The following is an exercise from Bruckner's Real Analysis:
Let $A_n$ be a sequence of Lebesgue measurable subsets of $[0, 1]$ and suppose that $\limsup _{n→∞} \mu(A_n) = 1$. Show that there is some subsequence with $\mu(\cap_{k=1}^{\infty} A_{n_k}) >0.$ [Hint: Arrange for $\sum_{k=1}^{\infty} (1−\mu(A_{n_k}))<1$.]
I have a kind of "proof" for that but :
1- Is my proof correct?
2- If so, what is the use of the hint that the book has given?
My attempt : For $A_n$ a sequence of Lebesgue measurable set we have $\limsup _{n→∞} \mu(A_n) \le \mu( \limsup _{n→∞} A_n)$, so by $\limsup _{n→∞} \mu(A_n) = 1$ we have $\mu( \limsup _{n→∞} A_n) \ge 1$ and because all $A_n \subset [0,1]$ so $\mu( \limsup _{n→∞} A_n) = 1$ which means the measure of all elements of $[0,1]$ that belongs to infinitely many $A_n\subset [0,1]$ is 1, and that infinitely many sets is a subsequence $A_n$ and $\mu(\cap_{k=1}^{\infty} A_{n_k}) = 1 >0.$