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 $f, g, h \in \mathfrak{L}^1$ and $x \in \mathbb{R}^n$. If $z \in \mathbb{R}^n$, then the endomorphism $\phi : \mathbb{R}^n \to \mathbb{R}^n$ given by $y \mapsto y + z$ 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, together with R2057: Fubini's theorem, yields \begin{equation} \begin{split} ((f * g) * h)(x) & = \int_{\mathbb{R}^n} (f * g)(y) h(x - y) \, \mu(d y) \\ & = \int_{\mathbb{R}^n} \Big( \int_{\mathbb{R}^n} f(z) g(y - z) \, \mu(d z) \Big) h(x - y) \, \mu(d y) \\ & = \int_{\mathbb{R}^n} f(z) \Big( \int_{\mathbb{R}^n} g(y - z) h(x - y) \, \mu(d y) \Big) \, \mu(d z) \\ & = \int_{\mathbb{R}^n} f(z) \Big( \int_{\mathbb{R}^n} g(\phi(y) - z) h(x - \phi(y)) |\mathsf{det}(D \phi)| \, \mu(d y) \Big) \, \mu(d z) \\ & = \int_{\mathbb{R}^n} f(z) \Big( \int_{\mathbb{R}^n} g(y) h(x - y - z) \, \mu(d y) \Big) \, \mu(d z) \\ & = \int_{\mathbb{R}^n} f(z) (g * h)(x - y) \, \mu(d z) \\ & = (f * (g * h))(x) \end{split} \end{equation} Since $x \in \mathbb{R}^n$ was arbitrary, the claim follows. $\square$