Measurable set and continuous function exercise Let $E \subset \mathbb R^n$ be a measurable set. Show that $f:\mathbb R_{\geq 0} \to \mathbb R$, given by $f(r)=|E \cap B(0,r)|$ is continuous.
So, given a fixed $r_0 \in \mathbb R_{\geq 0}$ and given $\epsilon$, I want to prove there is $\delta>0 : |r_0-x|<\delta \implies |f(r_0)-f(x)|<\epsilon$. We have $$|f(r_0)-f(x)|=||E \cap B(0,r_0)|_e-|E \cap B(0,x)|_e|$$
Since $|E \cap B(0,r)|_e \leq |B(0,r)|_e < \infty$ for any $r$, then $$||E \cap B(0,r_0)|_e-|E \cap B(0,x)|_e| \leq |E \cap B(0,r_0) \triangle E \cap B(0,x_0)|_e$$
Here I got stuck, I still didn't use the fact that $E$ is measurable. Any help is greatly appreciated.

Let $r_n\to r$ be increasing. Latter we need to do the case of decreasing.
Then $|f(r)-f(r_n)|\\=m(E\cap B_r)-m(E\cap B_{r_n})\\=m(E\cap\bigcup_{k=n}^{\infty}(B_{r_{k+1}}-B_{r_k}))\\=\sum_{k=n}^{\infty}m(E\cap(B_{r_{k+1}}-B_{r_k}))\to0$
The last limit is because $\sum_{k=-1}^{\infty}m(E\cap(B_{r_{k+1}}-B_{r_k}))$ is convergent, it is equal to $m(E\cap B_r)$ by $\sigma$-additivity, where we write $B_{-1}=\emptyset$.
Do, the analogous thing for decreasing $r_n\to r$. You will need to use that $m(E\cap B_{r_1})\leq m(B_{r_1})$ is finite.
