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If $S$ is a measurable Abelian group, and $\mu$ probability measure on $S$, how would you prove that $ f(x) = \mu (x + B) $ is measurable function for every fixed measurable set $B$?

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Write $\mu(x+B)=\int_S 1_{B}(y-x)\ \mu(dy)$.

By general measure theory, the map $x\mapsto\int_S f(x,y)\ \mu(dy)$ is measurable for any jointly measurable, bounded $f:S\times S\to \mathbb{R}$.

This result can be found, for instance, in Lemma 1.26 (p. 14) of Kallenberg's Foundations of Modern Probability (2nd edition). The proof uses a monotone class argument.

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    If you have a different measure theory te$x$t, $y$ou will probabl$y$ find this fact as part of the proof of Fubini's theorem.2011-04-03