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.

  • $\begingroup$ Sorry, but no, that does not make sense. (1) "such an $x_0$" as what? "Such an" indicates something satisfying a previously introduced condition, but you have not introduced any condition on points. If you mean "Let $x_0 \in \Bbb R$", say that. (2) "and interval". Again, no previous condition on intervals to explain your wording. But you are also fixing this interval, and it is evident in the remainder, the integral needs to depend on $\delta$. You cannot fix it before introducing $\delta$. (3) "for $x$ close enough to $x_0$." You do not reference $x$ again. Why introduce it? $\endgroup$ Commented Jul 18, 2023 at 18:52
  • $\begingroup$ The $x_0$ part is what I'm struggling with. Since Lebesgue differentiation theorem says the above is true for a.e. $x$ and the Interval needs to shrink to this $x$, I felt like I had to introduce an $x_0$ and and a new interval that shrinks to it and then repeat the argument. I had lost points on a homework question for not "choosing an x." $\endgroup$
    – signer59
    Commented Jul 18, 2023 at 19:26
  • $\begingroup$ For (3) I went back to limit definition to try to explain why this is a contradiction. The $|I \cap E| / |I|$ should be close to 0 or 1 depending on whether or not $x$ is in $E$. $\endgroup$
    – signer59
    Commented Jul 18, 2023 at 19:29
  • $\begingroup$ How do you define $\lim_{I \to x} F(I)$ for some function $F$ defined on intervals? (I am not looking for some wishy-washy description here. I am asking for the precise definition of this notation that you are using.) $\endgroup$ Commented Jul 18, 2023 at 19:37
  • $\begingroup$ And there is another issue that I overlooked before. In the equation $$\lim_{I \to x} \frac{1}{|I|} \int_{E} \chi_I(y) dy = \lim_{I \to x} \frac{|I \cap E|}{|I|} = \chi_I(x)$$ the final $\chi_I(x)$ is not defined. $I$ is a dummy variable of the limits. That is, it is only defined inside the limit expression, not outside. The first $\chi_I(x)$ inside the integral is defined because it is inside the limit. But the final $\chi_I(x)$ is outside of both limits, where no $I$ defined. Further, your statement misuses the theorem. It does not give this result. $\endgroup$ Commented Jul 18, 2023 at 20:03

1 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."

  • 1
    $\begingroup$ This makes a lot more sense now. We've shown $\lim_{I \to x} \frac{|I \cap E|}{|E|} = \chi_E(x)$ a.e. , so I can take $x_0=x$ (which is not in the set of measure zero for which the limit doesn't hold). After that, I was trying to apply Lebesgue Differentiation Theorem a second time to $x_0$ (which is where the second interval was coming from), but I can see now that isn't needed. Thank you so much for all of your help!! $\endgroup$
    – signer59
    Commented Jul 18, 2023 at 20:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .