showing that $f(x)=\mu(B(x,r))$ is measurable

Question

$\mu$ is a probability measure on $(X, B(X))$, and $r>0$. define $f(x)= \mu(B_r(x))$. I need to show that $f$ is measurable. I tried to show semi-continuity, but it failed. any other idea?

Answer 1

Lower semi-continuity is the way to go. Let me sketch you a proof:

let $x_k\rightarrow x$. Then for all $k$ big enough you have for a choosen $\varepsilon>0$ $$B_{r-\varepsilon}(x)\subseteq B_r(x_k).$$ Then monotonicity gives you $$\mu(B_{r-\varepsilon}(x))\leq \liminf_{k\rightarrow\infty}\mu(B_r(x_k)).$$ Now let $\varepsilon\rightarrow 0$ and use the continuity of a measure to obtain $$\lim_{\varepsilon\rightarrow 0}\mu(B_{r-\varepsilon}(x))=\mu(\bigcup_{\varepsilon>0}(B_{r-\varepsilon}(x))=\mu(B_r(x)).$$