I am attempting to prove that a sequence of events occurs infinitely often with probability 0, similarly to Borel-Cantelli. I know the series of probabilities of set differences converges. This is what I came up with, using Boole's inequality: $$\mathbb{P}(E_n\text{ i.o.}) \leq \lim_{n \rightarrow \infty} \sum_{m=n}^\infty \mathbb{P}(E_m) = \lim_{n \rightarrow \infty} \sum_{m=n}^\infty \mathbb{P}(E_m) - \mathbb{P}(E_m \cap E_{m-1})=\lim_{n \rightarrow \infty} \sum_{m=n}^\infty \mathbb{P} (E_m \setminus E_{m-1}) = 0$$ Is this a valid approach ?
Edit $$\sum_{n=1}^\infty \mathbb{P}(E_{n+1} \setminus E_n) < \infty$$ and $$\mathbb{P}(E_n) \rightarrow 0$$ as $n\rightarrow \infty$ are given.
We can use the "disjoint-ification" trick to find that \begin{align*} \bigcup_{k=n}^\infty E_k = E_n \cup \left[\bigcup_{k=n+1}^\infty \left( E_k \setminus \left(\bigcup_{j=n}^k E_j \right) \right) \right] \subseteq E_n \cup \left( \bigcup_{k=n+1}^\infty (E_k \setminus E_{k-1}) \right), \end{align*} and hence \begin{align*} \mathbb{P}\left( \bigcup_{k=n}^\infty E_k \right) &\le \mathbb{P}\left( E_n \cup \left( \bigcup_{k=n+1}^\infty (E_k \setminus E_{k-1}) \right) \right) \\ &\le \mathbb{P}(E_n) + \mathbb{P} \left( \bigcup_{k=n+1}^\infty (E_k \setminus E_{k-1}) \right) \\ &\le \mathbb{P}(E_n) + \sum_{k=n}^\infty \mathbb{P}(E_k \setminus E_{k-1}). \end{align*} By hypothesis, $\lim_{n \rightarrow \infty} \mathbb{P}(E_n) = 0$, and since $\sum_{k=1}^\infty \mathbb{P}(E_k \setminus E_{k-1}) < \infty$ we also have $\lim_{n \rightarrow \infty} \sum_{k=n}^\infty \mathbb{P}(E_k \setminus E_{k-1}) = 0$. Therefore we conclude $\lim_{n \rightarrow \infty} \mathbb{P}\left( \bigcup_{k=n}^\infty E_k \right) = 0$ and so $$ \mathbb{P}\left(\limsup E_n \right) = \mathbb{P}\left( \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty E_k \right) \le \lim_{n \rightarrow \infty} \mathbb{P}\left( \bigcup_{k=n}^\infty E_k \right) = 0 $$ as desired.