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!
On
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 $m(E)=0$.
We can assume that $m(E)<\infty$ by first considering the sets $E\cap[n,n+1]$, $n\in\mathbb{Z}$, which would satisfy the same properties. Suppose on the contrary that $m(E) = \alpha>0$. Then there exists an open set $V$ (which we could write as a countable union of disjoint intervals $V = \cup{I_k}$) such that $E\subset V$ and $m(V) < 2\alpha$. Now $$ m(E) = m(E\cap V) = m(E\cap(\cup{I_k})) = m(\cup{(E\cap I_k)}) = \sum{m(E\cap I_k)}.$$ Derive a contradiction.