1
$\begingroup$

Let $E$ be a locally compact polish space with Borel-$\sigma$-algebra denote by $\Sigma$. A measure $\mu$ is said to be inner regular, if for every set $A \inΣ$, $\mu(A) = \sup \{ \mu(K) | \text{ compact }K \subseteq A \}$. This means for every $\varepsilon>0, A \in \Sigma$ $\exists K$ compact s.t. $\mu(A\setminus K)< \varepsilon $.

Can I show that if $\mu(A)<\infty$ that $\mu(A)=\mu(K)$ for some compact K without demanding $K \subset A$? The Problem is that the limit of compact sets need not be compact. However this statement is very often assumed (for example it is used here in the 1st step of the answer:https://mathoverflow.net/questions/93088/atoms-of-regular-borel-measure )and I would also like to use it if it is true.

$\endgroup$
  • 1
    $\begingroup$ No; let $E$ be the open interval $(0,1)$ with Lebesgue measure. Then $\mu(E)=1$, but no compact subset has measure $1$. $\endgroup$ – Mike Earnest Jul 3 '18 at 0:06
  • $\begingroup$ @Yoh, you should use "\setminus" $\endgroup$ – mathworker21 Jul 3 '18 at 0:12
  • $\begingroup$ @MikeEarnest I also thought of the Lebesgue measure. However I am not demanding, that $K \subset A$, here $\mu([0,1])=1$. $\endgroup$ – user564337 Jul 3 '18 at 0:21
  • $\begingroup$ Here, you cannot use $K=[0,1]$, since the entire space is $(0,1)$. Another example is to let $E=\mathbb R$, with any absolutely continuous probability measure whose support is $\mathbb R$. Letting $A=\mathbb R$, $\mu(A)=1$, no $K$ has measure $1$. $\endgroup$ – Mike Earnest Jul 3 '18 at 0:24
  • $\begingroup$ Sorry, of course you are right. $\endgroup$ – user564337 Jul 3 '18 at 0:27
0
$\begingroup$

Here is a more interesting example where the counterexample is not the whole space. Let $E=\mathbb R$. Let $q_1,q_2,\dots,$ enumerate the rationals, and equip $E$ with the measure $$ \mu(A)=\sum_{i=1}^\infty 2^{-i}\cdot {\bf 1}(q_i\in A) $$ That is, letting $\delta(x)$ be the unit point mass at $x$, then $\mu=\sum_{i\ge 1} 2^{-i}\delta(q_i)$. You can show this is inner regular (in fact, sets $A$ can be approximated by finite subsets of rationals of the form $\{q_1,q_2,\dots,q_n\}\cap A$). Then, let $A=(0,1)$. You can show that there are no compact $K$ for which $\mu(K)=\mu(A)$.


The reason that the argument in https://mathoverflow.net/questions/93088/atoms-of-regular-borel-measure worked was because their set $A$ was an atom, meaning for every $B\subset A$, either $\mu(B)=\mu(A)$ or $\mu(B)=0$. When you let $\epsilon=\mu(A)/2$ in the definition of inner regularity, the resulting $K$ must fall in the former category.

$\endgroup$

Your Answer

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