Let $E \subseteq \Bbb{R}^n$.
Prove that $f_r(x)=m^*(E \cap B(x,r))$ is continuous $\forall r>0$.
where $m^*$ denotes the Lebesgue outer measure.
My idea is to prove that $f_r$ is locally Lipschitz.
I could use also the fact that :
If $E$ is Lebesgue measurable then $$m^*(A)=m^*(A \cap E)+m^*(A \setminus E) ,\forall A \subseteq \Bbb{R^n}$$
Can someone help me with this?
Thank you in advance.
Best Answer
We find that for all $x\in \Bbb R^n$ and $F\subset \Bbb R^n$ $$ m^*F = m^*(F\cap B(x,r))+m^*(F\setminus B(x,r)) $$ since $B(x,r)$ is measurable. Let $F=E\cap B(y,r)$ to obtain $$ f(y) =m^*(E\cap B(y,r)\cap B(x,r))+m^*\left(\left(E\cap B(y,r)\right)\setminus B(x,r)\right). $$ By interchanging $x$ and $y$, we also obtain $$ f(x) =m^*(E\cap B(x,r)\cap B(y,r))+m^*\left(\left(E\cap B(x,r)\right)\setminus B(y,r)\right). $$ This yields $$ f(x)-f(y) =m^*\left(\left(E\cap B(x,r)\right)\setminus B(y,r)\right)-m^*\left(\left(E\cap B(y,r)\right)\setminus B(x,r)\right), $$ hence $$\begin{eqnarray} f(x)-f(y)&\le& m^*\left(\left(E\cap B(x,r)\right)\setminus B(y,r)\right)\\ &\le&m^*\left( B(x,r)\setminus B(y,r)\right)\\ &\le& m^*\left( B(x,r+|x-y|)\setminus B(y,r)\right)\\ &=&m^*B(x,r+|x-y|)-m^*B(y,r)\\&=&\omega_n \left[(r+|x-y|)^n-r^n\right] \end{eqnarray}$$ where $\omega_n=m^*B(0,1)$ is the volume of the unit ball. By interchanging $x$ and $y$ we also find that $$ f(x)-f(y)\ge -\omega_n \left[(r+|x-y|)^n-r^n\right]. $$Therefore $$ |f(x)-f(y)|\le \omega_n \left[(r+|x-y|)^n-r^n\right] \stackrel{|x-y|\to 0}\longrightarrow 0, $$ which proves (uniform) continuity of $f$ as desired.
Note: Since $$ |f(x)-f(y)|\le \omega_n |x-y|\left[(r+|x-y|)^{n-1} +(r+|x-y|)^{n-2}r +\cdots +r^{n-1}\right], $$ your conjecture that $f$ is locally Lipschitz is correct.