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This is a homework problem that seems overly simple and because of that I am having doubts. The problem statement:

If $E \in \mathcal{L}$ and $m(E) > 0$, for any $\alpha < 1$ there is an open interval $I$ such that $m(E \cap I) > \alpha m(I)$.

Here, $\mathcal{L}$ denotes the space of Lebesgue measurable sets. We have the following theorem:

Theorem: If $E \in \mathcal{L}$, then $$\begin{align*} m(E) &= \inf\{m(U)\ \mid\ U \supset E,\ U\ \textrm{open}\}\\ &= \sup\{m(K)\ \mid\ K \subset E,\ K\ \textrm{compact}\} \end{align*}$$

My work:

By the above theorem, $m(E) = \sup\{m(K)\ \mid\ K \subset E\}$ for $K$ compact. Then, for some $K \subset E$, $m(K) > 0$, which implies $K$ is nonempty and hence $E \cap K$ is nonempty. Choose $I$ to be the interior of some connected component of some such $K$. Then, $\mu(I) > 0$, and since $\alpha < 1$, we have $\alpha m(I) < m(I)$. But since $I \subset K \subset E$, then $E \cap I = I$, so $\alpha m(I) < m(E \cap I)$.

Is this right?

Emily
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  • A compact set, even one of positive measure, might have empty interior. Instead, approximate $E$ outside by an open set $O$. Note you can write $O$ as a countable union of disjoint open intervals. – David Mitra Feb 14 '15 at 23:09
  • @DavidMitra So perhaps using the $U$ above, instead of the $K$? – Emily Feb 14 '15 at 23:09
  • Yes. One way to argue is given here. – David Mitra Feb 14 '15 at 23:11
  • @DavidMitra Thank you. Unfortunately, the argument in your answer uses concepts that are beyond my reach, but it gives me a basis from which to formulate an argument, I think. – Emily Feb 14 '15 at 23:17

1 Answers1

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Your proof is basically correct. It only need to change K from compact set to union of disjoint intervals because any Lebesgue measurable sets can be approached by union of disjoint intervals, see Theorem 12 on page 41 in Royden's Real Analysis.

Eugene Zhang
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