Elementary question about convolution If $f,g \in L^1(\mathbb{R})$, my textbook proves that the convolution $f*g$ is also in $L^1(\mathbb{R})$. But it doesn't say why, for any $x \in \mathbb{R}$, we have that $y\mapsto f(x-y)g(y)$ is in $L^1(\mathbb{R})$. I would be very happy to hear an explanation?
Thank you
 A: $$\int_\mathbb{R}\left[\int_\mathbb{R}|f(y-x)||g(x)|\,dx\right]\, dy=\int_{\mathbb{R}^2}|f(y-x)||g(x)|\,dx\, dy=\\=\int_\mathbb{R}\left[\int_\mathbb{R}|f(y-x)||g(x)|\,dy\right]\, dx=\int_\mathbb{R}\left[\int_\mathbb{R}|f(y-x)|\,dy\right]|g(x)|\, dx=\\=\int_\mathbb{R}\left[\int_\mathbb{R}|f(t)|\,dt\right]|g(x)|\, dx=\int_\mathbb{R}\|f\|_1|g(x)|\, dx=\|f\|_1\|g\|_1$$
Where cructial passages are justified by Tonelli's theorem.
Now, suppose that $\mathfrak{L}_1\left\{y\in\mathbb{R}:\int_\mathbb{R}|f(y-x)g(x)|\,dx=+\infty\right\}>0$.
If it were the case, the third term in that chain of inequalities would be $+\infty$, which goes against the hypothesis. So that set is negligible.
Hence, we can say that $\exists|f|*|g|\in L^1(\mathbb{R})$, since $|f|*|g|(y)$ is almost-everywhere real.
But $\forall y\ s.t.\ |f|*|g|(y)<+\infty\ \exists f*g(y)\in\mathbb{R}$.
$f*g$ is therefore well defined as an element of $L^0(\mathbb{R})$. Moreover, it satisfies
$$\int_\mathbb{R}|f*g(y)|\, dy=\int_\mathbb{R}\left|\int_\mathbb{R}f(y-x)g(x)\,dx\right|\, dy\le\int_\mathbb{R}\left[\int_\mathbb{R}|f(y-x)||g(x)|\,dx\right]\, dy=\|f\|_1\|g\|_1$$
Therefore $f*g\in L^1(\mathbb{R})$.
As far as I know, $f*g(y)$ needs not be defined $\forall y\in\mathbb{R}$, if $f\in L^1(\mathbb{R}),\ g\notin L^\infty(\mathbb{R})$.
