I am really unsure about my solution to question 2 in 4.6 and have nobody to talk about it. So may be someone in here can help me out.
I'll just recall the setting of the exercise.
We work on the measure space ([0,1], $\mathcal{B}([0,1]),m)$, where $\mathcal{B}$ is the Borel $\sigma$-algebra and $m$ the Lebesgue measure. Assume that for each $n \geq 1$, $A_n \subset [0,1]$ is measurable. Let $B$ be $\lim \sup$ of the $A_n$.
Questions 1 asks us to show that $B$ is measurable. That is the case because the collection of measurable subsets form a $\sigma$-algebra and $B$ is a countable intersection of a countable union of measurable sets.
Now question 2 wants us to prove the following: If $m(A_n) > \delta > 0$, for each $n$, then $m(B) \geq \delta$.
I did the following.
Let $B_n := \cup_{j \geq n} A_j$. Note that $\{B_n\}_{n \geq 1}$ is a decreasing sequence of sets. Thus: $$m(B)=m(\cap_{n \geq 1} B_n) = \lim_{n \rightarrow +\infty} m(B_n)=\lim_{n \rightarrow +\infty}m(\cup_{j\geq n}A_j)$$ Now since $A_n \subset \cup_{j\geq n} A_j$, we have that $m(A_n) \leq m(\cup_{j\geq n} A_j)$ and thus: $$m(B) = \lim_{n \rightarrow +\infty}m(\cup_{j\geq n}A_j) \geq \lim_{n \rightarrow +\infty}m(A_n) \geq \delta$$
Now, I don't see why we need to have the strict inequalities $m(A_n) > \delta > 0$. I feel like my argument also works if $m(A_n) \geq \delta \geq 0$...
There's probably an error in my whole arguement and I would appreciate your help :)
I haven't looked at your proof in detail, but the reason you say you're concerned it's wrong is no problem - in fact it's true that if $m(A_n)\ge\delta$ then $m(B)\ge\delta$.
So why not state the exercise that way? Who knows. But note that $m(A_n)>\delta$ does not imply $m(B)>\delta$...