$E$ measurable set and $m(E\cap I)\le \frac{1}{2}m(I)$ for any open interval, prove $m(E) =0$

Ran across this problem and need some help.

Let $E$ be a measurable subset of the real numbers and suppose that for any open interval $I$ one has $m(E\cap I)\le \frac{1}{2}m(I)$, where $m$ is the Lebesgue measure. Prove that $m(E)=0$.

Solution 1:

It is enough to prove that $m(E\cap [n,n+1])=0$ is for all $n\in\mathbb{Z}$. The same argument applies to all cases of $n$ so let's prove $m(F)=0$ where $F:=E\cap [0,1]$.

From the condition we have that $m(F)\leq 1/2$.

So, we should be able to cover $F$ with countably many intervals $I_n$ such that $m(\cup_n I_n)<1/2+1/4$. This is because the Lebesgue measure of a measurable set is (by definition) its outer measure and from the definition of outer measure the existence of such a sequence of intervals follows (recall that the outer measure is the infimum of the sums of the lengths of each covering by intervals).

Now

$$m(F)=m(\bigcup_n(F\cap I_n))\leq \sum_n m(F\cap I_n)\leq \frac{1}{2}\sum_n m(I_n)\leq 1/4+1/8$$

Again choose a new sequence of intervals covering $F$ such that their lengths add less than $1/4+1/8+1/16$. Repeating the argument we get that

$$m(F)\leq 1/8+1/16+1/32$$

Continuing in this way we get that, for all $n\in\mathbb{N}$,

$$m(F)\leq \sum_{k=n}^{2n}2^{-k}.$$

But the right-hand side tends to $0$ as $n\to\infty$ (because $\sum_{k=0}^{\infty}2^{-k}$ is convergent).

Therefore $m(F)=0$.

Solution 2:

Hint: Lebesgue Density Theorem.