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If we have that $X$ and $Y$ are measure spaces and we have measurable functions $f, g: X \to Y$, can one conclude that $f \times g : X \to Y\times Y$ given by $(f\times g)(x) = (f(x), g(x))$ is measurable?

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1 Answer 1

up vote 5 down vote accepted

You really need to specify the $\sigma$-algebra you are using on $Y\times Y$.

Assuming that $(X,S)$ and $(Y,T)$ are your two measure spaces, and you let $T\otimes T$ be the product $\sigma$-algebra on $Y\times Y$ (that is, the $\sigma$-algebra generated by all sets of the form $A\times B$ with $A,B\in T$), then yes.

It suffices to show that the inverse image of each set in a generating set of the $\sigma$-algebra $T\otimes T$ is in $S$. If $A,B\in T$, then $$x\in (f\times g)^{-1}(A\times B)\Longleftrightarrow f(x)\in A\text{ and }g(x)\in B\Longleftrightarrow x\in f^{-1}(A)\cap g^{-1}(B).$$

Since, by assumption, $f^{-1}(A)$ and $g^{-1}(B)$ lie in $S$, then so does their intersection, so $(f\times g)^{-1}(A\times B)\in S$. This proves that $f\times g$ is measurable.

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You are saying that it suffices to consider the sets in the generating set $T\times T$ instead of $T\otimes T$. Is it because $f^{-1}$ commutes with taking unions and intersections and $f^{-1}(T\times T)=S$? –  Jack Oct 14 '12 at 21:08

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