0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Can you include your definition of $\lambda_*$ (I'm guessing inner measure)? $\endgroup$ – Matthew C Nov 1 '18 at 13:29
  • $\begingroup$ outer measure sorry $\endgroup$ – Milos Tasic Nov 1 '18 at 13:33
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.