Let $X$ be a real random variable and $\{X_n\}_n$ be a random sample of $X$. My question is whether or not $$ \cfrac{1}{n}\sum_{i=1}^n \text{E}[|X-X_i|]\rightarrow 0 \text{ a.e.} $$
It is a result of the type of Strong Law of Large Numbers, but with the sum in other form. I'm stuck in this point.
Thank you in advance.
The claim is false. Take $\{X\}\cup(X_n)$ i.i.d. $N(0,1)$. If your claim is true then it would also hold with $X$ replaced by $-X$. This would imply that $ \cfrac{1}{n}\sum_{i=1}^n \text{E}[|X-(-X)|]\rightarrow 0 \text{ a.e.} $ by triangle inequality, but this is clearly false.
More simply, $E|X-X_i|$ is independent of $i$!