Let $\displaystyle E$ be a measurable subset of $\displaystyle [a,b]$. Let $\displaystyle \{ I_k \}$ be a sequence of open intervals in $\displaystyle [a,b]$ such that

$\displaystyle m(I_k \cap E) \geq \frac{2}{3}m(I_k)$, $\displaystyle k=1,2,3,...$.

Prove that $\displaystyle m((\cup_{k=1}^{\infty} I_k) \cap E) \geq \frac{1}{3}m(\cup_{k=1}^{\infty} I_k)$.

I've managed to prove below:

Suppose that $\displaystyle m((\cup_{k=1}^{\infty} I_k) \cap E) < \frac{1}{3}m(\cup_{k=1}^{\infty} I_k)$.

Then $\displaystyle m((\cup_{k=1}^{\infty} I_k) \cap E) < \frac{1}{3}m(\cup_{k=1}^{\infty} I_k)$

$\displaystyle \leq \frac{1}{3} \sum_{k=1}^{\infty} m(I_k)$

$\displaystyle \leq \frac{1}{3} \sum_{k=1}^{\infty} \frac{3}{2} m(I_k \cap E)$

$\displaystyle = \frac{1}{2} \sum_{k=1}^{\infty} m(I_k \cap E)$

Hence $\displaystyle m((\cup_{k=1}^{\infty} I_k) \cap E) < \frac{1}{2} \sum_{k=1}^{\infty} m(I_k \cap E)$.

I'm stuck here. How could I continue from here? Or is there another approach?

Thank you in advance.