Measurable Set Question (2)

**Markeur** · September 27th 2011, 10:15 AM

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

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

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

I've managed to prove below:
Suppose that $m((\cup_{k=1}^{\infty} I_k) \cap E) < \frac{1}{3}m(\cup_{k=1}^{\infty} I_k)$ .
Then $m((\cup_{k=1}^{\infty} I_k) \cap E) < \frac{1}{3}m(\cup_{k=1}^{\infty} I_k)$
$\leq \frac{1}{3} \sum_{k=1}^{\infty} m(I_k)$
$\leq \frac{1}{3} \sum_{k=1}^{\infty} \frac{3}{2} m(I_k \cap E)$
$= \frac{1}{2} \sum_{k=1}^{\infty} m(I_k \cap E)$
Hence $m((\cup_{k=1}^{\infty} I_k) \cap E) < \frac{1}{2} \sum_{k=1}^{\infty} m(I_k \cap E)$ .

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

Thank you in advance.

**InvisibleMan** · September 28th 2011, 04:16 AM

Deleted, I went wrong there.

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