This question is elementary and hence might be a duplicate. From Rudin, Real and Complex Analysis, page 57.
Let $\mu$ be a regular Borel measure on a compact Hausdorff space $X$: assume $\mu(X)=1$. Prove that there is a compact set $K\subset X$ (the carrier or support of $\mu$) such that $\mu(K)=1$ but $\mu(H)<1$ for every proper compact subset $H$ of $K$.
I could not follow Rudin's hint that If $K$ is the intersection of all compact set $K_{\alpha}$ with $K_{\alpha}=1$, then every open set in $V$ contains $K$ also contains some $K_{\alpha}$. I do not know how to prove this. In the simple scenario where $X$ has only finite (or countably) many such $K_{a}$'s, $\mu(K)=1$ is apparent by De Morgan's law. However I do not know what will be wrong if there is an open set $O$ containing $K$ but not any of the $K_{i}$, since $O-K_{i}$ could have zero measure.
I also do not know how to prove in the uncountable case that $\mu(K)=1$ where De Morgan's law gives uncountable union and $\bigcup_{\alpha \in A} \mu(O_{\alpha})$ except following Rudin's suggestion. So I am stuck. Also, even if I managed to show this property, what should I do with regularity to prove $\mu(H)<1$ for every proper compact set $H$ of $K$? Again $H$ can be differed by $K$ with a compact set of measure 0.
I tried to reduce the proof by checking the case of $K,K_{1},K_{2}$, where $\mu(K)=\mu(K_{1})=\mu(K_{2})=\mu(K_{1}\bigcup K_{2})=1$. But it is not clear to me why $K_{1}/K$ cannot contain an open set of 0 measure. So I feel I am stuck on something really elementary.
Best Answer
Let $\cal K$ the collection of compact sets of measure $1$; it's not empty as $X\in\cal K$. Let $K:=\bigcap_{K\in\cal K}K$: it's a compact set. Let $O$ an open set containing $K$. Assume that $C\cap O^c\neq\emptyset$ for all $C\in\cal K$. Then $\{C\cap O^c,C\in\cal K\}$ is a collection of compact subsets of $X$, with empty intersection. So $\{C^c\cup O,C\in\cal K\}$ is an open cover of the compact $X$, and we can extract a finite subcover $\{C_j^c\cup O,C_j\in\cal K, 1\leqslant j\leqslant n\}$. So $X=\bigcup_{j=1}^n(C_j^c\cup O)$ and $X=\left(\bigcap_{j=1}^nC_j\right)^c\cup O$. This gives $\bigcap_{j=1}^nC_j\subset O$. It's enough to conclude that $\mu(O)=1$, and by outer regularity that $\mu(K)=1$.
Indeed, $\mu(O)\leqslant\mu(X)\leqslant 1$, and we can show by induction that if $S_1,\dots,S_n$ are sets of measure $1$, so is their intersection (it's actually enough to do it for $n=2$).
If $H$ is a strict compact subset of $K$ and has measure $1$, then $H\in\cal K$, so $H$ would be strictly contained in $H$, a contradiction.