So I've been stuck on the following question for a while now and was hoping that someone could maybe give me a hint to the following question.
(please note that i mean outer measure for $\lambda^*$)
Suppose $A \subset \mathbb R$ is a Borel set with $\lambda(A) > 0$. Using the fact that $\lambda(A) = \lambda^*(A)$, show that for any $\epsilon > 0$ there exists a non-empty interval $I$ with $\lambda(A \cap I) \ge (1 - \epsilon) \lambda(I)$.
The result is trivial if $\epsilon \ge 1$, so assume that $1 - \epsilon > 0$.
You don't need $A$ to be a Borel set, or even Lebesgue measurable. Assume that $A \subset \mathbb R$ and that $\lambda^*(A) > 0$.
Suppose, to the contrary, that $\lambda^*(A \cap I) < (1-\epsilon) \lambda(I)$ for every interval $I$.
Let $\{I_k\}$ be an arbitrary cover of $A$ by bounded intervals. Then $$\lambda^*(A) = \lambda^* \left( A \cap \bigcup_k I_k \right) = \lambda^* \left(\bigcup_k (A \cap I_k) \right) \le \sum_k \lambda^*(A \cap I_k)$$ where the last inequality uses the subadditivity of the outer measure. According to the hypothesis made above, it follows that $$\lambda^*(A) < (1-\epsilon) \sum_k \lambda(I_k) = (1-\epsilon) \sum_k \ell(I_k).$$ Now take the infimum over all such coverings $\{I_k\}$ of $A$ to obtain $$\lambda^*(A) \le (1-\epsilon) \lambda^*(A)$$ by the definition of the Lebesgue outer measure.
This contradicts the hypothesis that $\lambda^*(A) > 0$.