Given a measurable group $(X,\mathbf{S},\mu)$ as in Halmos Measure Theory, $\S$59. That means that
- $(X,\mathbf{S},\mu)$ is a $\sigma$-finite measure space
- $\mu\ne0$
- $X$ is a group
- $\mathbf{S}$ and $\mu$ are invariant under left-translations
- The endomorphism $S$ on $X\times X$ defined by $(x,y)\mapsto (x,xy)$ is measurability-preserving.
Let $E\subseteq X$ be a measurable set such that $\mu(E)<\infty$. Is it true that for any $x\in G$ we have $\mu(Ex)<\infty$ where $Ex=\{yx\mid y\in E\}$ is a right-translation of $E$?
The reason I ask is that in $\S$60, Theorem A at the end of the proof there is a claim that $g(x^{-1})\nu(Ex^{-1})=f(x)$. This equality however only holds if $0<\nu(Ex^{-1})<\infty$. We are given that $0<\nu(E)<\infty$. The inequality $0<\nu(Ex^{-1})$ follows from $\S$59 Theorem D, but I don't see why $\nu(Ex^{-1})<\infty$.
The answer is yes. The following is a proof in a roundabout way (credit to Sébastien Gouëzel).
If we accept the proof of Theorem 60.A for now, let $R_x$ be right multiplication by $x$. Then $(R_x)_* \mu$ is a left-invariant measure, so by Theorem 60.B it is proportional to $\mu$, say equal to $c(x)\mu$ for some (finite) $c(x)$. In particular, if $E$ has finite measure for $\mu$, it also has finite measure for $(R_x)_* \mu$.
To fill the gap in Theorem 60.A, we will prove something weaker, namely that if $0 < \nu(E) < \infty$ then $\nu(Ex) < \infty$ holds for $\mu$-almost every $x$. This is sufficient to finish the proof of Theorem 60.A, since it implies that $\int g(x^{-1})\nu(Ex^{-1})d\mu(x)=\int f(x)d\mu(x)$.
We use the first part of the proof, stating that for any nonnegative measurable function $g$, any measurable set $E$, and any $\sigma$-finite left-invariant measures $\mu$ and $\nu$ you get the equality $$\mu(E) \int g(y) d\nu(y) = \int g(x^{-1}) \nu(Ex^{-1}) d\mu(x).\tag{$\star$}$$
Let $E$ such that $\nu(E)$ is finite, and let $g$ be the indicator function of a set $A$ with $\nu(A^{-1})<\infty$ and $\nu(A)<\infty$. Then $(\star)$ for the case that $\mu$ equals $\nu$ gives $\nu(E)\nu(A)=\int_{A^{-1}} \nu(Ex^{-1})d\nu(x)$. The left hand side is finite, so the right hand side is also finite. It follows that, on $A^{-1}$, the quantity $\nu(Ex^{-1})$ is $\nu$-almost everywhere finite. Now by using $\sigma$-finiteness of $\nu + \check\nu$ (where $\check\nu(B)=\nu(B^{-1})$), one obtains that $\nu(E x^{-1})$ is finite for $\nu$-almost every $x$.
We are almost done, but we still need to prove that $\nu(E x^{-1})$ is finite for $\mu$-almost every $x$, so it suffices to prove that $\mu\ll\nu$.
If $E$ is any measurable set with $\nu(E)=0$ then $\nu(Ex^{-1})=0$ (Theorem 59.D), so ($\star$) with $g=1$ reduces to $\mu(E)\nu(X)=0$. Since $\nu\ne0$, we get $\mu(E)=0$, finishing the proof.