1.28

Given $E \subset \mathbb{R}$, $m^*(E) > 0$, assume to the contrary that for any open interval $I$, $m^*(E \cap I) < \alpha m(I)$.

By the definition of the outer measure, we know that $\forall \epsilon > 0$ $\exists \lbrace Q_k \rbrace_{k=1}^\infty$ such that $E \subset \bigcup_{k=1}^\infty Q_k$, and: $$\sum_{k=1}^\infty |Q_k| \leq m^*(E) + \epsilon$$ Next, define $\lbrace \mathcal{O}_k \rbrace_{k=1}^\infty$ to be open intervals s.t. $Q_k \subset \mathcal{O}_k$ for all $k$, and $m(\mathcal{O}_k)) = m(Q_k) + \frac{\epsilon}{2^k}$. (Note, this can always be done by adding $\frac{\epsilon}{2^{k+1}}$ to each side of the interval $Q_k$.)

We constructed $\lbrace \mathcal{O}_k \rbrace_{k=1}^\infty$ with the following property in mind ($\dagger$): $$\sum_{k=1}^\infty m(\mathcal{O}_k) \leq \big(\sum_{k=1}^\infty m(Q_k)\big) + \epsilon \leq m^*(E) + 2\epsilon$$ Now, just observe that $E \subset \bigcup_{k=1}^\infty \mathcal{O}_k$, so, by ($\dagger$): $$m^*(E) \leq \sum_{k=1}^\infty m^*(E \cap \mathcal{O}_k) \leq \alpha \sum_{k=1}^\infty m(\mathcal{O}_k) \leq \alpha(m^*(E) + 2\epsilon)$$ So finally, we finally arrive at the result: $$m^*(E) \leq \alpha m^*(E)$$ Which, of course, gives that $m^*(E) = 0$... a contradiction.