Convergence in prob. + Uniform integrability => L1 convergence (proof with rate of convergence)

Question

We know that:

I am wondering if it is possible to be more precise and state that assuming the conditions of the theorem and adding that

• $P\left(\left|X_n-X\right| \geq \epsilon\right) \leq C/n$.

Then, we have $E|X_n - X| \leq C/n$.

In other words, redo the Vitali's proof by keeping the rate of convergence.

Answer 1

Counter-example: Let $X=0$ and $X_n=0$ with probability $1-\frac 1 n$, $X_n=n^{1/3}$ with probability $\frac 1n$. Then $EX_n^{2}=n^{-1/3}$ and boundeness of $EX_n^{2}$ implies that $(X_n)$ is uniformly integrable. $E|X_n-X|=n^{-2/3}$ and $n^{-2/3}\leq C/n$ does not hold for any $C$. Of course, $P|X_n-X|>\epsilon)=\frac 1n$.