# Riesz-Markov Theorem Proof question

In the proof of Riesz-Markov Theorem the inner regularity of the Radon measure was established for $\mu(\mathcal{O})< \infty$. The Theorem and the prove of this part are listed below. What about the case when $\mu(\mathcal{O}) = \infty$ How do we prove the inner regularity of the Radom measure in this case?

The Riesz-Markov Theorem: Let $X$ be a locally compact Hausdorff space and $I$ a positive linear functional on $C_c (X )$ (continuous on compact support). Then there is a unique Radon measure $\hat{\mu}$ on $\mathcal{B}(X)$, the Borel $\sigma$-algebra associated with the topology on $X$, for which $$I(f)= \int_X f d\hat{\mu} \text{ for all } f \in C_c(f)$$

Inner Regularity of Radom measure for $\mu(\mathcal{O})< \infty$:

Let $\mathcal{O}$ be open. Suppose $\mu(\mathcal{O})< \infty$. Let $\epsilon > 0$. We have to establish the existence of an open set $U$ that has compact closure contained in $\mathcal{O}$ and $\mu(U) > \mu(\mathcal{O}) - \epsilon$. Indeed, by the definition of $\mu$, there is a function $f_{\epsilon} \in C_c (X)$ that has support contained in $\mathcal{O}$ and for which $I(f_{\epsilon}) > \mu(\mathcal{O}) - \epsilon$. Let $K = supp f$. But $X$ is locally compact and Hausdorff and therefore has the locally compact separation property. Choose $U$ to be a neighborhood of $K$ that has compact closure contained in $\mathcal{O}$. Then $$\mu(U) \geq I(f_{\epsilon}) > \mu(O) - \epsilon$$ It remains only to show that if $\mathcal{O}$ is an open set of compact closure, then $\mu(\overline{O}) < \infty$. But $X$ is locally compact and Hausdorff and therefore has the locally compact extension property. Choose a function in $C_c(X)$ that takes the constant value 1 on $\overline{\mathcal{O}}$. Thus, since $I$ is positive, $\mu(\mathcal{O}) < I(f) < \infty$.

I imagine you're defining $\mu$ on open sets by $$\mu(U) = \sup \{I(f):\; f \in C_c(X),\; 0 \le f \le 1,\; \text{supp}(f) \subset U\}$$ BTW you left out the $0 \le f \le 1$ in your proof of the finite case.
In the case $\mu(\mathcal O) = \infty$, inner regularity would say that for any positive integer $N$ there is an open set $U$ with compact closure $\overline{U} \subseteq \mathcal O$ and $\mu(U) > N$. Take $f_N \in C_c(X)$ with support contained in $\mathcal O$, $0 \le f \le 1$ and $I(f_N) > N$, and continue as in the finite case.
• correct, that is how $\mu$ on open set was defined – Lucas Apr 7 '16 at 19:50