Let $\{A_n\}_{n\geq1}$ be a sequence events in $(\Omega, \mathcal{F}, P)$ such that
$$\sum_{n=1}^{\infty} P(A_{n}\cap A^c_{n+1}) < \infty$$
and $\lim_{n\to\infty} P(A_n) = 0.$ Show that $$ P(\limsup_{n\to\infty} A_n) = 0.$$
Show also that $\lim_{n\to\infty} P(A_n) = 0$ can be replaced by $\lim_{n\to\infty} P(\bigcap_{j\geq n}A_j) = 0$
So, me tried this in the following way,
Let $B_n = (A_{n}\cap A^c_{n+1}) \ \forall n \geq 1$, and $B = \limsup B_n$ and also let $A = \limsup A_n$.
We know that $B = \limsup (A_n\cap A^c_n) \subset (\limsup A_n) \cap (\limsup A^c_n)$
Now observe that $A \cap B^c \subset \liminf A_n $
Now from $\lim P(A_n) = 0$, We get that $P(\bigcap_{j\geq n} A_j) \leq P(A_n)$ and taking limit both side we get $$\lim_{n\to\infty}P(\bigcap_{j\geq n} A_j) \leq \lim_{n\to\infty}P(A_n) = 0$$ Therefore, $$\lim_{n\to\infty}P(\bigcap_{j\geq n} A_j) = 0 \\ \Rightarrow P(\liminf A_n) = 0 $$ So, $$P(A\cap B^c) \leq P(\liminf A_n) = 0 \\ \Rightarrow P(A\cap B^c) = 0 \\ \Rightarrow P(A) = P(A\cap B)\\ \Rightarrow P(A) \leq P(B) $$ and by First Borel-Cantelli Lemma $P(B) = 0 \Rightarrow P(A) = 0$. Hence the proof. So, Here I use the second fact of the question to prove first fact. Is there any possible way to prove the first fact without using the second fact? If possible what is that solution? Thanks in advance.