Problem involving Borel-Cantelli Lemma and Lim Sup of Sequences Let $(\Omega,\mathcal{F},P)$ be a probability space, and $A_n$, $B_n$ two sequences of elements in $\mathcal{F}$ such that, $\sum\limits_{n=1}^{\infty}P(A_n\cap B_{n+1}^c)<+\infty$ and, $\sum\limits_{n=1}^{\infty}P(B_n\cap A_{n+1}^c)<+\infty$, show that, if lim$_{n\rightarrow\infty}P(A_n)=0$, then, $P($limsup$_n\ A_n)=P($limsup$_n\ B_n)=0$.
I've seen a proof of a similar problem in $\lim_{n} P(U_n)=0,$ $\sum_{n}P(U_n \cap U_{n+q_n}^c)<+\infty \implies P(\limsup_nU_n)=0$ . Using this proof, I tried defining $U_n$ as:
$U_{4n}=A_{2n}$, $U_{4n+1}=A_{2n+1}$, $U_{4n+2}=B_{2n+1}$, $U_{4n+3}=B_{2n+2}$, $n\geq0$.
And $q_n$ as $q_n=2$. But it seems that the hypothesis of lim$_{n\rightarrow\infty} P(A_n)=0$ do not guarantee lim$_{n\rightarrow\infty}P(U_n)=0$. So I was left without ideas.
 A: To solve this, you need the following version of the Borel-Cantelli lemma: if $\mathbb P(U_n)\to 0$ and $\sum_{n=1}^\infty \mathbb P(U_n\setminus U_{n+1})<\infty$, then $\mathbb P(\limsup U_n)=0$ (see for instance here).
This corresponds to the case $A_n=B_n$ in the current question, which can consequently be considered as a generalization.
We will apply the previous version of the Borel-Cantelli lemma to $U_n=A_n\cup B_n$, which will give the result, as $\limsup_n A_n\subset\limsup_n U_n$ and $\limsup_n B_n\subset\limsup_n U_n$.
We thus have to check that $\mathbb P(U_n)\to 0$ and $\sum_{n=1}^\infty \mathbb P(U_n\setminus U_{n+1})<\infty$. Let us first check that $\mathbb P(U_n)\to 0$. Since $\mathbb P(A_n)\to 0$, it suffices to check that $\mathbb P(B_n)\to 0$. Now,
$$
\mathbb P(B_n)\leqslant \mathbb P\left(B_n\cap A_{n+1}^c\right)+\mathbb P\left( A_{n+1}\right)
$$
and both terms go to $0$ by assumption.
For the condition $\sum_{n=1}^\infty \mathbb P(U_n\setminus U_{n+1})<\infty$,  notice that $$\mathbb P\left(U_n\setminus U_{n+1}\right)\leqslant \mathbb P\left(A_n\setminus U_{n+1}\right)+\mathbb P\left(B_n\setminus U_{n+1}\right)\leqslant \mathbb P\left(A_n\setminus B_{n+1}\right)+\mathbb P\left(B_n\setminus A_{n+1}\right).$$
