Q:

Prove that both $ P (A_n - \limsup A_n) $ and $ P (\liminf A_n - A_n) $ converges to 0

in relation to Billingsley's Probability and Measure Exercise 4.5, part (b), where requires to prove 
$ P(A_n-\limsup A_n) \rightarrow 0 $ and $ P(\liminf A_n - A_n) \rightarrow 0 $, 
I think I should go about it by defining some monotone sequences with it to use continuity from above/below property of probability measure. 
I think $ \limsup A_n $ decreases as n progresses and $ \liminf A_n $ increases (is it correct?) because I don't know and cannot assume anything about  $A_n$ so I am completely stuck in creating something I think I can work with.
I would be much obliged if anyone could provide me with some guidance. 
Thanks!

A:

Assuming $A_n$ are events, let $B_n = \cup_{k\ge n} A_k$. The sequence $B_n$ decreases in the set theoretical sense and $\cap_n B_n = \lim\sup A_n$. It follows by $\sigma$-additivity that $P(B_n \setminus \lim\sup A_n)$ converges to $0$ when $n\to\infty$. But $A_n\subset B_n$ hence $P(A_n \setminus \lim\sup A_n)\le P(B_n \setminus \lim\sup A_n)$.
Similarly, let $C_n = \cap_{k\ge n} A_k$. This sequence of sets increases and one has $\lim\inf A_n = \cup_n C_n$. Again it follows by $\sigma$-additivity that $P(\lim\inf( A_n) \setminus C_n)$ converges to 0, but one has $C_n \subset A_n$, hence $P(\lim\inf( A_n) \setminus A_n)\le P(\lim\inf( A_n) \setminus C_n)$

