Show that for $L^1$ functions, the convolution is the product of the integrals

Question

So, basically, I need to prove that given two functions $f,g \in L^1$, then the convolution of f and g is the product of the integrals of these two functions. I know this has been asked before and I saw this post (why does the integral of convolution equal to the product of their integral separately?) but it doesn't use the hypothesis that these functions are $L^1$, and it seems weird to me that I'm told this hypothesis but I wouldn't need to use it, and also, as far as I understand, in order to use the Fubini's theorem, I need to know that $\int_{\mathbb{R^n}\times\mathbb{R^n}}|f(x-y)g(y)d(x,y)| < \infty$, and I guess I have to proof this statement first before using Fubini, but I don't know how to do that.

Answer 1

$$||f \ast g||_1 \leq \int \int |f(x-y)||g(y)|dydx$$ $$= \int |g(y)|\int |f(x-y)|dxdy$$ $$= \int ||f||_1|g(y)|dy=||f||_1||g||_1$$

We used Τonneli's theorem in the first equality.

Also by Fubini then you have the desired conclusion.