Prove $\mathbb{P}\left(\lim _{n \rightarrow \infty} X_{n}=0\right)=1.$ Let $(X_n)$ be a sequence of random variables. If $\sum_{n \geqslant 1} \mathbb{P}\left(\left|X_{n}\right|>\varepsilon\right)<\infty$ for each $\varepsilon>0$, prove $\mathbb{P}\left(\lim _{n \rightarrow \infty} X_{n}=0\right)=1$.
I have already tried this several times. Here's my last and most improved attempt.
\begin{align*}
        \sum_{n \geqslant 1} \mathbb{P}\left(\left|X_{n}\right|>\varepsilon\right)<\infty,\text{ }\forall\varepsilon>0\implies&\mathbb{P}(\left[|X_{n}|>\varepsilon\right]\text{ i.o.})=0\\
        \implies&X_{n}\overset{\underset{a.s.}{}}{\rightarrow}0\\
        \iff&\mathbb{P}\left(\lim _{n \rightarrow \infty} X_{n}=0\right)=1.
    \end{align*}
 A: You are missing one important step.
$\mathbb{P}(\left[|X_{n}|>\varepsilon\right]\text{ i.o.})=0$ implies  $\mathbb{P}(\bigcup_k \left[|X_{n}|>\frac 1 k\right]\text{ i.o.})=0$ which implies $X_n \to 0$ a.s.
It is important to let $\epsilon \to 0$ through a sequnce so that you don't have uncountably many sets of probability $0$.
A: It's not wrong but short of details. In particular, the first "$\Rightarrow$" needs "$\forall\epsilon$" on both sides, and this is subtle.
To be completely rigorous, we write the complement event of $\lim_{n\rightarrow\infty} X_n=0$ as $$\cup_{k=1}^\infty \cap_{n=1}^{\infty}\cup_{m=n}^\infty(|X_m|\ge \frac{1}{k})$$ (If you don't know yet, the translation from $\epsilon-\delta$ to the set operation is straightforward: $\forall\rightarrow \cap$ and $\exists\rightarrow\cup$ in a suitable way. This is also the standard way to show that $X_n\rightarrow 0$ is indeed measurable, hence its probability can be discussed.)
Note the first two set operations above are monotone, hence we may apply the measure $\mathbb P$ to have $$\mathbb P([\lim_{n\rightarrow\infty} X_n=0]^c)=\lim_k \lim_n \mathbb P(\cup_{m=n}^\infty(|X_m|\ge \frac{1}{k}))\le\lim_k\lim_n \sum_{m=n}^\infty\mathbb P(|X_n|\ge \frac{1}{k})=\lim_k 0=0$$
