Let $\mu$ be a probability measure on $(\mathbb R,\mathcal B)$ where $\mathcal B$ is the Borel $\sigma$-algebra on $\mathbb R$. Show that there is a unique closed set $C\subset \mathbb R$ such that $\mu(C)=1$ and for any closed set $D\subset C$, we would have $\mu(D)<1$.
My attempt is as follows.
Consider $C:=\cap\{A\subset\mathbb R|A\text{ closed},\mu(A)=1\}$. The intersection is non-empty since $\mathbb R$ is closed and $\mu(\mathbb R)=1$. Then $C$ being an arbitrary intersection of closed sets, is closed. Suppose $D\subset C$ is any closed set then if $\mu(D)=1$ then $C\subset D$ and thus $C=D$.
How does one show that $\mu(C)=1$?
$C^{c} =\cup \{A\subset \mathbb R|A \text {is open}, \mu(A)=0\}$. Since the real line is second countable, any union of open sets can be expressed as a countable union of sets from that family. Hence $C^{c}$ is a countable union of open sets of measure 0. So $\mu (C^{c})=0$ and $\mu (C)=1$. Proof of the result from topology I have used: let $X$ be a second countable space and $A$ be the union of a collection of open sets $\{U_i\}$. Let $\{V_n\}$ be a countable base for the topology. If $x \in A$ the there exists $i$ such that $x \in U_i$. Hence there exists $n$ such that $x \in V_n \subset U_i$. For each $n$ that arises this way there is a $U_i$, call it $U_{i_n}$ that contains $x$. The union of the sets $U_{i_n}$ equals $A$, so $A$ is the union of a countable sub-family of $\{U_i\}$.