Let $\mu$ be a Borel measure on $\mathbb{R}^n$ s.t. $\mu(K)< \infty$ for all compact $K$. Show that $\mu$ is the restriction of some Riesz measure on $\mathcal{B}$.

I try to prove it using the lemma: Let $\lambda$ be a Riesz measure,$\mu$ be a Borel measure on $\mathbb{R}^n$ s.t. $\lambda(G)=\mu(G)$ for all open $G$. Then $\lambda=\mu$ on $\mathcal{B}$.

Hence by Riesz representation theorem, let $\lambda$ be a Riesz measure associated with functional $h(f)=\int f d\mu$, then let $G$ be open set, we have $$\lambda(G)=\sup\{h(f)|f<G\}.$$ Then I can show $\lambda(G)\leq \mu(G)$. But I don't know how to show $\mu(G)\leq \lambda(G)$. I try to use Urysohn lemma to construct $K<f<G$, which shows for all compact $K\subset G$, $\mu(K)\leq \lambda(G)$. But I am not sure whether $G$ is inner regular with respect to $\mu$.


The problem is related with the following fact:

Given an open set $U \subset \mathbb R ^n $, there exists an exhaustion by compact sets, i.e. a sequence of compact sets $K_i$, s.t.

$\cup _{i=0}^{\infty} K_i = U$ and $\forall i \in \mathbb N : K_i \subset K_{i+1} $

Hence $\lim_{i\to\infty}\mu(K_i)=\mu(U)$. We get the inner regularity of any Borel measure for open set in $\mathbb{R}^n$.


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