Converging exponentially in probability implies convergence with probability one? When I read the paper On the Strong Universal Consistency of Nearest Neighbor Regression Function Estimate, the theorem 1 in it states something like

If for every $\epsilon > 0$ there exists some integer $n_0$ such that for $n>n_0$ $$\text{Pr}\{X_n>\epsilon\}\leq e^{-cn\epsilon^2}$$
  then $$X_n\rightarrow 0$$ with probability one as $n\rightarrow\infty$

It looks like that when $X_n$ converges in probability sufficiently fast (i.e., exponentially), then it converges with probability one. But I don't know where could I get proof this result or it just follows from some theorem I don't know. Can someone help?

Assume that for each positive $\varepsilon$, the series 
$$\sum_n\mathrm{Pr}\{|X_n|\gt \varepsilon \} \mbox{ is convergent}.$$
Then by the Borel-Cantelli lemma, we have for each positive $\varepsilon$, a set $\Omega_\varepsilon$  such that  $\mathbb P (\Omega_\varepsilon)=1$ and if $\omega\in\Omega_\varepsilon$, then there exists an integer $n(\omega)$ such that $\left| X_n(\omega)\right|\leqslant \varepsilon$ for $n\geqslant n(\omega)$. 
Taking $\varepsilon$ of the form $1/j$, we can find a subset $\Omega'$ of $\Omega$, of probability one, such that for each $j$ and each $\omega\in \Omega'$, there exists $n(\omega,j)$ such that if $n\geqslant n(\omega,j)$, the inequality $|X_n(\omega)|\leqslant  1/j$ takes place. This proves that $X_n (\omega)\to 0$ for each $\omega\in\Omega'$, hence that $X_n\to 0$ almost surely.
