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 $$\sum_{n=1}^\infty\mathbb P (E_n\setminus E_{n+1}) &lt;\infty$$ 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.
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 $\mathbb P(\liminf_n E_n)=0$.
Note that
$$
\bigcup_{n=m}^{\infty} E_n = \bigcup_{n=m}^{\infty} E_n\setminus E_{n+1}
$$
Let $\epsilon &gt; 0$, since $\sum P(E_n\setminus E_{n+1}) &lt; \epsilon$, then $\exists m\in \mathbb{N}$ such that
$$
\sum_{n=m}^{\infty} P(E_n\setminus E_{n+1}) &lt; \epsilon
$$
and so
$$
P(\bigcup_{n=m}^{\infty} E_n) &lt; \epsilon
$$
So if $E = \limsup E_n$, then $P(E) &lt;\epsilon$. This is true for any $\epsilon &gt;0$, so $P(E) = 0$.