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I am a graduate student at Iowa State University attempting to return after a five-year hiatus and take the Real/Complex Analysis qualifier on January 8 for potential reinstatement. Since the professors here are on hiatus, I hope you guys don't mind if I bombard the board with a few past questions from our quals here, like this one from Spring 2013:

Let $m$ denote Lebesgue measure on $\mathbb R$ and $M$ the $\sigma$-algebra of Lebesgue measurable subsets. Given $E \in M$ with $m(E) > 0$ and $\alpha \in (0, 1)$, prove that there is an interval $I$ such that $m(E \cap I) > \alpha \, m(I)$.

I've attempted to force the issue using the definition of outer measure but it doesn't seem to work. I've also attempted using the definition of Lebesgue measurability but that doesn't seem to work either - so I'm at a loss!

Thanks in advance!

-Darrin Rasberry

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3 Answers 3

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Here is an (simpler?) solution:

Suppose not. If $m(E) < \infty$. Then for each interval $I$ we have $m(E \cap I) \leq \alpha m(I)$. Let $\varepsilon > 0$. By the definition of the Lebesgue measure, there is some open set $G \supseteq E$ such that $m(G) < m(E) + \varepsilon$. Since every open subset of $\mathbb{R}$ is a (at most) countable union of open, disjoint intervals, we have $G = \bigcup_{k=1}^\infty (a_k,b_k)$. But then $$ m(E) = m(E \cap G) =\sum_{k=1}^\infty m(E \cap (a_k,b_k)) \leq \alpha \sum_{k=1}^\infty m((a_k,b_k)) = \alpha m(G) < \alpha (m(E)+\varepsilon) $$ Since $\varepsilon > 0$ was arbitrary, we have $m(E) \leq \alpha m(E)$. But then $m(E) = 0$ since $m(E) \geq 0$ and $0 < \alpha < 1$.

If $m(E) = \infty$, then there is disjoint union s.t. $E = \bigcup F_j$ where $m(F_j) < \infty$ since $E$ is $\sigma$-finite. Without loss of generality, suppose $m(F_1) = 0$ and $m(F_j) > 0$ for all $j > 1$. (By taking union operation for null sets, we can obtain as such.)

With $j > 1$, the previous argument implies that there exists an open interval $I_j$ such that $m(F_j \cap I_j) > \alpha m(I_j)$.

Let $I = \bigcup_{j=2}^{\infty} I_j$, then $$\begin{eqnarray} m(E \cap I) &\geq& m\left(\bigcup_{j=2}^{\infty} (F_j \cap I_j)\right)\\ &=& \sum_{j=2}^{\infty}m(F_j \cap I_j) \\ &>& \alpha\sum_{j=2}^{\infty}m(I_j) \\ &\geq& \alpha m(I) \\ \end{eqnarray}$$

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  • $\begingroup$ What if $m^*(E) = \infty$. How can we derive a similar result. $\endgroup$
    – Hamilton
    Apr 2, 2022 at 5:00
  • $\begingroup$ @user264745, approving edits that change the content of one's answer doesn't set a good precedent. Why not encourage $\text{@ areyouwithmydog}$ to write their own answer? $\endgroup$
    – PinkyWay
    Aug 17, 2022 at 20:14
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The Lebesgue density theorem states that for almost every point $x \in E$ with $m(E) > 0$, $$ \lim_{\epsilon\to 0} \frac{m(E \cap B_\epsilon(x))}{m(B_\epsilon(x))} = 1. $$

Fix such $x$ and pick $\epsilon$ such that $$ \frac{m(E \cap B_\epsilon(x))}{m(B_\epsilon(x))} > \alpha. $$

$I = B_\epsilon(x)$ is the desired interval.

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  • $\begingroup$ Thank you so much; that makes perfect sense geometrically. If the set has positive measure, we shrink the balls enough so that they will, for a specific x, all be in a "thick" part of E (unless x is in a subset of measure zero). As an aside, is this theorem usually included in texts doing standard measure theory development? I cannot find it in Royden, our typical text. Thanks again! $\endgroup$
    – Darrin
    Dec 21, 2013 at 2:10
  • $\begingroup$ @Darrin You're welcome. This theorem follows immediately from the Lebesgue differentiation theorem by considering the characteristic function of $E$. This is a standard theorem found in measure theory books. See Rudin's Real & Complex Analysis 7.10, 7.12, or Folland's Real Analysis 3.18. I don't have Royden's book, but I'd be surprised if it wasn't there under one form or another. $\endgroup$ Dec 21, 2013 at 2:20
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@AymanHourieh's answer is simple and direct. I will add this slightly longer (yet perhaps more basic) solution that does not rely on Lebesgue's density theorem.

We start with a basic claim on measurable sets (which is sometimes given as the definition for a measurable set):

Claim. Let $E \in M$ of finite measure, then for every $\epsilon>0$ there exists a finite union of disjoint intervals: $$A_\epsilon := \biguplus_{n=1}^{N}I_n$$ s.t. $m(E \triangle A_\epsilon) < \epsilon$.

Now, let $\epsilon:=(1-\alpha)m(E)$. And let $A_\epsilon$ from the claim. Then $$ m(E) = m(E\cap A_\epsilon) + m(E\setminus A_\epsilon) \\ \leq m(E\cap A_\epsilon) + m(E \triangle A_\epsilon) \\ < m(E\cap A_\epsilon) + \epsilon \\ = m(E\cap A_\epsilon) + (1-\alpha)m(E) $$ Then $$ m(E) < \frac{m(E\cap A_\epsilon)}{\alpha} \tag{1} $$ Similarly, $$ m(A_\epsilon) = m(E\cap A_\epsilon) + m(A_\epsilon \setminus E) \\ \leq m(E\cap A_\epsilon) + (1-\alpha)m(E) \\ \overset{\text{from (1)}}{<} m(E\cap A_\epsilon) + \frac{1-\alpha}{\alpha}m(E\cap A_\epsilon) \\ = \frac{m(E\cap A_\epsilon)}{\alpha} $$ We now note that $$ m(A_\epsilon ) = \sum_{n=1}^{N}m(I_n) \\ m(E \cap A_\epsilon ) = \sum_{n=1}^{N}m(E \cap I_n). $$ Therefore, $$ \sum_{n=1}^{N}m(I_n) < \frac{1}{\alpha}\sum_{n=1}^{N}m(E \cap I_n), $$ So there must exist $1\leq n \leq N$ s.t. $$ m(I_n) < \frac{1}{\alpha}m(E \cap I_n), $$ which completes the proof.

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  • $\begingroup$ What if $m(E)=\infty$? $\endgroup$
    – Twnk
    Mar 14, 2014 at 20:35
  • $\begingroup$ Then there exists $E'\subset E$ of finite positive measure, and we apply the claim to $E'$. $\endgroup$
    – doodle
    Mar 15, 2014 at 1:11

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