# Question about Convolution Operator

If $f \in L^{p}(\mathbb{R})$ and $g \in L^{1}(\mathbb{R})$, then why does $(f \ast g)(x) = \int_{\mathbb{R}}f(x - y)g(y)\, dy$ exist for almost every $x$?

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## 1 Answer

Consider first the case $p=1$, and apply Fubini's theorem to the map $F(x,y)=f(x-y)g(y)$. A complete proof is contained in Rudin's book Real and complex analysis. The case $p>1$ follows in a similar manner.

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