The problem goes like this:
"Let $A\subset \mathbb{R}$ be a measurable set with $0<m(A)<\infty$. Prove that $$\alpha := \text{sup} \{ \frac{m(A\cap I)}{m(I)}: I \text{ open interval}\}=1"$$
This seems obvious in my head, but i don't know where to start a mathematical proof of it. I will thank any hint or help.
On
The Lebesgue Density Theorem tells you that for almost all $x \in A$ we have $\frac {m(A\cap (x-r,x+r)} {2r} \to 1$. From this the result follows. Ref. for Lebesgue Density Theorem: https://en.wikipedia.org/wiki/Lebesgue%27s_density_theorem.
As the other poster noted, we must assume $m(A) > 0$. This solution is an application of the outer regularity of Lebesgue measure.
Assume for the sake of contradiction that there exists $\delta>0$ such that $\frac{m(A \cap I)}{m(I)} \leq 1-\delta$ for all intervals $I$. Let $\epsilon > 0$, then outer regularity of Lebesgue measure means we can find an open set $B \supseteq A$ such that $m(B) \leq m(A) + \epsilon$. Now recall that open sets in $\mathbb{R}$ are the countable union of (disjoint) open intervals, so that we can write $B = \bigcup_n I_n$. Note that the inequality is now written as $m(\bigcup_n I_n) \leq m(A) + \epsilon$. Since $\frac{m(A \cap I)}{m(I)} \leq 1-\delta$ for all intervals $I$, in particular it is true for these $I_n$, ie, $\frac{m(A \cap I_n)}{m(I_n)} \leq 1-\delta$, then rearranging and summing over all $n$, we get $\sum_n m(A \cap I_n) \leq (1-\delta) \sum_n m(I_n)$, or $m(A) \leq (1-\delta) m(\bigcup_n I_n)$. But this is a contradiction to the inequality we had above, $m(\bigcup_n I_n) \leq m(A) + \epsilon$, by choosing $\epsilon$ sufficiently small (this is where we use $m(A)>0$).