Haar change of variables under inversion

Transfer of regularity: if $\theta :G\to G$ is a homeomorphism and $\lambda$ a Radon measure, then ${\theta }_{\ast }\lambda (E):=\lambda ({\theta }^{-1}E)$ is a Radon measure. Indeed $\theta$ and ${\theta }^{-1}$ carry Borel sets to Borel sets by [F7], so ${\theta }_{\ast }\lambda$ is a Borel measure; it is finite on compact $K$ because ${\theta }^{-1}K$ is compact by [F8] and $\lambda$ is compact finite by [F6]; it is outer regular on Borel sets and inner regular on open sets by transporting the corresponding open and compact approximations along $\theta$ as in the definition of ${\theta }_{\ast }\lambda$.

The functional $P(f):={\int }_{G}f(x){\mathrm{\Delta }}_G(x^{-1})d\mu (x)$ on $C_c(G;\mathbb{R})$ is positive and real-linear: the integrand is continuous with compact support because ${\mathrm{\Delta }}_G\circ \mathrm{i}\mathrm{n}\mathrm{v}$ is continuous by [F2] and $f$ is compactly supported, so it is $\mu$-integrable by the compact finiteness of [F6]; positivity and linearity are those of the integral, and $f\ge 0$ gives $\int f(x){\mathrm{\Delta }}_G(x^{-1})d\mu (x)\ge 0$ because the density is positive. By [F5] there is a Radon measure $\sigma$ with $\int fd\sigma =P(f)$ for every $f\in C_c(G)$.

Identify the density on Borel sets. Put $w(x)={\mathrm{\Delta }}_G(x^{-1})>0$ and $\eta (E)={\int }_{E}wd\mu$. This is a nonzero Borel measure by [F9], finite on compact sets since $w$ is continuous. To prove outer regularity, only $\eta (E)<\mathrm{\infty }$ needs consideration. Partition $E$ into $E_n=E\cap \{2^n\le w<2^{n+1}\}$ for $n\in \mathbb{Z}$. Then $\mu (E_n)\le 2^{-n}\eta (E)<\mathrm{\infty }$. Given $ϵ>0$, choose positive ${ϵ}_n$ with ${\sum }_n{ϵ}_n<ϵ$. Outer regularity of $\mu$ and [A1] give open $O_n\supseteq E_n$ contained in $\{w<2^{n+2}\}$ with $\mu (O_n\setminus E_n)<2^{-n-2}{ϵ}_n$. Their union $O$ contains $E$ and satisfies $\eta (O\setminus E)\le {\sum }_n{2}^{n+2}\mu (O_n\setminus E_n)<ϵ$. Thus $\eta (E)={\inf }_{O\supseteq E\text{ open}}\eta (O)$; if $\eta (E)=\mathrm{\infty }$ the equality is automatic.

For an open $U$, the functions $s_m=2^{-m}{\sum }_{k=1}^{m{2}^m}{\mathbf{1}}_{U\cap \{w>k{2}^{-m}\}}$ increase to $w{\mathbf{1}}_U$. For each finite sum, inner regularity of $\mu$ on its open level sets lets one approximate its integral from below using compact subsets of these sets. Their finite union $K\subseteq U$ is compact, and $\eta (K)$ is at least the sum of their weighted measures, since their weighted indicators sum to at most $w{\mathbf{1}}_K$. This holds also for arbitrarily large finite lower bounds if one of the open sets has infinite measure. Monotone convergence [F9] gives $\eta (U)={\sup }_{K\subseteq U\text{ compact}}\eta (K)$. Therefore $\eta$ is Radon. By [F9] its $C_c$ integrals are $P$, so [F4] identifies $\eta =\sigma$ on all Borel sets. In particular $\sigma e0$ and $\int fd\sigma =\int fwd\mu$ for every nonnegative Borel $f$.

$\nu (E):=\mu (E^{-1})$ is a right Haar measure: by step 1.1 applied to the homeomorphism $\theta (x)=x^{-1}$ it is Radon and nonzero, and for Borel $E$ and $a\in G$, $\nu (Ea)=\mu ((Ea{)}^{-1})=\mu (a^{-1}{E}^{-1})=\mu (E^{-1})=\nu (E)$ by left invariance of $\mu$ and $(Ea{)}^{-1}=a^{-1}{E}^{-1}$.

$\sigma$ is right invariant. For $F\in C_c(G)$ and $a\in G$, using the Borel-level identity of [F1] with $H(x):=F(x{a}^{-1}){\mathrm{\Delta }}_G(x^{-1})\in C_c(G)$ (using real and imaginary parts if necessary) gives ${\int }_{G}F(x{a}^{-1})d\sigma (x)=P(x\mapsto F(x{a}^{-1}))={\int }_{G}F(x{a}^{-1}){\mathrm{\Delta }}_G(x^{-1})d\mu (x)={\mathrm{\Delta }}_G(a){\int }_{G}F(y){\mathrm{\Delta }}_G(a^{-1}{y}^{-1})d\mu (y)={\mathrm{\Delta }}_G(a){\mathrm{\Delta }}_G(a^{-1}){\int }_{G}F(y){\mathrm{\Delta }}_G(y^{-1})d\mu (y)=P(F)={\int }_{G}Fd\sigma$, where the third equality substitutes $x=ya$ and the fourth uses multiplicativity of ${\mathrm{\Delta }}_G$ from [F2]. Both $\sigma$ and its pushforward under $x\mapsto x{a}^{-1}$ are Radon by step 1.1, so [F4] upgrades this identity of $C_c$ integrals to $\sigma (Ea)=\sigma (E)$ for every Borel $E$.

Since $\sigma$ is right Haar by step 2.2, the measure ${\sigma }^{\mathrm{♯}}(E):=\sigma (E^{-1})$ is left Haar and Radon by step 1.1; so is $\mu$, so [F3] (whose choice hypothesis is discharged by [A1]) gives a scalar $\lambda >0$ with ${\sigma }^{\mathrm{♯}}=\lambda \mu$. Unwinding the definitions, for every $f\in C_c(G)$ we have $(\ast )$: ${\int }_{G}f(x^{-1}){\mathrm{\Delta }}_G(x^{-1})d\mu (x)={\int }_{G}fd{\sigma }^{\mathrm{♯}}=\lambda {\int }_{G}fd\mu$.

The scalar is one. Write $\mathrm{\Phi }f(x):=f(x^{-1}){\mathrm{\Delta }}_G(x^{-1})$; then $(\ast )$ reads ${\int }_G\mathrm{\Phi }fd\mu =\lambda {\int }_{G}fd\mu$ for all $f\in C_c(G)$. Since $\mathrm{\Phi }$ is an involution, $\mathrm{\Phi }(\mathrm{\Phi }f)(x)=\mathrm{\Phi }f(x^{-1}){\mathrm{\Delta }}_G(x^{-1})=f(x){\mathrm{\Delta }}_G(x){\mathrm{\Delta }}_G(x^{-1})=f(x)$ by [F2], applying $(\ast )$ to $\mathrm{\Phi }f$ gives ${\int }_{G}fd\mu ={\int }_G\mathrm{\Phi }(\mathrm{\Phi }f)d\mu =\lambda {\int }_G\mathrm{\Phi }fd\mu ={\lambda }^2{\int }_{G}fd\mu$; and some $f\in C_c(G)$ has ${\int }_{G}fd\mu \mathrm{\ne }0$, since otherwise the zero measure and $\mu$ would have the same $C_c$ integrals and [F4] would force $\mu =0$, contrary to [F6]. Hence ${\lambda }^2=1$, and $\lambda >0$ forces $\lambda =1$.

With $\lambda =1$, applying $(\ast )$ to the test function $g(x):=f(x){\mathrm{\Delta }}_G(x^{-1})$, which lies in $C_c(G)$ because ${\mathrm{\Delta }}_G\circ \mathrm{i}\mathrm{n}\mathrm{v}$ is continuous by [F2], gives ${\int }_{G}f(x^{-1})d\mu (x)={\int }_{G}g(x^{-1}){\mathrm{\Delta }}_G(x^{-1})d\mu (x)=\lambda {\int }_{G}f(x){\mathrm{\Delta }}_G(x^{-1})d\mu (x)={\int }_{G}f(x){\mathrm{\Delta }}_G(x^{-1})d\mu (x)$, after using ${\mathrm{\Delta }}_G(x){\mathrm{\Delta }}_G(x^{-1})=1$ in the first equality; so the two functionals $f\mapsto {\int }_{G}f(x^{-1})d\mu (x)$ and $f\mapsto {\int }_{G}f(x){\mathrm{\Delta }}_G(x^{-1})d\mu (x)$ agree on $C_c(G)$.

Step 5.1 shows that the functionals $f\mapsto \int fd\nu$ and $f\mapsto \int fd\sigma$ agree on $C_c(G)$, where $\nu =\mu \circ \mathrm{i}\mathrm{n}\mathrm{v}$ is the Radon measure of step 2.1 and $\sigma$ is the Radon measure of step 1.2 representing $P$. By [F4] the Radon measures $\nu$ and $\sigma$ are equal on all Borel sets. Since the integral of a nonnegative Borel function is determined by the measure it integrates against, for every nonnegative Borel $f$ one has ${\int }_{G}f(x^{-1})d\mu (x)={\int }_{G}fd\nu ={\int }_{G}fd\sigma ={\int }_{G}f(x){\mathrm{\Delta }}_G(x^{-1})d\mu (x)$, with extended nonnegative integrals. If $f$ is a complex Borel function with ${\int }_G{\mathrm{\Delta }}_G(x^{-1})\mathrm{\mid }f(x)\mathrm{\mid }d\mu (x)<\mathrm{\infty }$, applying this nonnegative identity to $\mathrm{\mid }f\mathrm{\mid }$ first shows ${\int }_G\mathrm{\mid }f(x^{-1})\mathrm{\mid }d\mu (x)<\mathrm{\infty }$. Apply the identity to the positive and negative parts of the real and imaginary parts of $f$ and combine them; each has finite weighted integral because it is bounded by $\mathrm{\mid }f\mathrm{\mid }$. This gives the asserted finite complex identity. ∎