# For every measurable $E$ with $0<m(E)< \infty$, there is $[a,b]$ such that $m(E \cap [a,b]) > \frac{1}{2} m(E)$

Let $$E$$ be any Lebesgue measurable set in $$\mathbb{R}$$ with $$0. I want to show that there exists a finite, nontrivial interval $$[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?

• Do you know anything about $\lim_{x\to\infty} m( E \cap [-x,x] )$? Commented May 19 at 0:47
• 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. Commented May 19 at 0:55
• You’re going to want $E$ to have finite measure. Consider $E=\Bbb R$, for example. Commented May 19 at 0:56
• @copper.hat I have edited the post now Commented May 19 at 1:35
• @GrigorHakobyan Note that $\mathbb{R} = \cup_n (-n,n+1]$, a disjoint union, now consider $\sum_n m(E \cap (n,n+1])$. Commented May 19 at 3:05

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.)
• 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) Commented May 19 at 1:29