Stein 3.7

$\textbf{Exercise 3.7}$ Prove that if a measurable subset $E$ of $[0,1]$ satisfies $$m(E \cap I) \geq \alpha m(I) \quad (\star)$$ for some $\alpha > 0$ and all intervals $I \in{[0,1]}$ then $m(E) = 1$

Strategy: Instead of appealing directly to corollary 1.5 , we consider $\chi_E$. If we can then show that $\chi_E=1$ a.e. for $E\in{[0,1]}$ then this implies that $m(E)=1$.

$\textit{Proof}$ : Since $I$ is an interval, then $m(I)>0$, so in $(\star)$ above, we can divide through to get $$0<\alpha \leq \frac{m(E \cap I)}{m(I)}$$ Consider $\chi_E$, which is of course integrable on [0,1]. Then, the Lebesgue differentiation theorem gives us that $$\lim_{x\in{I},m(I)\rightarrow0}\frac{1}{m(I)} \int_I\chi_E(x)dx=\chi_E \quad a.e. x\in{[0,1]}$$ Again, one of the key themes in these types of problems is to relate integration and measure. We may write $\int \chi_E = \int_E 1 = m(E)$. Further, we can write $\int_I \chi_E = \int_{I \cap E} 1 = m(E \cap I)$. So, our above expression becomes $$\lim_{x\in{I},m(I)\rightarrow0}\frac{1}{m(I)} \int_{(I \cap E)} 1 dx= \lim_{x\in{I},m(I)\rightarrow0}\frac{m(E \cap I)}{m(I)} = \chi_E \quad a.e. x\in{[0,1]}$$ $$\lim_{x\in{I},m(I)\rightarrow0}\frac{m(E \cap I)}{m(I)} \geq \alpha > 0$$ So, $$\chi_E > 0 \quad a.e.$$ However, characteristic functions take on only values of 0 and 1. So, $$\chi_E>0 \Longrightarrow \chi_E = 1 \quad a.e. \Longrightarrow \int \chi_E = 1 = m(E) \quad \blacksquare$$