(This is a homework question.)
Background information:
Let $\lambda^{d}$ be the Lebesgue measure on the $\sigma$-algebra $\Lambda^{d}$ of Lebesgue measurable subsets of $\mathbb{R}^{d}$. Note that $\lambda^{d}$ is translation invariant: $\lambda^{d} (A + v) = \lambda^{d} (A)$ for every $v \in \mathbb{R}^{d}$ and $A \in \Lambda^{d}$. Moreover, \begin{align} \lambda^{d} (A) &= \sup \{ \lambda^{d} (K) \; | \; K \subset A, \, K \text{ compact} \} \quad (1) \\ &= \inf{ \{ \lambda^{d} (U) \; | \; A \subset U, \, U \text{ open} \} } \qquad \; (2) \end{align}
for all $A \in \Lambda^{d}$. Let $E \in \Lambda^{1} $ such that $\lambda^{1} (E) > 0$.
The question: Show that for every $0 < \alpha < 1 $ there is an open interval $I$ such that $\lambda^{1} (E \cap I) > \alpha \lambda^{1} (I)$.
There is a hint, which is "use (1) and (2) to obtain a suitable compact $K \subset E$ with $\lambda^{1}(K) > 0$ and open interval $I \supset K $".
I find it hard to even get started with this exercise, to get some intuition of it. I know the definition of the Lebesgue pre-measure and outer measure, and that the Lebesgue measure $\lambda^{d}$ of $E$ is given by its is the Lebesgue outer measure $({\lambda^{d}})^{*} (E) = \lambda^{d} (E) $ iff for every $A \in \mathbb{R}^{d}$ we have $$ ({\lambda^{d}})^{*} (A) = ({\lambda^{d}})^{*} (A \cap E) + ({\lambda^{d}})^{*} (A \cap E^{\complement} ) \quad .$$ Furthermore, I know the definition of compactness. But why is it important that $K$ is compact? How is it useful to prove the aforementioned inequality?
Let's just write $m$ for $\lambda^1$ for simplicity of notation. Firstly, check that you can assume $0 < m(E) < \infty$ (otherwise replace $E$ by $E\cap [-n,n]$ for $n$ large enough). Now define $$ \beta := \sup \frac{m(E\cap I)}{m(I)} $$ where the supremum is taken over all open intervals $I$. Note that $\beta$ is well-defined and $\beta \leq 1$.
Now for any open cover $\{I_n\}$ of $E$ by countably many open intervals, note that $$ m(E) \leq \sum m(E\cap I_n) \leq \beta \sum m(I_n) $$ This is true for any such cover, and hence we get $$ m(E) \leq \beta m(E) \Rightarrow \beta = 1 $$
The result now follows from the definition of $\beta$.