Right translation scales left Haar measure

Fix $g\in G$ and define ${\nu }_g(E):=\mu (E{g}^{-1})$ for every Borel set $E$. The map ${\theta }_g(x):=xg$ is a homeomorphism of $G$ with inverse $x\mapsto x{g}^{-1}$, so it carries Borel sets to Borel sets by [F2] and compact sets to compact sets by [F3]. Hence ${\nu }_g$ is a nonzero Borel measure (it is nonzero because $\mu$ is, by [F1]) that is finite on compact sets: if $K$ is compact then $K{g}^{-1}$ is compact by [F3] and ${\nu }_g(K)=\mu (K{g}^{-1})<\mathrm{\infty }$ by [F1]. It is outer regular on Borel sets, since the homeomorphism gives a bijection $W\mapsto Wg$ from open supersets of $E{g}^{-1}$ to open supersets of $E$, and therefore ${\nu }_g(E)=\mu (E{g}^{-1})={\inf }_{W\supseteq E{g}^{-1}\text{ open}}\mu (W)={\inf }_{U\supseteq E\text{ open}}{\nu }_g(U)$, also when the value is infinite. Finally it is inner regular on open sets: for open $V$ the set $V{g}^{-1}$ is open, so inner regularity of $\mu$ gives ${\nu }_g(V)=\mu (V{g}^{-1})=\sup \{\mu (L):L\subseteq V{g}^{-1}\text{ compact}\}$, and $L\mapsto Lg$ is a bijection between the compact subsets of $V{g}^{-1}$ and those of $V$ with $\mu (L)={\nu }_g(Lg)$; hence ${\nu }_g(V)=\sup \{{\nu }_g(K):K\subseteq V\text{ compact}\}$. Thus ${\nu }_g$ is a Radon measure.

For $f\in C_c(G)$ define $N_g(f):={\int }_{G}f(xg)d\mu (x)$. By [F4] the function $x\mapsto f(xg)$ lies again in $C_c(G)$, so it is integrable against the compact-finite measure $\mu$ by [F1]; thus $N_g$ is a real-linear functional on $C_c(G;\mathbb{R})$, and it is positive, because $f\ge 0$ implies $f(xg)\ge 0$ for every $x$ and hence $N_g(f)\ge 0$.

The measure ${\nu }_g$ of step 1.1 represents $N_g$. Indeed, for the indicator of a Borel set $E$ the definitions give ${\nu }_g(E)=\mu (E{g}^{-1})={\int }_G{\mathbf{1}}_E(xg)d\mu (x)$; both sides are additive, so linearity extends the identity to nonnegative simple functions, [F6] extends it to all nonnegative Borel functions, and taking real and imaginary parts extends it to all of $C_c(G)$. In particular the Radon measure ${\nu }_g$ is the one attached to the positive functional $N_g$, and ${\nu }_g\mathrm{\ne }0$ by step 1.1.

${\nu }_g$ is left invariant: for $a\in G$ and Borel $E$, ${\nu }_g(aE)=\mu (aE{g}^{-1})=\mu (E{g}^{-1})={\nu }_g(E)$, using left invariance of $\mu$ in the middle step. With step 1.1 this makes ${\nu }_g$ a left Haar measure in the sense of [F1].

Since $\mu$ and ${\nu }_g$ are both left Haar measures, [F5] (whose choice hypothesis is discharged by [A1]) provides a scalar $c(g)>0$ with ${\nu }_g(E)=c(g)\mu (E)$ for every Borel set $E$. The scalar is unique: if also $c^{\mathrm{\prime }}\mu ={\nu }_g$ then $(c-c^{\mathrm{\prime }})\mu (E)=0$ for every Borel $E$, and some Borel set $E$ has $0<\mu (E)<\mathrm{\infty }$, since $\mu \mathrm{\ne }0$ together with outer and inner regularity of $\mu$ produces a compact set of positive finite measure by [F1].

Consequently, for every $f\in C_c(G)$, the representation in step 2.1 and the proportionality in step 3.1 give ${\int }_{G}f(xg)d\mu (x)=N_g(f)={\int }_{G}fd{\nu }_g=c(g){\int }_{G}fd\mu ,$ which is the asserted identity; the case $f=ix$ with $i$ the imaginary unit is handled by applying the real identity to $\mathrm{Re}f$ and $\mathrm{Im}f$. The scalar $c(g)$ is positive and unique by step 3.1, and $c(e)=1$ because ${\nu }_e=\mu$.