Let $\mu$ be a measure on the Borel $\sigma$-field of $\Bbb R$ such that $\mu (\Bbb R) = 1.$ Recall that the support of $\mu$ is the largest closed set $C$ such that for all open sets $U$ with $U \cap C \neq \varnothing$ we have $\mu (U) \gt 0.$ Assume that every continuous real-valued function is integrable with respect to $\mu.$ Prove that the support of $\mu$ is compact.
What I observe is that I need only to prove that $C$ is bounded because by Heine-Borel theorem then we are through.
How do I prove boundedness of $C\ $? Any help in this regard will be highly appreciated.
Thanks in advance.
I will flesh out the terse proof to which @user203940 pointed.
If the support of $\mu$ is not compact, then it is not bounded. This means that there is a countably infinite number of non-overlapping intervals that have positive $\mu$-measures.
Suppose that the support of $\mu$ is not compact. Let $\{(a_n,b_n):n\in\mathbb{N}\}$ be a collection of non-overlapping intervals that have positive $\mu$-measures. We assume without loss of generality that $0 < a_n < b_n < a_{n+1}$ for each $n\in\mathbb{N}$.
Let $f$ be a positive-valued and continuous real-valued function that is constant on each of these intervals. Let
Then \begin{equation} \int_{\mathbb{R}} f(x)d\mu(x) ~\geq~ \sum_{n\in\mathbb{N}}c_n\epsilon_n. \end{equation} We have not put too many restrictions on $f$. As $n$ grows, construct $f$ by interpolation between intervals so that $f(x) = \epsilon_n^{-1}$ for $x\in(a_n,b_n)$. Then $c_n = \epsilon_n^{-1}$.
We now have \begin{equation} \int_{\mathbb{R}} f(x)d\mu(x) ~\geq~ \sum_{n\in\mathbb{N}}c_n\epsilon_n = \infty. \end{equation}