If $f,g\in L^1({\mathbb{R}}^n)$, then $f\ast g$ exists almost everywhere, belongs to $L^1$, and $\mathrm{\parallel }f\ast g{\mathrm{\parallel }}_1\le \mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_1$

Use the stated countable choice and [L0] to choose Borel representatives $\tilde{f},\tilde{g}$ of $f,g$ and define [L0, L1, L2, L4, given, choose, algebra] $H(x,y):=\mathrm{\mid }\tilde{f}(x-y)\mathrm{\mid }\mathrm{\mid }\tilde{g}(y)\mathrm{\mid }\mathrm{.}$ By [L1], $H$ is measurable on ${\mathbb{R}}^{2n}$. For each $y$, ${\int }_{{\mathbb{R}}^n}H(x,y)dx=\mathrm{\mid }\tilde{g}(y)\mathrm{\mid }{\int }_{{\mathbb{R}}^n}\mathrm{\mid }\tilde{f}(x-y)\mathrm{\mid }dx=\mathrm{\parallel }\tilde{f}{\mathrm{\parallel }}_1\mathrm{\mid }\tilde{g}(y)\mathrm{\mid }$ by [L4].

Integrating the identity from step 1.1 in $y$ and applying [L2] gives [L2, step 1.1, algebra] ${\iint }_{{\mathbb{R}}^{2n}}H(x,y)dxdy=\mathrm{\parallel }\tilde{f}{\mathrm{\parallel }}_1\mathrm{\parallel }\tilde{g}{\mathrm{\parallel }}_1<\mathrm{\infty }\mathrm{.}$ Hence for almost every $x$, the section $y\mapsto \tilde{f}(x-y)\tilde{g}(y)$ is absolutely integrable, so $(f\ast g)(x)$ is defined there.

For those $x$, [L1, L2, L3, step 2.1, algebra] $\mathrm{\mid }(f\ast g)(x)\mathrm{\mid }\le \int \mathrm{\mid }\tilde{f}(x-y)\tilde{g}(y)\mathrm{\mid }dy$ by [L3]. Another application of [L2] then yields $\mathrm{\parallel }f\ast g{\mathrm{\parallel }}_1\le \iint \mathrm{\mid }\tilde{f}(x-y)\mathrm{\mid }\mathrm{\mid }\tilde{g}(y)\mathrm{\mid }dydx=\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_1\mathrm{.}$ By [L1], the resulting $L^1$ class is independent of the chosen Borel representatives.