Let $\lambda(A)$ be the Lebesgue measure of $A$. There exists an interval $I$ such that $\lambda(E \cap I) / \lambda(I) < 1-\epsilon$

(Not mentioned in title but $\epsilon$ is a number greater than $0$ and $E$ a Lebesgue measurable subset of $\mathbb{R}^n$.) I know a question equivalent to this one has been asked (here). But it was not answered. Also my thinking is a bit different. So I would love to get some help on this problem. So far I've got:

$$\frac{\lambda(E \cap I)}{\lambda(I)} > 1- \epsilon \iff \frac{\lambda(I)-\lambda(E\cap I)}{\lambda(I)} < \epsilon$$

And also, a theorem states that, for all $\epsilon$ there exists a closed set $F\subset E$ such that $\lambda(E\setminus F) < \epsilon$ so it would sufice to prove that there exists an interval $I$ such that $$\frac{\lambda(I)-\lambda(E\cap I)}{\lambda(I)}<\lambda(E\setminus F).$$ I'm not sure if this helps me much but any help is appreciated. What would you do?

Solution 1:

The statement is false. For example, $\mathbb{R}$ is Lebesgue measurable but $\frac{\lambda(I\cap \mathbb{R})}{\lambda(I)}=1\not< 1-\epsilon.$

(restriction to proper subset of $\mathbb{R}$ behave as $\mathbb{R}$ since $\mathbb{R}\setminus \{0\}$.)

I can prove it when $\overline{E}\neq \mathbb{R}.$

If $\epsilon \geq 1$, it does not make any sense. Thus, let's assume that $0<\epsilon<1$.

But it is trivial. Let $A=\overline{E}$ be the closure of $E$, then $A$ is Lebesgue measurable. Then $A^c$ is open. Note that $A^c$ is a countable disjoint union of open intervals. Just choose one of them,(Let's say $I$). Then clearly, $\lambda(E\cap I)=0$ and $\lambda(I)>0$ so clearly $$\frac{\lambda(E\cap I)}{\lambda(I)}=0<1-\epsilon . $$

If you want to prove that

when $E$ is positive Lebesgue measurable set then for any $\epsilon\in (0,1)$, there exists an interval $I$ such that $$\frac{\lambda(E\cap I)}{\lambda(I)}>1-\epsilon,$$

it is a little bit difficult. But the proof is as follow.

Let's assume that the statement is false. Then there exists such $\epsilon $ such that for any interval $I$$$\lambda(E\cap I)\leq (1-\epsilon)\lambda(I).$$

Note that $$\lambda(E)=\inf\left\{\lambda(U): E\subset U \text{ and } U \text{ is open.} \right\}.$$

Thus, there exists open set $U$ such that $E\subset U$ and $$ \lambda(U)<\lambda(E)+\frac{\epsilon}{1-\epsilon}\lambda(E) = \left(\frac{1}{1-\epsilon} \right)\lambda(E)$$

Note that $U$ is countable disjoint union of open intervals $\{I_n\}_{n\in\mathbb{N}}.$

Now observe that

$$\lambda(U)<\frac{1}{1-\epsilon} \sum_{n=1}^{\infty} \lambda(E\cap I_n)\leq \frac{1}{1-\epsilon} \sum_{n=1}^{\infty} (1-\epsilon)\lambda(I_n)=\lambda(U) $$

And note that $\lambda(U)<\lambda(U)$ does not make sense. Thus, it is a contradiction. Therefore, the statement is true.