On a compact support of a measure Let $\mu$ be a measure on the Borel-$\sigma$-field of $\mathbb{R}$ such that $\mu (\mathbb{R})=1$. Recall that the support of $\mu$ is the largest closed set $C$ such that  for all open sets $U$ with $U\cap C \neq \phi$, we have $\mu(U)>0$. Assume that every continuous real-valued function on $\mathbb{R}$  is integrable with respect to $\mu$. Prove that the support of $\mu$ is compact.
I think all we need to prove is that such a $C$ is bounded. Then by Heine-Borel we can conclude $C$ is compact. But merely the integrability of a continuous function w.r.t. $\mu$ is not going to help me to show that $C$ is bounded. So, what will be the best possible approach for that problem?
 A: Let $\left(a_n\right)_{n\in\mathbb Z}$ be a sequence of positive real numbers. Let $f_n$ be a continuous function supported in $(n-1,n+2)$ such that $0\leqslant f_n(x)\leqslant a_n$ for all $x$ and $f_n(x)=a_n$ for $x\in (n,n+1]$. Define $f\colon x\mapsto \sum_{n\in\mathbb Z}f_n(x)$ (the definition makes sense, since for all $x$, the numbers of terms in $\sum_{n\in\mathbb Z}f_n(x)$ which do not vanish is at most $3$). Since the convergence is uniform on compact sets, the function $f$ is continuous hence 
$$
+\infty\gt \int_{\mathbb R}f(x)\mathrm d\mu(x)=\sum_{n\in\mathbb Z}\int_{\mathbb R}f_n(x)\mathrm d\mu(x)\geqslant \sum_{n\in\mathbb Z}\int_{(n,n+1]}f_n(x)\mathrm d\mu(x).$$
This shows that for any sequence $\left(a_n\right)_{n\in\mathbb Z}$ of positive real numbers, the series $\sum_{n\in\mathbb Z}a_n\mu\left((n,n+1]\right)$ converges. A good choice allows to conclude.
A: If $\operatorname{supp}(\mu)$ wasn’t bounded, we would be able to find a sequence $(x_n)$ with $\lim_n x_n =\infty$ and a sequence of disjoint neighborhoods $(N_n)$ of $x_n$ such that for all $n \in \mathbb N$, $\mu(N_n)>0$.
We can find for each $n$ an interval $I_n$ of length $\delta_n$ centered on $x_n$ and included in $N_n$. From there, for each $n$, we can build a continuous function with support included in $I_n$ and of integral greater than $1$. The sum $f$ of those functions converges as all their supports are disjoint. And the integral of $f$ is unbounded in contradiction with our hypothesis.
