Showing that $\mathrm{support}(\mu\ast\nu)=\mathrm{support}(\mu)+\mathrm{support}(\nu)$ when $\mathrm{support}(\mu),\mathrm{support}(\nu)$ are compact

Question

Remark: By current question boils down to pretty simple measure theoretic machinery used when computing the convolution of two measures. You can safely jump to the end if you want. I am including more details here to also verify whether my reasoning prior to the troublesome part is sound. Thanks!

Let $\mu, \nu$ be two finite positive Borel measures on $\mathbb{R}^n$. Define the convolution $\mu\ast\nu$ between $\mu$ and $\nu$ as the measure which satisfies $\int_{\mathbb{R}^n}\varphi(x)d((\mu\ast\nu)(x)) = \int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\varphi(x + y)d\mu(x)d\nu(y)$ for all $\varphi\in C_0^+(\mathbb{R}^n)$, i.e. the set of continuous non-negative functions with a compact support on $\mathbb{R}^n$. While I have seen some definitions which directly define $\mu\ast\nu$ with indicator functions, a book I am reading requires continuity from the $\phi$s, so we are playing with that.

If $A, B\subset\mathbb{R}^n$, define the sumset $A + B$ as $A + B := \{a + b\mid a\in A, b\in B\}$.

I looking to prove that if the aforementioned $\mu$ and $\nu$ have compact supports, then $\mathrm{support}(\mu\ast\nu)=\mathrm{support}(\mu)+\mathrm{support}(\nu)$, when the support of a measure is defined as in Wikipedia

$$\mathrm{support}(\mu) := \{x\in\mathbb{R}^n\mid \forall N\in \tau(\mathbb{R}^n): x\in N\implies \mu(N) > 0\}$$

where $\tau(\mathbb{R}^n)$ is the standard topology of $\mathbb{R}^n$. In our case it suffices to consider balls with arbitrary center and radius.

As of now I have not proven either of the inclusions and I am confused in the direction $\mathrm{support}(\mu) + \mathrm{support}(\nu) \subset \mathrm{support}(\mu\ast\nu)$. To be specific, I am stuck on how to actually conclude that $\mu\ast\nu(N) > 0$ for a neighbourhood $N$ of a suitable point $x$ contined in $N$.

(Question:) That is, let $z\in\mathrm{support}(\mu) + \mathrm{support}(\nu) \subset \mathrm{support}(\mu\ast\nu) \Longleftrightarrow z = \tilde{x} + \tilde{y}, \tilde{x} \in \mathrm{support}(\mu), \tilde{y} \in \mathrm{support}(\nu)$ be any point and take $B(w, r)$ to be a ball with center $w$ and radius $r$ containing $z$. Let $\varphi_k$ be a continuous approximation of the characteristic function of $B(w,r)$. Then for any index $k$,

$$\int_{\mathbb{R}^n}\varphi_kd((\mu\ast\nu)(x))=\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\varphi_k(x + y)d\mu(x)d\nu(y)$$

From here you can perform a change of variables to get

$$\int_{\mathbb{R}^n}\varphi_kd((\mu\ast\nu)(x))=\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\varphi_k(x)d\mu(x - y)d\nu(y)$$

and it is all but clear to me how I should deal with the last integral when I am taking the measure of $\mu$ of a set which gets shifted over $\mathbb{R}^n$.

and use dominated convergence theorem to move a limit inside the integral to get

$$\int_{\mathbb{R}^n}\varphi_kd((\mu\ast\nu)(x)) = \int_{\mathbb{R}^n}\int_{\mathbb{R}^n}1_{B(w,r)}d\mu(x - y)d\nu(y)\rightarrow$$

$$\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}1_{B(w,r)}d\mu(x - y)d\mu(y)=\int_{\mathbb{R}^n}\mu(B(w, r) - y)d\nu(y)=\int_{\mathbb{R}^n}\mu(B(w - y, r))d\nu(y)$$

Answer 1

The solution to the OP's problem follows from the following general result:

For Borel measures $\mu$ and $\nu$ on $(\mathbb{R}^n,\mathscr{B}(\mathbb{R}^n))$, $$\overline{\operatorname{supp}(\mu)+\operatorname{supp}(\nu)}= \operatorname{supp}(\mu*\nu)$$

Here is a short proof:

We show first that \begin{align} \operatorname{supp}(\mu)+\operatorname{supp}(\nu)\subset \operatorname{supp}(\mu*\nu)\tag{1}\label{one} \end{align} Let $x_0\in\operatorname{supp}(\mu)$ and $y_0\in \operatorname{supp}(\nu)$. It is enough to show that $(\mu*\nu)\big(x_0+y_0+U)>0$ for any open neighborhood $U$ of $0$. Choose an open neighborhood $V$ of $0$ such that $V+V\subset U$. Then, $\mathbb{1}_{\{x_0+V\}}(x)\mathbb{1}_{\{y_0+V\}}(y)\leq\mathbb{1}_{\{x_0+y_0+U\}}(x+y)$. Integration with respect to $\mu\otimes\nu$ yields \eqref{one}, for $\mu(x_0+V)\nu(y_0+V)>0$.

We now show that \begin{align} \operatorname{supp}(\mu*\nu)\subset\overline{\operatorname{supp}(\mu)+\operatorname{supp}(\nu)}\tag{2}\label{two} \end{align} Suppose $z\in\operatorname{supp}(\mu*\nu)$. Let $X=\operatorname{supp}(\mu)$ and $Y=\operatorname{supp}(\nu)$. By Fubini-Tonelli's theorem, for any $\varepsilon>0$ \begin{align} 0<(\mu*\nu)\big(B(z;\varepsilon)\big)=\int_X\nu\big(Y\cap(B(z;\varepsilon)-x)\big)\,\mu(dx) \end{align} This means that for some $x\in X$, $\nu\Big(Y\cap(B(z;\varepsilon)-x)\big)>0$. This in turn implies that there exists $y\in Y\cap(B(z;\varepsilon)-x)$, that is $x+y\in B(z;\varepsilon)\cap(X+Y)$. Thus $z\in\overline{X+Y}$, and \eqref{two} follows.

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To complete the solution of the OP's question notice that is $X$ and $Y$ are compact subsets of $\mathbb{R}^n$, then $X+Y=\overline{X+Y}$.

The statement in the OP's posting also holds when either of the supports $X$ or $Y$ is compact. This last statement requires a little extra work.