I'm doing exercise in "REAL ANALYSIS" of Folland and got stuck on this problem. I got no clue on how to find the set $I$. Hope someone can help me solve this. Thanks so much
Suppose $m$ is Lebesgue measure and $L$ is its domain. If $E \in L$ and $m(E) \gt 0$, for any $\alpha < 1$, prove that there is an open interval $I$ such that $m(E \bigcap I) \gt \alpha m(I)$
You have $$ m(E)=\inf\{\sum_j b_j-a_j:\ E\subset\bigcup_j(a_j,b_j)\}. $$ So, given $\alpha\in(0,1)$, there exist intervals $\{(I_j)\}$ with $$ m(E)\geq\alpha\sum_jm(I_j). $$ Then $$ \sum_j m(E\cap I_j)\geq m(E)\geq\alpha\sum_jm(I_j). $$ For this last inequality to hold, it needs to hold for at least one $j$: for such $j$, $$ m(E\cap I_j)\geq\alpha m(I_j). $$
Let $E$ be a Lebesgue measurable set with $m(E)>0$, and let $f = \mathbf{1}_{E}$ be the indicator function of this set. For some $x\in E$, we have by the Lebesgue differentiation theorem that $$\lim_{{I\ni x}\atop{\left|I\right|\rightarrow 0}}\dfrac{m(E\cap I)}{m(I)}=\lim_{{I\ni x}\atop{\left|I\right|\rightarrow 0}}\dfrac{\int_{I}f}{\left|I\right|}=f(x)=1$$ where $I$ is an open interval containing the point $x$. Use the fact that the numerator is bounded from above by $\left|I\right|$ to show the existence of such an interval.
Hint: Use regularity of Lebesgue measure to find an open set with the desired property. Then use the fact that any open set is a disjoint union of intervals.
