Probability of Limsup of a bunch of events In "Probability Theory" by Athreya, I encountered the following question:
Let $\{A_n\}_{n=1}^\infty$ be a sequence of events on a common probability space. Suppose
$$\sum_{n=1}^\infty P(A_n\setminus A_{n+1}) < \infty \quad(1)$$
and $\lim_{n \to \infty} P(A_n) = 0$, show that 
$$P[\limsup A_n] = 0$$
Moreover show that one can replace $\lim_{n \to \infty} P(A_n) = 0$ with $$P(\liminf_{n \to \infty} A_n)=\lim_{n \to \infty} P(\bigcap_{j\geq n} A_j) = 0$$
My attempt:
I was able to show the first part. But the second part i.e. to show the replacement was kind of tricky. I tried various things including trying to show
$$\bigcup_{n=1}^\infty A_n = \bigcup_{n=1}^\infty \left\{(A_n\setminus A_{n+1})\cup \bigcap_{j \geq n}A_j\right\}$$
but I am not really sure if this is correct (For the result, a $\leq$ would suffice). This was inspired from the following identity (which I have proved for $m \geq 1$)
$$ \bigcup_{n=1}^m A_n = \bigcup_{n=1}^m \left\{(A_n\setminus A_{n+1})\cup \bigcap_{j = n}^{m+1}A_j\right\}$$
In trying to prove it for the $\infty$ case mentioned above, I ran into a technical difficulty, namely the switching of limits and infinite sums. Hence I am stuck here. Perhaps there is another way.
Kindly request clarifications if necessary and provide suitable hints (as opposed to whole answers). If I get it, I will provide the complete solution.
Update: Athreya gives a hint: He asks us to show that 
$$A \cap B^c \subset \liminf A_n$$
where $A = \limsup A_n$ and $B = \limsup B_n$. The result would then follow from Borel Cantelli Lemma as $(1) \Rightarrow P(B)=0 \Rightarrow P(A \cap B) = 0$ which would imply 
$$P(A) = P(A \cap B) + P(A \cap B^c) \leq 0 + P(\liminf A_n) =0$$
 A: Ok, according to this, courtesy Daniel Fischer,
$$\bigcup_{n=k}^\infty A_n = \bigcup_{n=k}^\infty \left\{(A_n\setminus A_{n+1})\cup \bigcap_{j \geq n}A_j\right\}$$
is correct (although it does not necessarily follow from the arguments I gave earlier). Proof is by logical induction:
Suppose $x \in A_{k_0}$ for some $k_0 \geq n$, then if $x \in A_{k}\setminus A_{k+1}$ for some $k\geq n$, then done. Suppose not, then note that $x \not\in A_{k_0}\setminus A_{k_0+1}$. But since $x \in A_{k_0}$, this means $x \in A_{k_0+1}$. Continuing in this fashion, we get $x \in A_{j}, \forall j \geq k_0$. Or to put it in another way, 
$$x \in \bigcap_{j \geq k_0}A_j$$
Since $\bigcap_{j \geq k_0}A_j \subset \bigcup_{n=k}^\infty \bigcap_{j \geq n}A_j $, we thus have
$$\bigcup_{n=k}^\infty A_n \subset \bigcup_{n=k}^\infty \left\{(A_n\setminus A_{n+1})\cup \bigcap_{j \geq n}A_j\right\}$$
We can prove the other way also although that is not required for the main proof. Hence we have
$$ P\left[\bigcup_{n=k}^\infty A_n\right]\leq P\left[\bigcup_{n=k}^\infty \left\{(A_n\setminus A_{n+1})\cup \bigcap_{j \geq n}A_j\right\}\right] \\
\leq \sum_{n=k}^\infty P\left[A_n \setminus A_{n+1}\right] + P\left[\bigcup_{n=k}^\infty \bigcap_{j \geq n}A_j\right] \\
= \sum_{n=k}^\infty P\left[A_n \setminus A_{n+1}\right] + P[\liminf A_n]$$
Take limits to get the result. $\blacksquare$
