Property of a positive Lebesgue measure set in $\mathbb{R}^n$

Question

Let $E\subset\mathbb{R}^n$ be a positive Lebesgue measure set. Can we always find a $x\in E$ such that for any $r>0$, $B(x,r)\cap E$ is positive Lebesgue measure in $\mathbb{R}^n$?

($B(x,r)$ denotes the ball of radius $r$ with centre at $x$ in $\mathbb{R}^n$)

Answer 1

For ae. $x \in E$ we have $\lim_n {1 \over m B(x,{1 \over n}) } \int_{B(x,{1 \over n})} 1_E(t)dt = 1_E(x)$.

Pick any $x$ for which we have equality, then ${m (E \cap B(x,{1 \over n})) \over m B(x,{1 \over n}) } \to 1$.

In particular, for some $N$, if $n \ge N$ then $m (E \cap B(x,{1 \over n})) \ge {1 \over 2}{m B(x,{1 \over n}) } >0$.

Since $r \mapsto mB(x,r)$ is increasing we see that $m (E \cap B(x,r)) >0$ for all $r>0$.

Alternative:

We can prove this directly. Since the Lebesgue measure is inner regular, we can assume that $E$ is compact.

Now suppose that for all $x \in E$ there is some $r>0$ such that $m(E \cap B(x,r)) =0 $.

The $B(x,r)$ form an open cover of $E$ and hence there is a finite subcover $B(x_k,r_k)$. Then $mE \le \sum_k m (E \cap B(x,r)) = 0$.

Hence there is some $x$ such that for all $r >0$ we have $m(E \cap B(x,r)) >0 $.