Lebesgue Differentiation Theorem Qualifying exam Question

Question

I'm studying for a real analysis qualifying exam and want to make sure the wording of my answer makes sense, particularly the portion in bold.

Prove that if $0 < \epsilon < 1$, there is no Lebesgue measurable set $E \subset \mathbb{R}$ satisfying $$ \epsilon < \frac{|I \cap E|}{|I|} < 1- \epsilon$$ for every interval $I$.

Proof: Fix $0 < \epsilon < 1$. By contradiction. Let $E$ be a Lebesgue measurable set satisfying $$ \epsilon < \frac{|I \cap E|}{|I|} < 1- \epsilon$$ for every interval $I$. By the Lebesgue Differentiation Theorem $$\lim_{I \to x} \frac{1}{|I|} \int_{I} \chi_E(y) dy = \lim_{I \to x} \frac{|I \cap E|}{|I|} = \chi_E(x) $$ a.e. where $\chi_E$ is the characteristic function. Choose such an $x_0$ and interval centered at the $x_0$. Then for $0 < \delta < \epsilon$, and for $x$ close enough to $x_0$ either $$\frac{|I \cap E|}{|I|} < \delta$$ or $$\frac{|I \cap E|}{|I|} > 1- \delta $$ which is a contradiction.

Answer 1

Let me address this in more detail. First you have your application of the Lebesgue differentiation theorem (corrected): $$\lim_{I \to x} \frac{|I \cap E|}{|I|} = \chi_E(x)\quad\text{a.e.}$$

The "a.e." on the end indicates that the limit is not true for every value of $x$, but the set of all $x$ for which it is false has measure $0$, so for most values of $x$, it is true.

Choose such an $x_0$ and interval centered at the $x_0$.

This is not at all clear. As I said in the comments, "such an $x_0$" indicates that $x_0$ is to be picked according to some previously established criterion. But you have not established any criterion to refer to. And also, you say to choose an interval. But you don't want to choose any interval here. You need your interval to be dependent on $\delta$, which you haven't introduced yet. Talking about choosing an interval here is just confusing.

Now the rest of your argument is based on the limit above, thus you need the point $x$ where the limit is taken to be one of the points where the equation is true. So what you should have said is

• "Let $x = x_0$ be a point where the limit converges to $\chi_E(x_0)$."

With no mentions of an "interval" yet, as it is too early. The phrasing $x = x_0$ indicates that $x_0$ is to play the part of a particular value your earlier variable $x$ will take on.

Then for $0<\delta <\epsilon$, and for $x$ close enough to $x_0$

The rest of your calculation does not involve $x$. It is based on the limit $\lim_{I \to x_0}$, so you need to be talking about $I$ here, not an unused $x$. You could say

• "Let $0 < \delta < \epsilon$, and let $I$ be an interval about $x_0$ such that $\left|\frac{|I \cap E|}{|I|} - \chi_E(x_0)\right| < \delta.$"

but you don't actually need $\delta$. You can finish this just using $\epsilon$:

• "Let $I$ be an interval about $x_0$ such that $\left|\frac{|I \cap E|}{|I|} - \chi_E(x_0)\right| < \epsilon$."

Then finish the argument:

"Either $x_0 \in E$ or not. If $x_0 \in E$, then $\chi_E(x_0) = 1$ and $$1-\epsilon < \frac{|I \cap E|}{|I|}.$$ If $x_0 \notin E$, the $\chi_E(x_0) = 0$ and $$\frac{|I \cap E|}{|I|} < \epsilon$$ In either case, $I$ does not satisfy $\epsilon < \frac{|I \cap E|}{|I|} < 1- \epsilon$, a contradiction."