TITLE: Proving a set is Lebesgue Measurable
QUESTION [5 upvotes]: Measure is a serious weak point of mine, and I cannot figure out this problem:
Let $E \subset \mathbb{R}$ be Lebesgue measurable. Suppose that for all open intervals $I$, we have $m(E\cap I) \leq \frac{1}{2} m(I)$, where $m$ denotes Lebesgue measure. Prove that $m(E)=0$.
Any help is greatly appreciated!

REPLY [0 votes]: Since $E \subset \mathbb{R}$ is Lebesgue measurable, there exists pairwise disjoint interval $I_i$ such that
$$
E\subset \bigcup_{i=1}^{\infty} I_i\quad\text{ and }\quad m(E)=\inf{ \sum_{i=1}^{\infty} m(I_i)}\tag{1}
$$
Since
$$
\quad E=E\cap\bigcup_{i=1}^{\infty} I_i=\bigcup_{i=1}^{\infty} (E\cap I_i)
$$
By Lebesgue measure
$$
m(E)\leqslant\sum_{i=1}^{\infty} m(E\cap I_i)\leqslant \frac1{2}\sum_{i=1}^{\infty}m(I_i)
$$
By (1), we have
$$
m(E)\leqslant \frac1{2}m(E)
$$
Thus  $m(E)=0$.
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