# Intersection of all compact sets of measure 1 in a measure space

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!

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.

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 $\mu$!!