Let $f,g\in L^{1}(\mathbb R^d).$ The convolution of $f$ and $g$ is the function $f\ast g$ defined by $$(f\ast g)(x)=\int_{\mathbb R^d}f(x-y)g(y)\,dy.$$ Show that $(f\ast g)(x)$ is well-defined for a.e. $x\in\mathbb R^d$. That is, show that for a.e. $x\in\mathbb R^d$, the function $y\mapsto f(x-y)g(y)$ is integrable (in particular, show that it is measurable).
$\textbf{My Attempt:}$ The function $f^{\star}(x,y)=\vert f(x-y)\vert$ is measurable in $\mathbb R^{d}\times\mathbb R^{d}.$ Also, the function $g^{\star}(x,y)=\vert g(x)\vert$ is measurable in $\mathbb R^d\times\mathbb R^d.$ It follows that $h:\mathbb R^d\times\mathbb R^d\to\mathbb R$ defined by $h(x,y)=\vert f(x-y)g(y)\vert$ is measurable in $\mathbb R^d\times\mathbb R^d,$ since it is a product of measurable functions. Since $h$ is also non-negative, Tonelli's theorem and translation invariance of the Lebesgue integral imply that \begin{equation}\begin{split}\int_{\mathbb R^d}\int_{\mathbb R^d}\vert h(x,y)\vert\,dy\,dx &=\int_{\mathbb R^d}\left(\int_{\mathbb R^d}\vert f(x-y)\vert\vert g(y)\vert\,dx\right)\,dy\\&=\int_{\mathbb R^d}\left(\int_{\mathbb R^d}\vert f(x)\vert\,dx\right)\vert g(y)\vert\,dy\\&=\left(\int_{\mathbb R^d}\vert f(x)\vert\,dx\right)\left(\int_{\mathbb R^d}\vert g(y)\vert\,dy\right)\\&=\|f\|_{1}\|g\|_{1}\\&<\infty.\end{split}\end{equation} $\color{red}{\text{This implies that } \int_{\mathbb R^d}\vert f(x-y)\vert\vert g(y)\vert\,dy<\infty \text{ for a.e } x\in\mathbb R^d}.$ It follows that the function $y\mapsto f(x-y)g(y)$ is integrable for a.e. $x\in\mathbb R^d$, and hence measurable.
Is my argument correct?
My main concern is with the claim in red.
Thank you for your time,