1
$\begingroup$

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 $\mu (U_{\epsilon} \backslash K_{\epsilon}) < \epsilon$
$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$
  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.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.