Generalization of weak law of large numbers In Exercise 5.2.12 of Chung Kai-lai's A Course in Probability Theory, there is a question:
Let $\{X_n\}$ be pairwise independent with a common distribution function $F$ (namely, identically distributed) such that
(i) $\displaystyle \int_{|x|\leq n}x \, dF(x)=o(1)$,
(ii) $\displaystyle n \int_{|x|>n} \, dF(x)=o(1)$.
Then $\displaystyle\frac{\sum_{j=1}^{n}X_{j}}{n}\rightarrow 0 $ in probability.
From condition (i) we can get $\operatorname E(X_j)=0$. My idea is to get a bound of $\operatorname E(X_j^2)$ but failed, and another thought is to try to construct an equivalent sequence of random variables to $\{X_n\}$ by truncating $X_n$ at $n$ as the author did in this section but also failed. The problem is that I don't know how to apply condition (ii) and find no way to understand it. Can someone give me some hint on the interpretation of (ii) or some method to get the result? Thanks in advance!
 A: I will sketch how to use the technique of truncation to prove the claim. For every $j$, let $\bar X_j$  denote $X_j\cdot1_{|X_j|\le n}$ (the truncation of $X_j$ to height $n$), let $\bar S_n$ denote $\sum_{j=1}^n \bar X_j$, and let $\varepsilon > 0$. By the union bound,
\begin{align*}
\mathbb P(|S_n|\ge \varepsilon n) &\le \mathbb P(S_n\ne \bar S_n) + \mathbb P(|\bar S_n|\ge \varepsilon n)\\
&\le \sum_{j=1}^n\mathbb P(X_j\ne \bar X_j) + \mathbb P(|\bar S_n|\ge \varepsilon n)\\
&= n\int_{|x|>n}dF(x) + \mathbb P(|\bar S_n|\ge \varepsilon n).
\end{align*}
Hopefully now it is clearer how to use the assumption (ii), and what its nature is. To finish, it suffices to prove that $\mathbb P(|\bar S_n|\ge \varepsilon n) \xrightarrow{n\to\infty}0$. By the union bound again,
\begin{align*}
\mathbb P(|\bar S_n/n|\ge \varepsilon) &\le \mathbb P(|\bar S_n/n-\mathbb E[\bar X_1]|\ge \varepsilon /2) + \underbrace{\mathbb P(|\mathbb E[\bar X_1]| \ge \varepsilon/2)}_{=\ 0\ \text{if $n\gg 1$ by (i)}}.
\end{align*}
By Chebyshev's inequality and the pairwise independence of the $\bar X_j$ (exercise!),
\begin{align*}
\mathbb P(|\bar S_n/n-\mathbb E[\bar X_1]|\ge \varepsilon /2) \le \frac{4}{\varepsilon^2n}\mathbb E\big[\bar X_1^2\big]
\end{align*}
Applying the second lemma below finishes the proof that $\mathbb P(|\bar S_n|\ge\varepsilon n) \xrightarrow{n\to\infty}0$.
Lemma 1. If $Y\ge 0$ and $\displaystyle\lim_{y\to\infty}y\,\mathbb P(Y>y)=0$, then $$\frac{1}{n}\int_0^n y\,\mathbb P(Y>y)\,dy\xrightarrow{n\to\infty} 0.$$
Lemma 2. $\displaystyle \frac{1}{n}\mathbb E[\bar X_1^2] \xrightarrow{n\to\infty}0$.
Proof sketch. For any r.v. $Y\ge 0$, (by Fubini-Tonelli) $$\mathbb E\big[Y^2\big] = \int_0 ^\infty 2y\,\mathbb P(Y>y)\,dy.$$ Apply this equation to $Y = |\bar X_1|$, and use condition (ii) and Lemma 1. $\square$
