$x\mapsto \mu(A\cap \bar{B}(x, r))$ is $\mu$-measurable function

Question

I have to show that the map $x\mapsto \mu(A\cap \bar{B}(x, r))$ is $\mu$-measurable for every $A\subset \ X$ and $r>0$, where $\mu$ is Radon measure. I managed to do this when $A$ is $\mu$-measurable set. How to generalize this for possibly non-measurable sets?

Answer 1

Let us assume that $\mu(A)<\infty$. Since $\mu$ is a Radon measure, you can find an $S\subset X$ Borel measurable such that for all Borel measurable sets $B\subset X$ we have $$\mu(A\cap B)=\mu(S\cap B).$$ Then you can use your proof, by using $\tilde{A}:=A\cap B(0,R)$ instead of $A$ for $R$ big enough.

Let me show how to construct such an $S$. Since $\mu$ is Borel regular, you find an $S\subset X$ Borel measurable such that $\mu(S)=\mu(A)$ and $A\subset S$. Now let $B\subset X$ be an arbitrary Borel Set. Since $A\cap B\subset S\cap B$ and $B$ is measurable (and $\mu(S)<\infty$) we have $$\mu(A\cap B)\leq\mu(S\cap B) = \mu(S)-\mu(S\setminus B).$$ Since $A\setminus B\subset S\setminus B$ we further conclude $$\mu(A\cap B)\leq\mu(S\cap B) = \mu(S)-\mu(S\setminus B)\leq \mu(A) - \mu(A\setminus B) = \mu(A\cap B),$$ because $B$ is measurable.