Does the function $f(x) = \sum_{n=1}^\infty P(|X_n| > x)$ have any special properties? Consider independent random variables $X_1, X_2,\ldots$. 
By the Borel-Cantelli lemma, if $X_n\to_{a.s.}0$ then for any $x > 0$
$$
f(x) = \sum_{n=1}^\infty P(|X_n| > x) < \infty.
$$
Question: Does the function $f:\mathbb R_{++} \to \mathbb R$ have a name and does it have any interesting properties?
 A: I do not know whether this function has a name. Denote 
$$
f_N(x)=\sum_{n=1}^N\Pr\left(\left\lvert X_n\right\rvert\gt x\right); \quad R_N(x)=\sum_{n\geqslant N}\Pr\left(\left\lvert X_n\right\rvert\gt x\right).
$$
We know that $\left(\sup_{n\geqslant N}\left\lvert X_n\right\rvert\right)_{N\geqslant 1}$ converges to $0$ in probability and we can quantify it via the inequalities
$$
\Pr\left(\sup_{n\geqslant N}\left\lvert X_n\right\rvert\gt x\right)\leqslant R_N(x)\leqslant f(x).
$$
Some properties of $f$.


*

*The function $f$ is non-increasing. 

*Since $f(x)\geqslant f_N(x)$, it follows that $\liminf_{x\to 0^+}f(x)\geqslant \sum_{n=1}^N\Pr\left(X_n\neq 0\right)$ hence 
$\liminf_{x\to 0^+}f(x)\geqslant \sum_{n=1}^{+\infty}\Pr\left(X_n\neq 0\right)$ and since $f(x)\leqslant \sum_{n=1}^{+\infty}\Pr\left(X_n\neq 0\right)$, we get that $\lim_{x\to 0^+}f(x)=\sum_{n=1}^{+\infty}\Pr\left(X_n\neq 0\right)$. From the Borel-Cantelli, we see that the last term is finite if and only if almost surely, only finitely many $X_n$ are not zero.

*The convergence 
$$
\tag{C}\lim_{x\to +\infty}f(x)=0
$$
holds. Indeed, fix $N$. Then for all $x\geqslant 1$, 
$$
f(x)=f_N(x)+R_N(x)\leqslant f_N(x)+R_N\left(1\right)
$$
hence for all $N$, 
$$
\limsup_{x\to +\infty}f(x)\leqslant R_N\left(1\right)
$$
and the right hand side can be made arbitrarily small. 

*The convergence in (C) can be arbitrarily slow. Let $\left(\delta_n\right)_{n\geqslant 1}$ be a sequence of positive numbers which decreases to $0$. Let $\left(X_n\right)_{n\geqslant 1}$ be an independent sequence such that $\Pr(X_n=n)=\delta_n-\delta_{n+1}=:p_n$ and $\Pr(X_n=0)=1-p_n$. Then 
$f(N-1)=\sum_{n\geqslant N}p_n=\delta_N$.

