Existence of left and right Haar measures

Let $I$ be the integral in [F1]. At the outer-subadditivity step in its RMK construction, use the equivalent supremum in [F6] over $0\le f\le 1$ with $\mathrm{supp}f\subseteq U$. If $U={\bigcup }_n{U}_n$, this support has a finite subcover with distinct indices. A subordinate partition from [F7] gives $f={\sum }_{j}f{\phi }_j$, each term compactly supported in its assigned $U_{n_j}$ and bounded by one there, since the partition sums to one on the support of $f$. Hence $I(f)\le {\sum }_n\rho (U_n)$ and taking the supremum gives open subadditivity. For arbitrary sets $E_n$ with finite sum of outer contents, choose open supersets with errors $ϵ2^{-n-1}$ under AC. Their union gives ${\mu }^{\ast }(\bigcup E_n)\le \sum {\mu }^{\ast }(E_n)+ϵ$; infinite sums need no estimate. Letting $ϵ\downarrow 0$ supplies the required subadditivity. This corrects the inference from $f\le 1_U$ to support containment, which is not valid by itself.

With that construction step justified, [F2] represents $I$ by a Radon measure $\mu$. It is nonzero because some $I(f)>0$. A homeomorphism $T$ takes compact sets to compact sets and bijects open sets and Borel sets. Therefore $T_{\ast }\mu (E)=\mu (T^{-1}E)$ is finite on compact sets and inherits outer regularity and open inner regularity by transporting the approximating open and compact sets.

For $T(x)=ax$, integration against $T_{\ast }\mu$ gives $\int f(ax)d\mu (x)=I(L_{a^{-1}}f)=I(f)$ for every real $f\in C_c(G)$. The pushforward integral identity follows first for indicators and simple functions and then by monotone approximation of nonnegative measurable functions and positive/negative parts. RMK uniqueness now gives $T_{\ast }\mu =\mu$, hence left invariance. For inversion, put $\nu (E)=\mu (E^{-1})$. Since $(Ea{)}^{-1}=a^{-1}{E}^{-1}$, $\nu (Ea)=\mu (a^{-1}{E}^{-1})=\nu (E)$. It is nonzero and Radon by step 2.1, hence right Haar.