Haar measure on an infinite product of compact groups

The tuple of identities belongs to $G$. Coordinatewise multiplication and inversion are continuous by [F5]; distinct tuples differ in one coordinate whose Hausdorff neighbourhoods separate them. Under the assumed AC, Tychonoff gives compactness and [F1] gives $\mu$. For a finite $F\subseteq I$, the projection ${\pi }_F$ has a continuous section $s_F$ filling all other coordinates with identities.

For any Borel $B\subseteq G$, outer approximation of $G\setminus B$ and complementation give compact inner approximation of $B$, since $\mu (G)=1$. Set $\nu (E)=\mu ({\pi }_F^{-1}E)$. A compact $K\subseteq {\pi }_F^{-1}E$ projects to compact ${\pi }_{F}K\subseteq E$ with $\nu ({\pi }_{F}K)\ge \mu (K)$. Thus $\nu$ is inner regular on every Borel set; taking complements gives outer regularity. It is a Radon probability. A translation in the finite subproduct lifts by $s_F$, so invariance of $\mu$ makes $\nu$ invariant. [F1] identifies $\nu$ with the normalized Haar measure there.

If $h\in C(G;\mathbb{R})$ and $ϵ>0$, choose finitely many basic open cylinders covering $G$ on each of which $h$ oscillates by $<ϵ$. Let $F$ include their finitely many specified coordinates. Any $x$ and $s_F{\pi }_{F}x$ belong together to one covering cylinder, hence $\mathrm{\mid }h(x)-h(s_F{\pi }_{F}x)\mathrm{\mid }<ϵ$. Two Radon probabilities with equal finite marginals therefore have integrals of $h$ differing by at most $2ϵ$, by the uniform bound and total mass one. Letting $ϵ\downarrow 0$ and applying [F3] proves equality on the full Borel sigma algebra.

In the finite group $C_2^F$, invariance gives every singleton the same mass; their $2^{\mathrm{\mid }F\mathrm{\mid }}$ masses sum to one, so each is $2^{-\mathrm{\mid }F\mathrm{\mid }}$. Step 2.1 therefore gives $2^{-r}$ to a cylinder fixing $r$ coordinates of ${\prod }_I{C}_2$. The empty cylinder has mass $1$. If $I=\mathrm{\varnothing }$, the product itself is the one-point group and its measure is point mass one.