Let $E$ be any Lebesgue measurable set in $\mathbb{R}$ with $0<m(E)< \infty$. I want to show that there exists a finite, nontrivial interval $[a,b]$ such that $m(E \cap [a,b]) > \frac{1}{2} m(E)$

As $E$ is Lebesgue measurable, then there is an open set $\mathcal{O} \subset \mathbb{R}$ with $E \subset \mathcal{O}$ such that $$m_*(\mathcal{O} - E) < \varepsilon$$

And we remind ourselves that

$m(E) = m_*(E) = \inf \displaystyle\sum_{i=1}^\infty |Q_i|$ for $Q_i$ are closed cubes such that $E \subset \displaystyle\bigcup_{i=1}^\infty Q_i$

Now, take the open subset $\mathcal{O}$ where $E \subset \mathcal{O}$ with $m_*(\mathcal{O} - E) < \varepsilon$. As $\mathcal{O}$ is open, we can write $\mathcal{O}$ as a disjoint countable union of open intervals: $\mathcal{O} = \displaystyle\bigcup_{i=1}^\infty(a_i,b_i)$. Now, let $m[(a_i,b_i)] = \ell_i$. Then since the $(a_i,b_i)$'s are disjoint,

\begin{align} m(\mathcal{O}) = m \left( \bigcup_{i=1}^\infty (a_i,b_i) \right) = \sum_{i=1}^\infty m[(a_i,b_i)] = \sum_{i=1}^\infty \ell_i \end{align}

I can see that this approach is falling apart quickly because I need to construct/find a connected closed interval $[a,b]$ whereas by my approach, the $(a_i,b_i)$'s may have a positive distance between each other. If the $(a_i,b_i)$'s were right next to each other, you could simply take the closure of $\cup_i (a_i,b_i)$ and pick a closed interval $[a,b]$ inside such a closure. But the $(a_i,b_i)$'s may be too far apart. What can be done?

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    $\begingroup$ Do you know anything about $\lim_{x\to\infty} m( E \cap [-x,x] )$? $\endgroup$ Commented May 19 at 0:47
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    $\begingroup$ The result is not true, $\mathbb{R}$ is measurable, but there is no interval $I$ such that $mI > {1 \over 2} m \mathbb{R}$. You need to assume that $E$ has finite measure. $\endgroup$
    – copper.hat
    Commented May 19 at 0:55
  • $\begingroup$ You’re going to want $E$ to have finite measure. Consider $E=\Bbb R$, for example. $\endgroup$ Commented May 19 at 0:56
  • $\begingroup$ @copper.hat I have edited the post now $\endgroup$ Commented May 19 at 1:35
  • $\begingroup$ @GrigorHakobyan Note that $\mathbb{R} = \cup_n (-n,n+1]$, a disjoint union, now consider $\sum_n m(E \cap (n,n+1])$. $\endgroup$
    – copper.hat
    Commented May 19 at 3:05

2 Answers 2


In this question, we have that, if $0 < m(E) < \infty$, for any $\epsilon > 0$, there exists $M > 0$ such that $$ m(E \setminus [-M, M]) < \epsilon $$ Now, choose $\epsilon = \dfrac{1}{2}m(E) > 0$, together with the equality $$ m(E) = m(E \cap [-M, M]) + m(E \setminus [-M ,M]) $$ we get $$ m(E \cap [-M, M]) > \dfrac{1}{2}m(E) $$


Hint: Consider a function $\phi: \mathbb{R} \to [0, \infty]$ defined by $\phi(x) = m(E \cap B(0,x))$ where $B(0,x)$ is the ball centred at $0$ with radius $x$. First show that $\phi$ is continuous, and then study its value at $0$ and then for very large $x$. Use the IVT to conclude.

(This will fail if $E$ has infinite measure.)

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    $\begingroup$ Note that "then [study its value] for very large $x$" is actually equivalent to the question asked here. If we give a hint, make sure it's for the right problem (this seems to be the hint for a strictly stronger problem of showing $\phi(x) = r\cdot m(E)$ has a solution for any $0 < r < 1$, for which the given problem is just a single step) $\endgroup$ Commented May 19 at 1:29

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