The Borel Cantelli Lemma states that for a sequence of events $(A_n)$ with finite sum $\sum_{k=1}^{\infty}P(A_k)< \infty$ the probability of infinite many events happening is $0$, thus $P(\lim\sup_{n \to \infty} A_n)=0.$
Before getting to my question, here are two different proofs that make sense to me:
First one:
Let $B_m=\bigcup_{k \geq m}A_k$, then $\lim_{m \to \infty} B_m=\lim\sup_{n \to\infty}A_n$ from above. Hence, by the continuity, subadditivity of measures $$P(\lim\sup_{n \to \infty} A_n)=P(\lim_{m \to \infty} B_m)=\lim_{m \to \infty}P( \bigcup_{k \geq m}A_k)\leq \lim_{m \to \infty}\sum_{k=m}^{\infty}P(A_k)=0.$$
Second one:
$P(\lim\sup_{n \to \infty} A_n)=0$ is equivalent to $\sum_{k=1}^{\infty}1_{A_k}(\omega) < \infty$. So we need to show that $P(\sum_{k=1}^{\infty}1_{A_k}(\omega))\geq \infty)=0$. First of all, by linearity of the expectation we have $$E\bigl (\sum_{k=1}^{\infty}1_{A_k}\bigr )=\sum_{k=1}^{\infty}E(1_{A_k})=\sum_{k=1}^{\infty}P(A_k) < \infty$$ Thus, by using Markov's inequality we obtain $$P(\sum_{k=1}^{\infty}1_{A_k}(\omega) \geq \epsilon) \leq {1 \over \epsilon}\sum_{k=1}^{\infty}P(A_k) \rightarrow0$$ for $\epsilon \to \infty$ which completes the proof.
Question:
If both proofs are correct (please check), does any of those two proofs require something that is not needed in the other proof?