Show that $m(E)=1$.

Question

Let $E$ be measurable and $E\subset [0,1]$ .If for every $a,b$ with $0<a<b<1$ we have $m(E\cap (a,b))>\frac{b-a}{4}$ then show that $m(E)=1$.

MY TRY::

Consider the sequence $a_n=\frac{1}{n},b_n=1-\frac{1}{n}$ Then $\cup_{n=3}^\infty (a_n,b_n)=(0,1)$

Now $E=E\cap (0,1)$

Then $m(E)=m(E\cap (\cup_{n=3}^\infty (a_n,b_n))=m(\cup _{n=3}^\infty (E\cap (a_n,b_n))=\sum _{n=3}^\infty m(E\cap (a_n,b_n))$

$\ge\sum_{n=1}^\infty \dfrac{1-\frac{2}{n}}{4}$

How to show that $\sum_{n=1}^\infty {1-\frac{2}{n}}>4$

If I am able to show this I will be done.

Any help.

Answer 1

Unfortunately, your idea is not going to work, no matter if that sums is correct or not.

Your crucial mistake is

$$m(\cup _{n=3}^\infty (E\cap (a_n,b_n))=\sum _{n=3}^\infty m(E\cap (a_n,b_n))$$

That formula for a measure of a union is valid only if the parts under the union are disjoint! This is absolutely not the case in your construction, as $(a_n,b_n) \subset (a_{n+1},b_{n+1})$.

That means you are measering the union of sets where each is a subset of the next, so for example, $$\cup _{n=3}^4 (E\cap (a_n,b_n))=\cup _{n=4}^4 (E\cap (a_n,b_n))= E\cap (a_4,b_4)$$

and thus

$$m(\cup _{n=3}^4 (E\cap (a_n,b_n))= m(E\cap (a_4,b_4)) < m(E\cap (a_3,b_3)) + m(E\cap (a_4,b_4))$$

I can't, at the moment, give a solution to the problem, but the way to do it must IMO go through increasinly smaller intervals, not, as you did, increasingly larger.

Answer 2

As already noted, this is immediate from the Lebesgue Differentiation Theorem. It's also easy to give a much more elementary proof.

If $I$ is an interval let $|I|$ denote the length of $I$. Let $F=[0,1]\setminus E;$ we need to show $m(F)=0$.

Huge Hint: Suppose $F\subset\Bbb R$ is measurable, $\lambda>0$ and $m(F\cap I)\le \lambda |I|$ for every interval $I$. If $\alpha>m(F)$ then $m(F)\le\lambda\alpha$.

Proof. Since Lebesgue measure is outer regular, or just "by definition of Lebesgue measure" depending on how it was defined, there exists a sequence $I_j$ of intervals such that $$F\subset\bigcup I_j$$and $$\sum|I_j|<\alpha.$$ Now $$F=\bigcup(F\cap I_j),$$hence $$m(F)\le\sum m(F\cap I_j)\le\dots..$$

It should be easy to finish the proof of the Huge Hint, and then to deduce that $m(F)=0$ if $\lambda<1$.