For 𝑘 ∈ ℕ, let 𝐸_𝑘 be a measurable subset of [0,1] with Lebesgue measure \(\rm \frac{1}{k^2}\) . Define 

\(\rm E=\cap_{n=1}^{\infty}\cup_{k=n}^{\infty}\) and \(\rm F=\cup_{n=1}^{\infty}\cap_{k=n}^{\infty}E_k\)

Consider the following statements:

𝑃: Lebesgue measure of 𝐸 is equal to zero.

𝑄: Lebesgue measure of 𝐹 is equal to zero.

Then 

Concept:

(i) \(\cap_{n=1}^{∞}\cup_{k=n}^{∞}\) = \(\lim\sup E_k\) and \(\cup_{n=1}^{∞}\cap_{k=n}^{∞}E_k\)= \(\lim\inf E_k\)

(ii) If {E_i} is sequence of measurable sets, \(m\left(\cup_{i=1}^{∞}E_i\right)\) < ∞ and \(\lim E_i\) exist then m(\(\lim E_i\)) = lim(m(E_i))

Explanation:

For 𝑘 ∈ ℕ,t 𝐸𝑘 is a measurable subset of [0,1] with Lebesgue measure \(\rm \frac{1}{k^2}\) .

Now,  m(\(\lim_{k\to\infty} E_k\)) = \(\lim_{k\to\infty}\)(m(Ek)) = \(\lim_{k\to\infty}\)\(\rm \frac{1}{k^2}\) = 0

Hence limit superior and limit inferior are both equal to 0 

Therefore, Lebesgue measure of \(\rm E=\cap_{n=1}^{\infty}\cup_{k=n}^{\infty}\) = \(\lim\sup E_k\) is 0

and Lebesgue measure of \(\rm F=\cup_{n=1}^{\infty}\cap_{k=n}^{\infty}E_k\)= \(\lim\inf E_k\) is 0

Hence both 𝑃 and 𝑄 are TRUE 

Option (1) is correct