Suppose $\{X_n\}$ is a uniformly integrable sequence of independent random variables with zero mean. Prove that $1/n \sum\limits_{i=1}^n X_i \rightarrow0 $ in probability.
I tried to follow the proof of law of large numbers and transform it in order to prove this statement, but all such proofs use variance which may be not finite. Help please!
Uniform integrability suggests a truncation argument. For $R$ a fixed positive number, define $X'_j:=X_j\chi_{\{|X_j|\leqslant R\}}-\mathbb E(X_j\chi_{\{|X_j|\leqslant R\}})$ and $X''_j:=X_j\chi_{\{|X_j|\gt R\}}-\mathbb E(X_j\chi_{\{|X_j|\gt R\}})$. Then $S_n:=\sum_{j=1}^nX_j=\sum_{j=1}^n(X'_j+X''_j)$. We have $$\mathbb E|S_n|\leqslant \mathbb E\left|\sum_{j=1}^nX'_j\right|+n\max_{1\leqslant j\leqslant n}\mathbb E|X_j|\chi_{\{|X_j|\gt R\}}\leqslant\mathbb E \left|\sum_{j=1}^nX'_j\right|+n\sup_{k\in\mathbb N}\mathbb E(|X_k|\chi_{\{|X_k|\gt R\}}).$$ For the first term, we can use Cauchy-Schwarz inequality.