# If $f,g$ integrable then $f(x-y)g(y)$ integrable for almost every $x$

I am trying to prove that for two integrable functions $f,g: \mathbb{R}^n \rightarrow \mathbb{R}$ the function $y \mapsto f(x-y)g(y)$ is integrable for almost every $x$. By using the holder inequality I reduced this to showing that if a function is integrable then also its square is integrable but after browsing a bit I found this so I guess this leads nowhere. Any hints are welcomed.

• This is called the convolution. It is very important. Try to derive some of the other properties found at en.wikipedia.org/wiki/Convolution Dec 28, 2015 at 3:00
• The function $x\mapsto\sqrt x$ on the interval $(0,1)$ is integrable but its square is not. ${}\qquad{}$ Dec 28, 2015 at 3:44

Hint: consider $\int\int |f(x-y)g(y)| \, dy \, dx$, and use Tonelli's theorem to reverse the order of integration. If you can show this integral is finite, then $\int |f(x-y)g(y)| \, dy$ is finite for almost every $x$.
• Tonelli says $$\int\int |f(x-y)g(y)| \, dy \, dx = \int\int |f(x-y)g(y)| \, dx \, dy.$$ The inside integral is plainly finite if $f$ is integrable and $g(y)$ is finite. But how does that imply the outside integral is finite? ${}\qquad{}$ Dec 28, 2015 at 6:42
• Once you've switched the order of integration, $|g(y)|$ is just a constant in the inner integral, so you may bring it outside the (inner) integral (but still inside the outer integral)...
• But you still have the factor $y\mapsto \int |f(x-y)|\,dx$, so the integrability of $g$ is not enough at that point. ${}\qquad{}$ Dec 28, 2015 at 17:00
• You also need to use the translation-invariance of the Lebesgue integral and the integrability of $f$.
• aha! I guess I was distracted by other aspects of the problem. ${}\qquad{}$ Dec 28, 2015 at 22:27