# Let $E\subset\mathbb{R}^{n}$. Prove that if $m(E)>0$, then there exists $x\in E$ such that, for all $\delta >0$, $m\big(E\bigcap O(x,\delta)\big)>0$.

Let $E\subset\mathbb{R}^{n}$. Prove that if $m(E)>0$, then there exists $x\in E$ such that, for all $\delta >0$, $m\big(E\bigcap O(x,\delta)\big)>0$.

PS: Please give a proof using Cantor's intersection theorem.

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I think this is Lebesgue's density theorem: en.wikipedia.org/wiki/Lebesgue%27s_density_theorem –  Paul Mar 10 '12 at 13:01
@Paul: This is much weaker than Lebesgue's density theorem, which says for almost all $x\in E$ instead of there exists $x\in E$, and for those almost all $x\in E$, gives for all $c<1$ there exists $\delta>0$ such that $m(E\cap O(x,\delta))>cm(O,\delta)$. –  Jonas Meyer Mar 10 '12 at 17:55
@Jonas: Oh yes, you are right. Thanks for your explanation. –  Paul Mar 11 '12 at 10:58

We can assume that $m(E)\in\mathbb R_{>0}$. By regularity of Lebesgue measure, we can find a compact $K\subset E$ such that $m(K)>m(E)/2$. Assume that for all $x\in K$ we can find a $\delta_x>0$ such that $m(O(x,\delta_x))=0$. Then the family $\{O(x,\delta_x)\}$ is an open cover of $K$, and we can find an integer $N$ and $x_1,\ldots,x_N\in K$ such that $K\subset\bigcup_{j=1}^NO(x_j,\delta_{x_j})$. So $K$ has measure $0$, which is a contradiction.

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Proof. $0<mE=m\left(\bigcup_{k=1}^{\infty}\left(E\bigcap[-k,k]^n\right)\right) \leqslant\sum_{k=1}^{\infty}m\left(E\bigcap[-k,k]^n\right)$, so there exists a $k_0$ such that $m\left(E\bigcap[-k_0,k_0]^n\right)>0$.
Let $E_0=E\bigcap[-k_0,k_0]^n$, $E_1=[-k_0,k_0]^n-E_0$. By definition of Lebesgue outer measure, for all $\varepsilon>0$ we have a sequence of open intervals, $\{I_k\}$, covering $E_1$, satisfying $\sum_{k=1}^{\infty}mI_k<mE_1+\varepsilon$. Then $G=\bigcup_k I_k\supset E_1$ is an open set and $mG<mE_1+\varepsilon$.
Now we got a closed set $F=[-k_0,k_0]^n-G\subset E_0$ and $E_0-F=E_0-G^C=E_0\bigcap G$ $\subset$ $G-E_1$. Therefore $mE_0-mF=m(E_0-F)\leqslant m(G-E_1)=mG-mE_1<\varepsilon$. Let $\varepsilon$ be small enough so that $mF>0$. Cut $F$ into two closed sets $F\bigcap (-\infty,0]$ and $F\bigcap [0,\infty)$, one of which must have a positive Lebesgue measure by (sub)additivity, then cut it in the same way. Keeping doing this, We got a decreasing nested sequence of non-empty, bounded and closed subsets of $F$, all with positive Lebesgue measures. By Cantor's intersection theorem, this sequence $\{F_k\}$ has non-empty intersection $X$. Take an $x\in X$, we assert for all $\delta>0$, $m(F\bigcap O(x,\delta))>0$. Otherwise, there exists $\delta>0$ s.t. $m(F\bigcap O(x,\delta))=0$, implying $mF_k=0$ for $k$'s big enough to make $F_k\subset O(x,\delta)$, which is a contradiction.