# A Borel measure defines semi-continuous function?

Let $X$ be a metric space with outer measure $\mu$, which is assumed to be a Borel measure, i.e., all Borel sets are measurable. For a fixed subset $A\subset X$ (not necessarily measurable, but you can assume to be measurable first) and a fixed $\rho >0$, we define $$\alpha(x)=\mu (A\cap \bar{B}(x,\rho)), x\in X,$$ $$\beta(x)=\mu (A\cap B(x,\rho)), x\in X,$$ where $B(x,\rho)$ is the open ball with center $x$ and radius $\rho$ and $\bar{B}(x,\rho)$ is the closed one.

Then what can we say about the function $\alpha$ and $\beta$? I'm concerning the semi-continuous property, and there is a conclusion says that $\alpha$ is upper semi-continuous and $\beta$ is lower semi-continuous, but I cannot prove either. Could any one give a proof or give some hint? Thanks in advance!

## 1 Answer

Let $d$ be the metric.

Hint 1: If $d(x,y)<\epsilon$, then $$\overline{B}(y,\rho-\epsilon) \subseteq \overline{B}(x,\rho) \subseteq \overline{B}(y,\rho+\epsilon)$$

Hint 2: $$\overline{B}(x,\rho) = \bigcap_{n \in \mathbb N} \overline{B}(x,\rho+1/n)$$

I will let you do open balls.