}\mu_{P}\,r^{\mu_{P}-1}\left(\int_{U}g(\varphi^{-1}(u))|h_{E,\varphi}(u)|\,du\right)\,dr$
$\displaystyle=$
$\displaystyle\int_{U}g(\varphi^{-1}(u))|h_{E,\varphi}(u)|\,du$

By combining (50), (51) and (53), we have shown that
$(\varphi^{-1})^{*}g=g\circ\varphi^{-1}$ is Lebesgue measurable on
$U=\varphi(\mathcal{U})$ and

$\int_{S}g(\eta)\sigma_{P}(d\eta)=\int_{U}g(\varphi^{-1}(u))|h_{E,\varphi}(u)|\,du.$

Finally, if $(\mathcal{U},\varphi)\in\mathcal{A}_{+}(S)$, an appeal to
Proposition 5.2 and (49) of Lemma 5.5 guarantees that

$|h_{E,\varphi}(u)|=h_{E,\varphi}(u)=(d\sigma_{P})_{\varphi^{-1}(u)}(\partial_{u^{1}},\partial_{u^{2}},\dots,\partial_{u^{d^{\prime}}})>0$

for all $u=(u^{1},u^{2},\dots,u^{d^{\prime}})\in U$ and thus

$\displaystyle\int_{S}g(\eta)\sigma_{P}(d\eta)$ $\displaystyle=$
$\displaystyle\int_{U}g(\varphi^{-1}(u))h_{E,\varphi}(u)\,du$ $\displaystyle=$
$\displaystyle\int_{U}g(\varphi^{-1}(u))\,h_{E,\varphi}(u)\,du^{1}\wedge
du^{2}\wedge\cdots\wedge du^{d^{\prime}}$ $\displaystyle=$
$\displaystyle\int_{U}(\varphi^{-1})^{*}(g\cdot d\sigma_{P}),$

as desired. ∎

######  Remark 3.

In studying the proof of Lemma 5.6, we deduce the (slightly) more general
statement: Given any chart $(\mathcal{U},\varphi)\in\mathcal{A}(S)$ and $g\in
L^{1}(S,\Sigma_{P},\sigma_{P})$ for which
$\operatorname{Supp}(g)\in\mathcal{U}$, $(\varphi^{-1})^{*}g$ is Lebesgue
measurable on $U=\varphi(\mathcal{U})$ and

$\int_{S}g(\eta)\sigma_{P}(d\eta)=\begin{cases}\displaystyle\int_{U}(\varphi^{-1})^{*}(g\,d\sigma_{P})&\mbox{
if }(\mathcal{U},\varphi)\in\mathcal{A}_{+}(S)\\\ &\\\
-\displaystyle\int_{U}(\varphi^{-1})^{*}(g\,d\sigma_{P})&\mbox{ if
}(\mathcal{U},\varphi)\in\mathcal{A}_{-}(S)\coloneqq\mathcal{A}(S)\setminus\mathcal{A}_{+}(S).\end{cases}$

###### Proof of Theorem 5.4.

In view of Proposition 5.1 and the discussion following its proof, it remains
to prove the assertions in the last two sentences in the statement of the
theorem. Given any $F\in\Sigma_{P}$ and chart
$(\mathcal{U},\varphi)\subseteq\mathcal{A}(S)$, we have
$\chi_{\varphi(F\cap\mathcal{U})}=(\varphi^{-1})^{*}\left(\chi_{F\cap\mathcal{U}}\right)$
is Lebesgue measurable on $U=\varphi(\mathcal{U})$ and so it follows (Exercise
4.6.2 of [19]) that $F\in\mathcal{L}(S)$. Consequently,
$\mathcal{B}(S)\subseteq\Sigma_{P}\subseteq\mathcal{L}(S)$.

Let $g\in L^{1}(S,\Sigma_{P},\sigma_{P})$, which is necessarily Lebesgue
measurable on $S$ in view of the results of the previous paragraph. Now, let
$\\{(\mathcal{U}_{j},\varphi_{j})\\}\subseteq\mathcal{A}_{+}(S)$ be a
countable atlas on $S$ and let $\\{\kappa_{j}\\}$ be a smooth partition of
unity subordinate to the cover $\\{\mathcal{U}_{j}\\}$. For each
$j\in\mathbb{N}$, observe that $\kappa_{j}g\in L^{1}(S,\Sigma_{P},\sigma_{P})$
and has $\operatorname{Supp}(\kappa_{j}g)\subseteq\mathcal{U}_{j}$. By virtue
of Lemma 5.6, we have $(\varphi_{j}^{-1})^{*}(\kappa_{j}g\,d\sigma_{P})$ is
integrable on $U_{j}=\varphi(\mathcal{U}_{j})$ and

$\int_{S}\kappa_{j}(\eta)g(\eta)\sigma_{P}(d\eta)=\int_{U_{j}}(\varphi_{j}^{-1})^{*}(\kappa_{j}g\,d\sigma_{P})$
(54)

for each $j\in\mathbb{N}$. With the help of Proposition 5.2, it is easy to see
that $g\,d\sigma_{P}$ and $\kappa_{j}g\,d\sigma_{P}$ (for $j\in\mathbb{N}$)
are Lebesgue measurable $d^{\prime}$-forms on $S$. In view of (4.4.6) of [19],
(54) ensures that, for each $j\in\mathbb{N}$, $\kappa_{j}g\,d\sigma_{P}$ is
integrable (in the sense of forms) on $S$ and

$\int_{S}\kappa_{j}(\eta)g(\eta)\sigma_{P}(d\eta)=\int_{S}\kappa_{j}g\,d\sigma_{P}.$

By the monotone convergence theorem, it follows that

$\sum_{j=1}^{\infty}\left|\int_{S}\kappa_{j}g\,d\sigma_{P}\right|=\sum_{j=1}^{\infty}\left|\int_{S}\kappa_{j}(\eta)g(\eta)\sigma(d\eta)\right|\leq\sum_{j=1}^{\infty}\int_{S}\kappa_{j}(\eta)|g(\eta)|\sigma_{P}(d\eta)=\|g\|_{L^{1}(S,\Sigma_{P},\sigma_{P})}<\infty.$

Therefore, in view of the construction on p. 242 of [19], we conclude that the
$d^{\prime}$-form $g\,d\sigma_{P}$ is integrable and

$\int_{S}g(\eta)\sigma_{P}(d\eta)=\sum_{j=1}^{\infty}\int_{S}\kappa_{j}(\eta)g(\eta)\sigma_{P}(d\eta)=\sum_{j=1}^{\infty}\int_{S}\kappa_{j}g\,d\sigma_{P}=\int_{S}g\,d\sigma_{P}$

by virtue of the dominated convergence theorem; this is (47).

Finally, given that $\absolutevalue{\gradient P}$ is continuous and non-
vanishing on the compact set $S$,

$C_{1}:=\inf_{S}\frac{1}{|\gradient P|}\quad\mbox{and}\quad
C_{2}:=\sup_{S}\frac{1}{\absolutevalue{\gradient P}}$

are both positive real numbers. For each
$F\in\Sigma_{P}\subseteq\mathcal{L}(S)$, (47) guarantees that

$\sigma_{P}(F)=\int_{S}\chi_{F}(\eta)\sigma_{P}(d\eta)=\int_{S}\chi_{F}\,d\sigma_{P}$

By virtue of Proposition 5.2, it follows that

$C_{1}\operatorname{Vol}_{S}(F)=C_{1}\int_{S}\chi_{F}\,d\operatorname{Vol}_{S}\leq\int_{S}\frac{\chi_{F}}{|\gradient
P|}d\operatorname{Vol}_{S}=\int_{S}\chi_{F}\,d\sigma_{P}=\sigma_{P}(F)$

and

$\sigma_{P}(F)=\int_{S}\chi_{F}d\sigma_{P}=\int_{S}\frac{\chi_{F}}{|\gradient
P|}\,d\operatorname{Vol}_{S}\leq
C_{2}\int_{S}\chi_{F}\,d\operatorname{Vol}_{S}=C_{2}\operatorname{Vol}_{S}(F)$

where we have used the definition of the Riemannian volume measure on $S$,
c.f., [1]. In short, there are positive constants $C_{1}$ and $C_{2}$ for
which

$C_{1}\operatorname{Vol}_{S}(F)\leq\sigma_{P}(F)\leq
C_{2}\operatorname{Vol}_{S}(F)$ (55)

for all $F\in\Sigma_{P}$. In particular, (47) holds for all
$F\in\mathcal{B}(S)$ and so it follows that the completions of the
$\sigma$-algebra $\mathcal{B}(S)$ with respect to $\sigma_{P}$ and
$\operatorname{Vol}_{S}$ coincide. We know, however, that
$\operatorname{Vol}_{S}$, which is defined on $\mathcal{L}(S)$, is a Radon
measure (Proposition 1.5 [1, Chapter XII]) and, by virtue of Proposition 4.10,
it follows that $\Sigma_{P}=\mathcal{L}(S)$ and (55) holds for all $F$ in this
common $\sigma$-algebra. Thus $\sigma_{P}$ and $\operatorname{Vol}_{S}$ are
mutually absolutely continuous and the theorem is proved. ∎

We immediately obtain the following corollary which allows us to compute the
Lebesgue integral with respect to $\sigma_{P}$ in coordinates.

######  Corollary 5.7.

Let $g\in L^{1}(S,\mathcal{L}(S),\sigma_{P})$. Then, given any countable (or
finite) atlas
$\\{(\mathcal{U}_{j},\varphi_{j})\\}\subseteq\mathcal{A}_{+}(S)$, smooth
partition of unity $\\{\kappa_{j}\\}$ subordinate to $\\{\mathcal{U}_{j}\\}$,
and $E\in\operatorname{Exp}(P)$,

$\int_{S}g(\eta)\sigma_{P}(d\eta)=\sum_{j}\int_{S}\kappa_{j}g\,d\sigma_{P}=\sum_{j}\int_{U_{j}}\kappa_{j}(\varphi^{-1}(u))g(\varphi^{-1}(u))h_{E,\varphi_{j}}(u)\,du$

where, for each $j$,
$U_{j}=\varphi_{j}(\mathcal{U}_{j})\subseteq\mathbb{R}^{d^{\prime}}$ and

$h_{E,\varphi_{j}}(u)=\operatorname{det}(E\varphi^{-1}(u)|D_{u}\varphi^{-1})$

for $u=(u^{1},u^{2},\dots,u^{d^{\prime}})\in U_{j}$.

Acknowledgement: We thank Laurent Saloff-Coste for helpful discussions.

##  Appendix A Appendix

This appendix amasses some facts about continuous one-parameter subgroups of
$\operatorname{Gl}(\mathbb{R}^{d})$ and, in particular, those which are
contracting.

######  Proposition A.1 (see Section 8 of [21]).

Let $E,G\in\operatorname{End}(\mathbb{R}^{d})$ and
$A\in\operatorname{Gl}(\mathbb{R}^{d})$. Also, let $E^{*}$ denote the adjoint
of $E$. Then, for all $t,s>0$, the following statements hold:

$\bullet$ $1^{E}=I$ |  $\bullet$ $t^{E^{*}}=(t^{E})^{*}$ |  $\bullet$ $(t^{E})^{-1}=t^{-E}$ |  $\bullet$ If $EG=GE$, then $t^{E}t^{G}=t^{E+G}$
---|---|---|---
$\bullet$ $(st)^{E}=s^{E}t^{E}$ |  $\bullet$ $At^{E}A^{-1}=t^{AEA^{-1}}$ |  $\bullet$ $\operatorname{det}\left(t^{E}\right)=t^{\tr E}$ |

We recall from the introduction that a continuous one-parameter group
$\\{T_{t}\\}$ is said to be contracting if $\lim_{t\to 0}\|T_{t}\|=0$. The
notion of a contracting group (or semigroup) is closely related to notion of
asymptotic stability of solutions to linear systems [5]. By virtue of the
Banach-Steinhaus theorem, we have the following useful characterization of
contracting groups.

######  Proposition A.2.

Let $\\{T_{r}\\}$ be a continuous one-parameter group. Then $\\{T_{r}\\}$ is
contracting if and only if

$\lim_{t\to 0}|T_{r}x|=0$ (56)

for all $x\in\mathbb{R}^{d}$.

###### Proof.

It is clear that (56) is a necessary condition for $\\{T_{r}\\}$ to be
contracting. We must therefore prove (56) also sufficient. To this end, we
assume that the continuous one-parameter group $\\{T_{r}\\}$ satisfies (56).
By virtue of the continuity of $\\{T_{r}\\}$ and (56), we have

$\sup_{0<r\leq 1}|T_{r}x|<\infty$

for each $x\in\mathbb{R}^{d}$. From the Banach-Steinhaus theorem, it follows
that $\|T_{r}\|\leq C$ for all $0<r\leq 1$. Now, suppose that $\\{T_{r}\\}$ is
not contracting. In this case, one can find a sequence
$\\{\eta_{n}\\}\subseteq\mathbb{S}$ and a sequence $r_{n}\rightarrow 0$ for
which $\lim_{n}|T_{r_{n}}\eta_{n}|>0$. But because the unit sphere is compact,
$\\{\eta_{n}\\}$ has a convergence subsequence $\eta_{n_{k}}\rightarrow\eta$
with $|\eta|=1$ . Observe that, for all $n$,

$|T_{r_{n}}(\eta-\eta_{n})|\leq C|\eta-\eta_{n}|$

and so it follows that

$\lim_{k\rightarrow\infty}|T_{t_{r_{k}}}\eta|=\lim_{k\rightarrow\infty}|T_{r_{n_{k}}}\eta_{n_{k}}|>0,$

a contradiction. ∎

######  Lemma A.3.

Let $\\{T_{r}\\}\subseteq\mbox{Gl}_{d}(\mathbb{R})$ be a continuous one-
parameter group and let $E\in\operatorname{End}(\mathbb{R}^{d})$ be its
generator, i.e., $T_{r}=r^{E}$ for all $r>0$. If $\\{T_{r}\\}$ is contracting,
then $E\in\operatorname{Gl}(\mathbb{R}^{d})$ and there is a positive constant
$C$ for which

$\|T_{r}\|\leq C+r^{\|E\|}$

for all $r>0$.

###### Proof.

If for some non-zero vector $\eta$, $E\eta=0$, then $r^{E}\eta=\eta$ for all
$r>0$ and this would contradict our assumption that $\\{T_{r}\\}$ is
contracting. Hence $E\in\operatorname{Gl}(\mathbb{R}^{d})$ and, in particular,
$\|E\|>0$. From the representation $T_{r}=r^{E}$, it follows immediately that
$\|T_{r}\|\leq r^{\|E\|}$ for all $r\geq 1$ and so it remains to estimate
$\|T_{r}\|$ for $r<1$. Given that $\\{T_{r}\\}$ is continuous and contracting,
the map $r\mapsto\|T_{r}\|$ is continuous and approaches $0$ as $r\rightarrow
0$ and so it is necessarily bounded for $0<r\leq 1$. ∎

######  Proposition A.4.

Let $\\{T_{r}\\}_{r>0}\subseteq\operatorname{Gl}(\mathbb{R}^{d})$ be a
continuous one-parameter contracting group. Then, for all non-zero
$x\in\mathbb{R}^{d}$,

$\lim_{r\rightarrow 0}|T_{r}x|=0\hskip 14.22636pt\mbox{ and }\hskip
14.22636pt\lim_{r\rightarrow\infty}|T_{r}x|=\infty.$

###### Proof.

The validity of the first limit is clear. Upon noting that
$|x|=|T_{1/r}T_{r}x|\leq\|T_{1/r}\||T_{r}x|$ for all $r>0$, the second limit
follows at once. ∎

######  Proposition A.5.

Let $\\{T_{r}\\}_{r>0}$ be a continuous one-parameter contracting group. There
holds the following:

  1. (a)

For each non-zero $x\in\mathbb{R}^{d}$, there exists $r>0$ and
$\eta\in\mathbb{S}$ for which $T_{r}\eta=x$. Equivalently,

$\mathbb{R}^{d}\setminus\\{0\\}=\\{T_{r}\eta:r>0\mbox{ and
}\eta\in\mathbb{S}\\}.$

  2. (b)

For each sequence $\\{x_{n}\\}\subseteq\mathbb{R}^{d}$ such that
$\lim_{n}|x_{n}|=\infty$, $x_{n}=T_{r_{n}}\eta_{n}$ for each $n$, where
$\\{\eta_{n}\\}\subseteq\mathbb{S}$ and $r_{n}\rightarrow\infty$ as
$n\rightarrow\infty$.

  3. (c)

For each sequence $\\{x_{n}\\}\subseteq\mathbb{R}^{d}$ such that
$\lim_{n}|x_{n}|=0$, $x_{n}=T_{r_{n}}\eta_{n}$ for each $n$, where
$\\{\eta_{n}\\}\subseteq\mathbb{S}$ and $r_{n}\rightarrow 0$ as
$n\rightarrow\infty$.

###### Proof.

In view of Proposition A.4, the assertion (a) is a straightforward application
of the intermediate value theorem. For (b), suppose that
$\\{x_{n}\\}\subseteq\mathbb{R}^{d}$ is such that $|x_{n}|\rightarrow\infty$
as $n\rightarrow\infty$. In view of (a), take $\\{\eta_{n}\\}\subseteq S$ and
$\\{r_{n}\\}\subseteq(0,\infty)$ for which $x_{n}=T_{r_{n}}\eta_{n}$ for each
$n$. In view of Lemma A.3,

$\infty=\liminf_{n}|x_{n}|\leq\liminf_{n}\left(C+r_{n}^{M}\right)|\eta_{n}|\leq
C+\liminf_{n}r_{n}^{M},$

where $C,M>0$ and therefore $r_{n}\rightarrow\infty$. If instead
$\lim_{n}x_{n}=0$,

$\infty=\lim_{n\rightarrow\infty}\frac{|\eta_{n}|}{|x_{n}|}=\lim_{n\rightarrow\infty}\frac{|T_{1/r_{n}}x_{n}|}{|x_{n}|}\leq\limsup_{n}\|T_{1/r_{n}}\|\leq\limsup_{n}(C+(1/r_{n})^{M})$

from which we see that $r_{n}\rightarrow 0$, thus proving (c). ∎

######  Proposition A.6.

Let $\\{T_{r}\\}$ be a continuous contracting one-parameter group. Then for
any open neighborhood $\mathcal{O}\subseteq\mathbb{R}^{d}$ of the origin and
any compact set $K\subseteq\mathbb{R}^{d}$, $K\subseteq T_{r}(\mathcal{O})$
for sufficiently large $r$.

###### Proof.

Assume, to reach a contradiction, that there are sequences
$\\{x_{n}\\}\subseteq K$ and $r_{n}\rightarrow\infty$ for which $x_{n}\notin
T_{r_{n}}(\mathcal{O})$ for all $n$. Because $K$ is compact, $\\{x_{n}\\}$ has
a subsequential limit and so by relabeling, let us take sequences
$\\{\zeta_{k}\\}\subseteq K$ and $\\{t_{k}\\}\subseteq(0,\infty)$ for which
$\zeta_{k}\rightarrow\zeta$, $t_{k}\rightarrow\infty$ and $\zeta_{k}\notin
T_{t_{k}}(\mathcal{O})$ for all $k$. Setting $s_{k}=1/t_{k}$ and using the
fact that $\\{T_{r}\\}$ is a one-parameter group, we have
$T_{s_{k}}\zeta_{k}\notin\mathcal{O}$ for all $k$ and so
$\liminf_{k}|T_{s_{k}}\zeta_{k}|>0$, where $s_{k}\rightarrow 0$. This is
however impossible because $\\{T_{r}\\}$ is contracting and so

$\lim_{k\rightarrow\infty}|T_{s_{k}}\zeta_{k}|\leq\lim_{k\rightarrow\infty}|T_{s_{k}}(\zeta_{k}-\zeta)|+\lim_{k\rightarrow\infty}|T_{s_{k}}\zeta|\leq
C\lim_{k\rightarrow\infty}|\zeta_{k}-\zeta|+0=0$

in view of Lemma A.3. ∎

######  Proposition A.7.

Let $\\{T_{r}\\}\subseteq\operatorname{Gl}(\mathbb{R}^{d})$ be a continuous
one-parameter group with generator $E$. If $\\{T_{r}\\}$ is contracting, then
$\tr E>0$.

###### Proof.

The supposition that $\\{T_{r}\\}$ is a contracting group implies that
$r^{E}\to\mathbf{0}$ as $r\to 0$ in the operator-norm topology on
$\operatorname{End}(\mathbb{R}^{d})$; here $\mathbf{0}$ is zero
transformation. Because the determinant is a continuous function from
$\operatorname{End}(\mathbb{R}^{d})$, equipped with operator-norm topology,
into $\mathbb{R}$, we have

$0=\operatorname{det}(\mathbf{0})=\operatorname{det}\left(\lim_{r\to
0}r^{E}\right)=\lim_{r\to 0}\operatorname{det}\left(r^{E}\right)=\lim_{r\to
0}r^{\tr E}$

in view of the preceding proposition. Therefore, $\tr E>0$. ∎

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