# Measurable function

How can I show that $f(x-y)g(y)$ is measurable on $\mathbb{R}^{2n}$ if $f,g$ are measurable on $\mathbb{R}^n$?

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Answer: by writing this function as a composition of functions known to be measurable. For instance, you could first explain why the functions $(x,y)\mapsto x-y$ and $(x,y)\mapsto (x-y,y)$ and $(x,y)\mapsto (f(x),g(y))$ are all measurable. (And the tag (homework) would seem mandatory here.)
Is it true that every composition of two measurable function is measurable? That's why I'm asking this question. e.g. it is not true that $f\circ g$ is measurable whenever $f$ is measurable and $g$ is continuous. – user8484 Apr 12 '11 at 13:31
The standard argument seems to be the following. There exist Borel measurable functions $f_0$ and $g_0$ such that $f(x)=f_0(x)$, $g(x)=g_0(x)$ almost everywhere. Now prove that $f_0(x-y)g_0(y)$ is Borel measurable and $f(x-y)g(y)=f_0(x-y)g_0(y)$ for almost all $(x,y)\in\mathbb{R}^{2n}$. See for instance theorem 7.14 in W. Rudin's Real & Complex Analysis.