In this paper, I found that if $\lbrace X_i \rbrace$ is an independenly indentically distributed (i.i.d.) sequence and $E(|X_i|)<\infty$, then $\bar{X}_{N}=\sum_{i=1}^N\frac{X_i}{N}$ converge to $E(X_i)$ almost surely, i.e. $P(\lim_{N\to\infty}\bar{X}_N=E(X_i))=1$.
I have this counter-example in my mind: $X_i\overset{i.i.d.}{\sim} N\times Bernoulli(\frac{1}{N})$, i.e. each $X_i$ takes the value $N$ with probability $\frac{1}{N}$ and $0$ otherwise. Obviously $E(|X_i|)=E(X_i)=1<\infty$. Note that $Var(X_i)=O(N)$ and hence $Var(\bar{X}_{N})=O(1)>0$. Therefore, $P(\lim_{N\to\infty}\bar{X}_N=E(X_i))\neq 1$.
I am wondering whether one should replace $E(|X_i|)<\infty$ with the condition that $\lbrace X_{i,N}\rbrace$ to be uniformly integrable: $\lim_{c\to\infty}\sup_N \sup_{1\leq i\leq N} E(|X_{i,N}|I(X_{i,N}>c))=0$, where $I(\cdot)$ is the indicator function equal to $1$ when $X_i>c$ and $0$ otherwise, to rule out my counter-example.