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Let $\mu$ and $\nu$ be $s$-finite Borel measures on $\mathbb{R}^{d}$ having Borel-measurable densities $f$ and $g$ respectively with respect to Lebesgue measure $m$ on $\mathbb{R}^{d}$. Prove that the Borel-measurable function $f \ast g$ is a density for $\mu \ast \nu$ with respect to $m$.
My attempt: We have $$f \ast g(x) = \int_{\mathbb{R}^d} f(x-y)g(y)dy.$$ Since $f$ is a density for $\mu$ with respect to $m$, for any $A \in \text{Borel}(\mathbb{R}^{d})$, we have $$\mu(A) = \int_{A}fdm.$$ Since $g$ is a density for $\nu$ with respect to $m$, for any $A\in \text{Borel}(\mathbb{R}^{d})$, we have $$\nu(A) = \int_{A}gdm.$$ We have $$(\mu \ast \nu)(A) = (\mu \otimes \nu)(\{(x,y) \in \mathbb{R}^{d} \times \mathbb{R}^{d}: x + y \in A\}).$$ So we need to show for any $A \in \text{Borel}(\mathbb{R}^d)$, $$(\mu \ast \nu)(A) = \int_{A} \int_{\mathbb{R}^d}f(x-y)g(y)dy.$$ We rewrite $$(\mu \ast \nu)(A) = \int_{\mathbb{R}^{d}}\int_{\mathbb{R}^d} 1_{A}(x+y)d\mu(x)d\nu(y).$$ But I really don't know how to go from here to show $$\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^d} 1_{A}(x+y)d\mu(x)d\nu(y) = \int_{A} \int_{\mathbb{R}^d}f(x-y)g(y)dy.$$

Miranda
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1 Answers1

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So to make the final deduction, a brief sketch is as follows.

Notice that by an approximation argument and an application of Monotone convergence, one can show for any non-negative Borel measurable $h:\mathbb{R}^d\to \mathbb{R}_{\ge 0}$, $$\displaystyle \int_{\mathbb{R}^d} h(x)d\mu(x) = \displaystyle \int_{\mathbb{R}^d} h(x)f(x)dm(x)\,, $$ and similarly for $\nu$.

Now, by Fubini-Tonelli and the above, the function $$y\mapsto \displaystyle\int_{\mathbb{R}^d}\mathbf{1}(x+y\in A)d\mu(x) \,,$$ is measurable and equals $y\mapsto \displaystyle\int_{\mathbb{R}^d}\mathbf{1}(x+y\in A)f(x) dm(x)$.

Thus, applying the first step to this function, we obtain for all $A\subseteq \mathbb{R}^d$ Borel, $$\displaystyle\int_{\mathbb{R}^d}\displaystyle\int_{\mathbb{R}^d}\mathbf{1}(x+y\in A)d\mu(x)d\nu(y) = \displaystyle\int_{\mathbb{R}^d}\displaystyle\int_{\mathbb{R}^d}\mathbf{1}(z\in A)f(z-y)g(y)d m(z)d m(y) = \displaystyle\int_{A}\displaystyle\int_{\mathbb{R}^d}f(z-y)g(y)d m(z)d m(y)\,,$$ where the last equality follows from a change of variables $z\mapsto x+y$, (essentially follows from the translation invariance of the Lebesgue measure) which is a special case of a much more general result that is useful to be aware of, see this post. Now since this equality holds for arbitrary Borel sets $A$, you obtain the desired form for the density of the convolution measure.

Pantelis Tassopoulos
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