Question

For \(k \in \mathbb{N}\), let \(E_k\) be a measurable subset of \([0,1]\) with Lebesgue measure \(\frac{1}{k^2}\).
Define \[ E = \bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} E_k \quad \text{and} \quad F = \bigcup_{n=1}^{\infty} \bigcap_{k=n}^{\infty} E_k \] Consider the following statements:
P: Lebesgue measure of \(E\) is equal to zero.
Q: Lebesgue measure of \(F\) is equal to zero.
Then

• A. both P and Q are TRUE
• B. both P and Q are FALSE
• C. P is TRUE but Q is FALSE
  (D) Q is TRUE but P is FALSE
• D.

Hint:

The Borel-Cantelli Lemma is a powerful tool. Remember the condition: if the sum of measures is finite, the measure of \(\limsup\) is zero. This intuitively means that the probability of belonging to infinitely many \(E_k\) is zero if the sets become small enough quickly.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The sets E and F are the limit superior (\(\limsup\)) and limit inferior (\(\liminf\)) of the sequence of sets \(\{E_k\}\). The question is a direct application of the Borel-Cantelli Lemma from measure theory.
- \(E = \limsup E_k\) is the set of points that are in infinitely many of the sets \(E_k\).
- \(F = \liminf E_k\) is the set of points that are in all but a finite number of the sets \(E_k\).
Step 2: Key Formula or Approach:
First Borel-Cantelli Lemma: Let \((X, \mathcal{M}, \mu)\) be a measure space and \(\{A_k\}\) be a sequence of measurable sets. If the sum of their measures is finite, i.e., \(\sum_{k=1}^{\infty} \mu(A_k)<\infty\), then the measure of the limit superior of these sets is zero, i.e., \(\mu(\limsup A_k) = 0\).
Step 3: Detailed Calculation:
Analysis of Statement P:
We are given a sequence of measurable sets \(E_k\) with Lebesgue measure \(m(E_k) = 1/k^2\).
Let's check the condition for the First Borel-Cantelli Lemma: \[ \sum_{k=1}^{\infty} m(E_k) = \sum_{k=1}^{\infty} \frac{1}{k^2} \] This is a p-series with \(p=2>1\), so the series converges. (Specifically, it converges to \(\pi^2/6\)).
Since \(\sum_{k=1}^{\infty} m(E_k)<\infty\), the lemma applies directly.
The set \(E\) is the limit superior of the sequence \(\{E_k\}\).
Therefore, by the First Borel-Cantelli Lemma, the Lebesgue measure of \(E\) is zero. \[ m(E) = m(\limsup E_k) = 0 \] Thus, P is TRUE.
Analysis of Statement Q:
The set \(F\) is the limit inferior of the sequence \(\{E_k\}\).
There is a general relationship between the limit inferior and limit superior of sets: \[ \liminf E_k \subseteq \limsup E_k \] This means that \(F \subseteq E\).
We have already established that \(m(E) = 0\).
By the monotonicity property of measures, if \(A \subseteq B\), then \(m(A) \le m(B)\).
Since \(F \subseteq E\), we must have \(m(F) \le m(E)\).
Substituting \(m(E) = 0\), we get \(m(F) \le 0\).
Since Lebesgue measure is non-negative, this implies \(m(F) = 0\).
Thus, Q is TRUE.
Step 4: Final Answer:
Both statements P and Q are TRUE.
Step 5: Why This is Correct:
P is a direct application of the First Borel-Cantelli Lemma, as the sum of the measures of the sets converges. Q follows from P because the limit inferior is always a subset of the limit superior, and any subset of a measure-zero set also has measure zero.