Probability of $\limsup$ of a sequence of sets (Borel-Cantelli lemma) Let $(E_n)$ be a sequence of events in a probability space such that $$\lim_{n\to\infty}\mathbb P(E_n)=0.$$ I am trying to prove that if $$\sum_{n=1}^\infty\mathbb P (E_n\setminus E_{n+1}) <\infty$$ then $$\mathbb P\left(\limsup_{n\to\infty} E_n\right)=0.$$
For an example I am using the sequence of intervals $[0, 1/n]$ which goes to $0$ as $n$ increases to infinity. Then I can see how the probability of infinite sum of the intersection of the $E_n$ and the complement of $E_{n+1}$  is less than infinity and the probability of $\limsup_{n\to\infty} E_n$ equals zero. 
Can anyone help me figure out how to prove this in general using any sequence? Especially to show how $\mathbb P\left(\limsup_{n\to\infty} E_n\right)=0$?
Thank you in advance for your thoughts. 
 A: An application of the Borel-Cantelli lemma gives that $\mathbb P(\limsup_n E_n\setminus E_{n+1})=0$. This means that there is a set $\Omega'$ of probability $1$ such that if $\omega\in\Omega'$, then $\omega\in E_n^c\cup E_{n+1}$ for each $n\geqslant N(\omega)$. Notice that if for some $n\geqslant N(\omega)$ we have $\omega\in E_n$, then $\omega\in E_k$ for each $k\geqslant n$ (this can be seen by induction on $k$). Hence $\limsup_n E_n\subset \liminf_n E_n$ (up to a subset of measure $0$).  The assumption $\mathbb P(E_n)\to 0$ show that $\mathbb P(\liminf_n E_n)=0$.
A: If $A_n=E_n - E_{n+1}$, then
$$\text{limsup}_n \ E_n \ = \ \text{liminf}_n \ E_n \ \bigcup \ \text{limsup}_n \ A_n.$$
To see this, pick $\omega$ in LHS, then there exist a subsequence $ E_{n_k}$ such that $ \omega \in E_{n_k}, \ \forall k$.
Now, or $ \omega \in \text{liminf}_n \ E_n,$ for which $ \ \omega \in $RHS, or $\omega \notin \text{liminf}_n \ E_n$. But then there exist a subsequence $ E_{n_ k}$ such that $ \omega \notin E_{n_k}, \ \forall k$.
This implies that there exist a third subsequence $ E_{n_ k}$ such that $ \omega \in E_{n_k} \text{and} \ \omega \notin E_{n_k+1},\ \forall k$.
So $\omega \in \text{limsup}_n \ A_n \subset $ RHS.
With Fatou and Borel-Cantelli, we see that RHS has probability 0.
A: Note that
$$
\bigcup_{n=m}^{\infty} E_n = \bigcup_{n=m}^{\infty} E_n\setminus E_{n+1}
$$
Let $\epsilon > 0$, since $\sum P(E_n\setminus E_{n+1}) < \epsilon$, then $\exists m\in \mathbb{N}$ such that
$$
\sum_{n=m}^{\infty} P(E_n\setminus E_{n+1}) < \epsilon
$$
and so
$$
P(\bigcup_{n=m}^{\infty} E_n) < \epsilon
$$
So if $E = \limsup E_n$, then $P(E) <\epsilon$. This is true for any $\epsilon >0$, so $P(E) = 0$.
