Let $\mathcal{B}^n$ Borel sets of $\mathbb{R}^n$ and $\mu:\mathcal{B}^n\to[0,\infty]$ a measure. Define the support of $\mu$ as: \begin{equation} \text{supp}(\mu)=\{c\in\mathbb{R}^n:\forall U\subseteq\mathbb{R^n}\, \text{open set such that}\, c\in U \, \text{we have}\, \mu(U)>0\} \end{equation}
Prove that $\text{supp}(\mu)$ is closed.
Prove that $c\in\text{supp}(\mu)\iff$ forall $\phi\in C^0_c(\mathbb{R}^n)$ such thet $\phi(x)\ge 0$ forall $x\in\mathbb{R}^n$ and $\phi(c)>0$ we have: \begin{equation} \int_{\mathbb{R}^n}\phi(x)\,\mathrm d\mu(x)>0 \end{equation}
SOLUTION
Let $\{c_n\}$ a sequence in $\text{supp}(\mu)$ such that $c_n\to c$, then let $c\in U\subseteq\mathbb{R}^n$ an open set. By definition there exists $\epsilon>0$ such that $B_\epsilon(c)=\{x\in\mathbb{R}^n:d(x,c)<\epsilon\}\subseteq U$. Now by definition of a convergent sequence for any $\delta>0$ there exists $\bar{n}\ge 1$ such that for any $n\ge\bar{n}$ we have that $c_n\in B_\delta(c)$. The conclusion follows from the arbitrariety of $\delta$ because taking $0<\delta<\epsilon$ we have that $B_\delta(c)\subset B_\epsilon(c)\subseteq U$ and then for the monotonicity of $\mu$: \begin{equation} \mu(U)\ge\mu(B_\delta(c))>0 \end{equation} because $B_\delta(c)$ contains all $c_n$ for $n\ge\bar{n}$ and $c_n\in\text{supp}(\mu)$ forall $n$
($\Longrightarrow$) we have for Hypothesis that $\mu(U)>0$ forall $U$ open set with $c\in U$. Then since $\phi$ is continuous with compact support we have that $\phi$ is non-zero only on a compact set $K$ and the fact that $\phi(c)>0$ implies that there exists $\epsilon>0$ such that $\phi(z)>0$ forall $z\in B_\epsilon(c)$, now: \begin{equation} \int_{\mathbb{R}^n}\phi(x)\,\mathrm d\mu(x)=\int_K\phi(x)\,\mathrm d\mu(x)\ge\int_{B_\epsilon(c)}\phi(x)\,\mathrm d\mu(x)>0 \end{equation} Where the last follows from the fact that $\phi(x)>0$ on $B_\epsilon(c)$ and $\mu(B_\epsilon(c))>0$.
($\Longleftarrow$) suppose by contradiction that $c$ does not belong to $\text{supp}(\mu)$, then there exist $B_\delta(c)$ open bowl such that but $\mu(B_\delta(c))=0$. We consider a function $\phi$ such that is zero outside the compact set $\overline{B_\delta(c)}$ and is zero also on $\partial B_\delta(c)$ \begin{equation} \phi(x)= \begin{cases} 1-\dfrac{1}{e^{|x-c|^2-\delta}} \, x\in B_\delta(c)\\ 0 \, \text{otherwise} \end{cases} \end{equation} Then $\phi$ satisfies the Hypothesis and: \begin{equation} \int_{\mathbb{R}^n}\phi\,\mathrm d\mu(x)=\int_{\partial B_\delta(c)}\phi(x)\,\mathrm d\mu(x)+\int_{B_\delta(c)}\phi(x)\,\mathrm d\mu(x)=0 \end{equation} Because the second one is over a set of measure $0$ and in the first one the function is null on that set. Hence we get a contradiction.