# 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.)

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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
en.wikipedia.org/wiki/… – Did Apr 12 '11 at 13:35
@user8484: so you're speaking of Lebesgue measurable functions? You should definitely specify that in your question, then. For Borel measurable functions this is indeed trivial, as Didier points out. However, you're right: The composition of Lebesgue measurable functions need not be measurable. In the situation you're interested in, this is true but rather subtle (essentially this is because you're composing a measurable function with an absolutely continuous function). You can find a complete proof in Hewitt and Ross, Abstract harmonic analysis I, Theorem (20.10), p.291. – t.b. Apr 12 '11 at 13:40
Right, that's why I am confused. Now, I think that the everything is clear now. Thank you. – user8484 Apr 12 '11 at 14:21

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.

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