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.