Radon Measure Problem

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A measure $\mu$ on ($\mathbb{R}, \mathcal{B}_{\mathbb{R}})$ is called a Radon Measure if $\mu (K) < \infty $ for every compact subset $K$ of $\mathbb{R}$.

We need to prove the following:

  1. $\mu$ is a $\sigma$-finite measure
  2. For any $B \in \mathcal{B}_{\mathbb{R}}$, and $\epsilon >0$, there exists an open set $U_{\epsilon}$ and a closed set $K_{\epsilon}$ such that $K_{\epsilon} \subseteq B \subseteq U_{\epsilon}$ and $\mu (U_{\epsilon} \backslash K_{\epsilon}) < \epsilon$
2

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  1. Let $(K_n)_{n=1}^\infty$ be an increasing sequence of compact sets, converging to $\mathbb{R}$, set for instance: $K_n = [-n,n]$ then $\bigcup_{n=1}^\infty K_n = \mathbb{R}$. Since $K_n$ is compact and $\mu$ is Radon we have $\mu(K) <\infty$, which is the definition of a $\sigma$-finite measure.
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This is an outline of the proof of 2). [ 1) has already been answered by Jonas]. It is enough to show that the approximation property in 2) holds for Borel sets contained in $[-n,n]$, for each $n$. Consider the collection of all Borel sets $B$ contained in $[-n,n]$ which has the approximation property in 2). [This means that for every $\epsilon >0$ there must exist $K_\epsilon$ and $U_\epsilon$ satisfying the stated properties]. If $B$ is a closed set we can take $K_\epsilon =B$ and $U_\epsilon =\{x:d(x,B)<1/N\}$ with $N$ sufficiently large. [ $\{x:d(x,B)<1/N\}$ is open decreases to $B$]. Now verify that the class of sets we are considering is a sigma algebra. When you show that it contains countable unions you need the following observation: if we have closed sets $K_n$ with $K_n \subset A_n$ for each $n$ then $\cup K_n \subset \cup A_n$, but $\cup K_n$ is not necessarily closed. To over come this you have to observe that $\mu (\cup_1^{n} K_n) \to \mu (\cup_1^{\infty} K_n)$ so we can use the finite union in place of the infinite union. Once you fill in the details of this argument you get a sigma algebra which contains all closed subsets of $[-n,n]$ hence all Borel subsets of $[-n,n]$. If you need help with any particular step please let me know.