# A property of the Borel measure

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 $$\lambda(A \cap I) \ge (1 - \epsilon) \lambda(I)$$.

• Can you include your definition of $\lambda_*$ (I'm guessing inner measure)? – Matthew C Nov 1 '18 at 13:29
• outer measure sorry – Milos Tasic Nov 1 '18 at 13:33

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 $$\lambda^*(A) > 0$$.