A function defined in a separable metric space

Question

Let $X$ be a separable metric space and $\{U_n:n\in\mathbb{N}\}$ a basis for the topology on $X$. For each $A\subseteq X$ define $$\mu_n(A)=\left\{\begin{array}{cc} 1, & \text{if }U_n\cap A\neq\emptyset \\ 0, & \text{if }U_n\cap A=\emptyset \end{array}\right.$$ $$\mu(A)=\sum_{n=1}^{\infty}2^{-n}\mu_n(A)$$

I have to show that $\mu(A)=\mu(\overline{A})$, where $\overline{A}$ denotes the closure of $A$. I managed to show that if $A\subseteq B$ then $\mu(A)\leq\mu(B)$, so we have that $\mu(A)\leq\mu(\overline{A})$. I'm having trouble proving that $\mu(\overline{A})\leq\mu(A)$. I'll appreciate any help you can give me.

Answer 1

Let $x \in u_n \cap \overline A$. Since $u_n$ is open, there is a $\delta > 0$ for which $B_\delta(x) \subset u_n$. Now take a sequence $\{ x_k \} \subset A$ for which $x_k \to x$.

There is a $N \in \mathbb{N}$ for which $x_m \in B_{\delta}(x)$ for all $m > N$, hence $x_m \in u_n$ for all $m > N$. Therefore $u_n \cap A$ is nonempty.

We have just demonstrated that $u_n \cap \overline A \neq \emptyset$ implies that $u_n \cap A \neq \emptyset$.

You should be able to wrap up the rest on your own.

Answer 2

Fix $n\in\mathbb N$, and suppose $x_0\in\overline A\cap U_n$. Since $U_n$ is open, there is some $x\in A\cap U_n$. Thus $\mu_n(A)=\mu_n(\overline A)$, and since $n$ was arbitrary, it follows that $\mu(A)=\mu(\overline A)$.