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 $s \in \Bbb R_{\ge 0}$. For all $r > s$, 
$$0 \le f(r) - f(s) = |E\cap [B(0,r)\setminus B(0,s)]| \le |B(0,r) \setminus B(0,s)| = C_n(r^n - s^n),$$
where $C_n$ is a constant depending only on $n$. Since $r^n - s^n \to 0$ as $r \to s$, it follows that $f(r) \to f(s)$ as $r\to s^{+}$. A similar argument shows that $f(r) \to f(s)$ as $r\to s^{-}$. Hence $f$ is continuous at $s$. Since $s$ was arbitrary, $f$ is continuous.
