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The problem:

Let $X = [0,1]$ and $\mathcal{B}$ be the Borel subsets of $X$. Let $\mu:\mathcal{B}\to[0,1]$ be a probability measure on $X$. Suppose that $\mu$ is regular, i.e., for all $B\in\mathcal{B}$ we have $$\mu(B) = \inf\{\mu(O) | B\subset O, O\ \operatorname{is\ open}\}$$ and $$\mu(B) = \sup\{\mu(K) | K\subset B, K\ \operatorname{is\ compact}\}.$$

Let $K = \cap_\alpha K_\alpha$ where $\{K_\alpha\}_\alpha$ is the collection of all compact subsets of $X$ with $\mu(K_\alpha) = 1$. Show that for every open set $O\subset X$ with $K\subset O$ there is $\alpha$ such that $K\subset K_\alpha\subset O$.

Now we know that each $K_\alpha$ satisfies $K\subset K_\alpha$, by definition, so all we need to show is that for any open set $O\subset X$ with $K\subset O$, there exists some $\alpha$ such that $K_\alpha \subset O$.

It is fairly easy to show that there is some compact set $J$ satisfying $$ K\subset J\subset O,$$ But what I'm not seeing is how we may conclude $\mu(J) = 1$.

I should add, part (ii) of this problem asks us to show $\mu(K) = 1$, which is fairly easy from the part above, but it means for this part I cannot use it combined with monotonicity of the measure to show $\mu(J) = 1$, which is about the only idea I've come up with.

Anyone have a hint at why $J$ must have measure 1, simply because it contains $K$, before knowing $K$ has measure 1?

Thanks!

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  • $\begingroup$ Just for your information: The hypotheses of this exercise are redundant. A Borel probability measure on $[0,1]$ is automatically inner and outer regular in the sense described. $\endgroup$
    – t.b.
    Oct 23, 2011 at 1:28
  • $\begingroup$ The title is wrong... you are intersecting only the compact sets! $\endgroup$
    – André Caldas
    Oct 23, 2011 at 1:31
  • $\begingroup$ There is such a $J$!!! Just take $J = K$!! $\endgroup$
    – André Caldas
    Oct 23, 2011 at 1:55
  • $\begingroup$ @t.b.: Fair enough, though we haven't discussed a measure being regular in class, so I'm guessing he just put it there for our benefit. $\endgroup$
    – Alex
    Oct 23, 2011 at 2:03
  • $\begingroup$ @AndréCaldas: True, I'll fix the title. Also, while $K$ would work, we aren't expected to prove $\mu(K) = 1$ until the second part of the problem, so I'm guessing the intention was for us to come up with an argument that does not involve that :) $\endgroup$
    – Alex
    Oct 23, 2011 at 2:06

2 Answers 2

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If $O$ is an open set containing $K$, then the collection of compact sets $\{O^c\cap K_{\alpha}\}_{\alpha}$ has empty intersection (here $^c$ denotes complement). Now use the finite intersection property of compact sets.

Since you put the homework tag I tried to be a little vague, but let me know if anything is unclear and I'll give detail.

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  • $\begingroup$ Nice! So, there is no need to use measure theoretic arguments!! :-) $\endgroup$
    – André Caldas
    Oct 23, 2011 at 1:27
  • $\begingroup$ @André Thanks! Well, I guess in order to conclude you do need to know that the union of finitely many sets of measure zero has measure zero. :) $\endgroup$
    – Nick Strehlke
    Oct 23, 2011 at 1:33
  • $\begingroup$ You have a set $K$ which is the intersection of compact sets $K_\alpha$. If $K \subset O$, for an open $O$, then there is a $\alpha$ such that $K \subset K_\alpha \subset O$. No need to use measure... as you demonstrated! But if you want to show that $\mu(K) = 1$, then you need the fact that the union of countable many sets of measure zero has measure zero. $\endgroup$
    – André Caldas
    Oct 23, 2011 at 1:53
  • $\begingroup$ Actually, the "part (i)" is topology, while "part (ii)" is measure theory. $\endgroup$
    – André Caldas
    Oct 23, 2011 at 1:55
  • $\begingroup$ @André Hmmm, well here's what I had in mind: The finite intersection argument shows that there exist finitely many indices $\alpha_1,\dots,\alpha_n$ such that $K_{\alpha_1}\cap\cdots\cap K_{\alpha_n}\cap O^c=\varnothing$. Using the fact the $\mu(K_{\alpha_1}^c\cup\cdots\cup K_{\alpha_n}^c)=0$, we get $K_{\alpha_1}\cap\cdots\cap K_{\alpha_n}=K_{\alpha}$ for some $\alpha$. This gives $K\subset K_{\alpha}\subset O$. To show that $\mu(K)=1$, I would then use the regularity of $\mu$ and the arbitrariness of $O$. (So much for vagueness!:)) $\endgroup$
    – Nick Strehlke
    Oct 23, 2011 at 2:23
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Let $\mathcal{U}$ be a countable base for the topology of $[0,1]$. Then, for each $K_\alpha$, there is a $\mathcal{U}_\alpha \subset \mathcal{U}$ such that $K_\alpha^c = \bigcup_{U \in \mathcal{U}_\alpha} U$. It is clear that every set in $\mathcal{U}_\alpha$ has measure $0$.

Now, let $\mathcal{U}_K = \bigcup_\alpha \mathcal{U}_\alpha \subset \mathcal{U}$. Then, $$ K^c = \bigcup_{U \in \mathcal{U}_K} U $$ is a countable union of sets with measure $0$. Therefore, $\mu(K) = 1$. In particular, $K = K_\alpha$ for some $\alpha$.

Notice that this $K$ is exactly the support for $\mu$!!

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