Let $M = (\mathbb{R}^n, \mathcal{L}, \mu)$ be a D1744: Lebesgue measure space.
Let $\mathfrak{L}^1 = \mathfrak{L}^1(M)$ be the D2401: Set of absolutely integrable functions on $M$.

Let $f, g \in \mathfrak{L}^1$ and $x \in \mathbb{R}^n$. The endomorphism $\phi : \mathbb{R}^n \to \mathbb{R}^n$ given by $y \mapsto x - y$ is differentiable with $|\mathsf{det}(D \phi)| = 1$. Applying a change of variables in the sense of R2143: Differentiable change of variables for signed Lebesgue integral to this endomorphism yields \begin{equation} \begin{split} (f * g)(x) & = \int_{\mathbb{R}^n} f(y) g(x - y) \, \mu(d y) \\ & = \int_{\mathbb{R}^n} f(\phi(y)) g(x - \phi(y)) |\mathsf{det}(D \phi) | \, \mu(d y) \\ & = \int_{\mathbb{R}^n} f(x - y) g(y) \, \mu(d y) \\ & = (g * f)(x) \end{split} \end{equation} Since $x \in \mathbb{R}^n$ was arbitrary, the claim follows. $\square$