Discussion
For \(k \in \mathbb{N}\), let \(E_{k}\) be a measurable subset of [0,1] with Lebesgue measure \(\frac{1}{k^{2}}\). Define
\[ \begin{aligned} E &= \bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}E_{k} \text{ and} \\ F &= \bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}E_{k} \end{aligned} \]
Consider the following statements:
P: Lebesgue measure of E is equal to zero.
Q: Lebesgue measure of F is equal to zero.
Then
Answer: A
Both P and Q are TRUE.
E is the lim sup of the sets \(E_{k}\), the points lying in infinitely many of them. Since
\[ \begin{aligned} \sum_{k=1}^{\infty} m(E_{k}) &= \sum_{k=1}^{\infty}\frac{1}{k^{2}} \\ &= \frac{\pi^{2}}{6} \lt \infty, \end{aligned} \]
the Borel-Cantelli lemma gives m(E) = 0. Directly, \(E \subseteq \bigcup_{k \ge n}E_{k}\), whose measure is at most \(\sum_{k \ge n}1/k^{2} \to 0\).
F is the lim inf, the points lying in all but finitely many \(E_{k}\). Every such point lies in infinitely many, so \(F \subseteq E\) and m(F) = 0 as well. The other options would need the series of measures to diverge.
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