# On almost sure convergence of sequence of random variables $\{X_n\}$ such that $\forall p>0, E(|X_n-X|^p)\to 0$ as $n\to \infty$

Let $$(\Omega,\mathcal F, P)$$ be a probability measure space. If $$\{X_n\}_{n=1}^\infty$$ is a sequence of random variables on that probability measure space such that for a random variable $$X$$ on it, $$\lim_{n\to\infty} E(|X_n-X|^p)=0,\forall p>0$$, then is it true that $$X_n\to X$$ a.s. ?

If this is not true in general, what happens if we also assume $$X_n$$ s are independent ?

• Certainly can't be so if the $X_n$'s are independent outside of some edge cases like $X$ a.s. constant. – enthdegree Oct 31 '18 at 23:11

Let $$\left(A_n\right)_{n\geqslant 1}$$ be a sequence of independent events such that $$\Pr\left(A_n\right)=1/n$$. Define $$X_n:=\mathbf 1_{A_n}$$ and $$X=0$$. Then $$\mathbb E\left\lvert X_n-X\right\rvert^p=\Pr\left(A_n\right)=1/n$$ and by the second Borel-Cantelli lemma, $$\limsup_{n\to +\infty}\left\lvert X_n-X\right\rvert=1$$.
There is a standard example of sequence $$\{X_n\}$$ such that $$X_n \to 0$$ in probability and $$0\leq X_n \leq 1$$ but not almost surely. By DCT $$E|X_n-0|^{p} \to$$ for all $$p$$. For the second part let $$\{Y_n\}$$ be an independent sequence such that $$Y_n$$ has same distribution as $$X_n$$ for each $$n$$. Then $$P\{Y_n >\epsilon\} =P\{X_n >\epsilon\} \to 0$$, so we still have $$E|Y_n-0|^{p} \to$$ for all $$p$$. By Borel Cantelli lemma, if $$Y_n \to 0$$ almost surely then $$\sum P\{Y_n >\epsilon \} <\infty$$ for each $$\epsilon$$. But then $$\sum P\{X_n >\epsilon \} <\infty$$ so $$X_n \to$$ almost surely, which is a contradiction.