$E \subset [0,1]$, $m(E \cap I) \ge \alpha m(I)$ for some $\alpha \in (0,1)$ and all intervals $I$, then $m(E) = 1$

Question

Currently, I'm solving exercise 7 of chapter 3 in Stein's Real Analysis.

For reference, the Corollary 1.5 of chapter 3 is as follows:

Corollary 1.5 Suppose E is a measurable set in $\mathbb R^d$. Then,

(i) almost every $x \in E$ is a point of density of $E$

(ii) almost every $x \notin E$ is not a point of density of $E$

And, $x$ is a point of Lebesgue density of $E$ if $\lim_{m(B) \to 0;x\in B}\frac{m(B \cap E}{m(B)} = 1$

.....................

I started this problem with defining $F = [0,1] - E$ which is measurable.

My goal is to show that $m(F) = 0$ and thus $m(E) = 1$.

Since $E \cap F = \phi$, $(I \cap E) \cap (I \cap F) = \phi$.

$E \cup F = [0,1]$ leads to $I = (I \cap E) \cup (I \cap F)$

Then, $(I \cap F) = I - (I \cap E)$ and $m(I \cap F) = m(I) - m(I \cap E) \le (1-\alpha)m(I)$

So, $\frac{m(I \cap F)}{m(I)} \le 1-\alpha$ and this holds for all intervals $I \subset [0,1]$

This leads to $\lim_{m(I) \to 0; x \in I}\frac{m(I \cap F)}{m(I)} \le 1-\alpha$.

(It is trivial that $0 < \alpha < 1)$

///

This is what I've done and I've found it difficult to connect this result to $m(F) = 0$.

Any comments about my trial or other better approaches would be grateful. Thank you.

Answer 1

Initially observe that $ E $ has positive measure. Then use Lebesgue's differentiation theorem to obtain that $ \mathrm{1}_E \geq \alpha $ pointwise almost everywhere in $[0,1] $. Now if $ m(E)<1 $ we would have $ m([0,1] \setminus E )>0. $ But then we could find $ x \in [0,1] $ so that $ x \not \in E $ and $ \mathrm{1}_E(x) \geq \alpha $ which implies that $ \alpha \leq 0 $ which cannot happen from premise of the problem.

Answer 2

The limit you proved shows $x$ is not a point of density of $F$. That is, $F$ has no points of density. So apply (i) to $F$.

Answer 3

Take $F= [0,1] \backslash E$. Assume that $\beta \colon= \mu(F) > 0$. Let $\epsilon > 0$. We have $(1+\epsilon) \beta > \beta$. There exists (by the definition of (exterior) measure) a countable family of intervals $(I_n)$ such that $\cup_n I_n \supset F$ and $\sum \mu(I_n) < (1+\epsilon) \beta$. We get $$ \sum \mu(I_n) < (1+\epsilon) \beta = (1+\epsilon) \mu(F)\le (1+\epsilon) \sum \mu(F\cap I_n)$$ and so, for at least one of the $I_n$'s we have $$\mu(I_n) \le (1+\epsilon) \mu(F\cap I_n)$$