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?
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.