Convergence of the empirical average of iid random variables in probability implies integrability Let $(X_n)_{n\in\mathbb{N}}$ be non-negative i.i.d. random variables and assume that $S_n=(1/n)\sum_{k=1}^{n}X_{k}$ converge to a real random variable $X$ in probability. 
How can I prove that the expectation of $X_{k}$, $k\in\mathbb{N}$, are finite?
Intuitively (by the weak law of large numbers), the random variable $X$ should be constant equal to $\mathbb{E}[X_1]$. Is this right? would this fact be helpful?
 A: I think I got the answer.
Claim: Let $(X_{n})_{n\in\mathbb{N}}$ be i.i.d. non-negative RVs with $\mathbb{E}[X_1^{+}]=\infty$. Then $S_{n}=(1/n)\sum_{k=1}^{n}X_{k}$ converges to $\infty$ a.s.
So suppose that the $S_{n}$ with $(X_{n})_{n\in\mathbb{N}}$ i.i.d. and non-negative converges to a real random variable $X$ in probability but $\mathbb{E}[X_{1}]=\mathbb{E}[X_{1}^{+}]=\infty$. The latter assumption and the claim assert that $S_{n}$ converges to $\infty$ a.s. and hence in probability. This contradicts to the fact that $S_{n}$ converges to $X$.
The claim follows in particular by the strong law of large numbers for integrable random variables:
Let $X_{n}^{\leq d}:=X_{n}I_{\left\{X_{n}\leq d\right\}}$, where $d>0$ and $I$ denotes the indicator function. $(X_{n}^{\leq d})_{n\in\mathbb{N}}$ are i.i.d. and integrable. Thus by the strong law of large numbers we have for $d>0$:
\begin{equation*}
\frac{1}{n}\sum_{k=1}^{n}X_{k}\geq \frac{1}{n}\sum_{k=1}^{n}X^{\leq d}_{k}\rightarrow\mathbb{E}[X^{\leq d}_{1}]\quad\text{as $n\rightarrow\infty$}\quad\text{a.s.}
\end{equation*} 
By monotone convergence Theorem, $\mathbb{E}[X^{\leq d}_{1}]$ converges to $\mathbb{E}[X_{1}]$ and hence diverges to $\infty$.
