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Let $f \in C_0^\infty $, $g \in L^1 $ . Then $$ \int_{\mathbb R^n} \int_{\mathbb R^n} f(x-y)g(y) dy dx = \int_{\mathbb R^n}\int_{\mathbb R^n} f(x-y)g(y) dx dy $$holds? If so, why? ($f,g : \mathbb R^n \to \mathbb R $)

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There is something called Fubini's theorem which gives the general setting . – Theorem Jun 10 '12 at 11:36
up vote 0 down vote accepted

This depends on the finiteness of the integral. Fubini's theorem can be applied:

Let $X$ and $Y$ be complete measure spaces. If $$\int_{X\times Y} |f(x,y)| \ \mathrm{d}(x,y) < \infty$$ then $$\int_X \int_Y f(x,y) \ \mathrm{d}y \ \mathrm{d}x = \int_{X\times Y} f(x,y) \ \mathrm{d}(x,y) = \int_Y \int_X f(x,y) \ \mathrm{d}x \ \mathrm{d}y$$

Since $f \in C^\infty_0$ and $g \in L^1$ I think it's safe to say that the integral in question converges.

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