The problem
I want to check truthfulness about equivalence :
$$E(|X_n-X|) \rightarrow 0 \Leftrightarrow X_n \rightarrow X\;in\;probability.$$
What I've Done
"$\Rightarrow$" Using Chebyshev's inequality i get :
$P(|X_n-X|\ge \varepsilon) \le \frac{E(|X_n-X|)}{\varepsilon}\rightarrow0 \Rightarrow X_n\rightarrow X$ in probability.
"$\Leftarrow$"
A'im not sure about this implication, my intuition tell's me that it's truth but i cant figure out any prove of it.
The convergence on the left is called $L^1$-convergence. You can find something more about convergence of random variables on wikipedia.
$L^1$-convergence implies convergence in probability, but the other implication is false: consider $X_n = 2^{n} \mathbb{1}_{[1/2^{n+1},1/2^{n}]}$ and $X = 0$ on the space of probability $([0,1],\mu)$. Then:
$$P(|X_n - X| > \epsilon) = P(X_n > \epsilon) = P(2^{n} \mathbb{1}_{[1/2^{n+1},1/2^{n}]} > \epsilon) = \mu([1/2^{n+1},1/2^{n}]) \rightarrow_{n \rightarrow \infty} 0$$
so $X_n \rightarrow X$ in probability, but $E(|X_n - X|) = E(X_n) = 1/2$, so there isn't $L^1$-convergence.