[proofplan] We show $(f * g)(x) = 0$ for $x \notin \overline{\operatorname{supp}(f) + \operatorname{supp}(g)}$ by verifying that the integrand $f(x-y)\,g(y)$ vanishes for $\mathcal{L}^n$-a.e. $y$: either $y \notin \operatorname{supp}(g)$ (so $g(y) = 0$) or $x - y \notin \operatorname{supp}(f)$ (so $f(x-y) = 0$). When both supports are compact, the Minkowski sum is compact (hence closed), and the closure is unnecessary. [/proofplan] [step:Show the integrand vanishes a.e. for $x \notin \overline{A + B}$] Let $A := \operatorname{supp}(f)$ and $B := \operatorname{supp}(g)$. Suppose $x \notin \overline{A + B}$. Since $\overline{A + B}$ is closed, there exists $\delta > 0$ with $B(x, \delta) \cap (A + B) = \varnothing$, so $x \neq a + b$ for all $a \in A$, $b \in B$. For any $y \in \mathbb{R}^n$, consider the integrand $f(x-y)\,g(y)$. Either $y \notin B$, in which case $g(y) = 0$ $\mathcal{L}^n$-a.e. (since $g$ vanishes a.e. outside its support), or $y \in B$, in which case $x - y \notin A$ (because $x - y \in A$ and $y \in B$ would give $x = (x-y) + y \in A + B$, contradicting $x \notin A + B$), so $f(x-y) = 0$ $\mathcal{L}^n$-a.e. In either case, $f(x-y)\,g(y) = 0$ for $\mathcal{L}^n$-a.e. $y$. [/step] [step:Conclude $\operatorname{supp}(f * g) \subseteq \overline{A + B}$] Since the integrand vanishes a.e. for every $x \notin \overline{A + B}$: \begin{align*} (f * g)(x) = \int_{\mathbb{R}^n} f(x-y)\,g(y) \, d\mathcal{L}^n(y) = 0. \end{align*} Hence $\operatorname{supp}(f * g) \subseteq \overline{A + B}$. [/step] [step:Remove the closure when both supports are compact] If both $A$ and $B$ are compact, then $A + B$ is compact: it is the image of the compact set $A \times B$ under the continuous map $(a, b) \mapsto a + b$. A compact subset of $\mathbb{R}^n$ is closed, so $\overline{A + B} = A + B$, and the closure is unnecessary. [/step]