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

Question

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 ?

Answer 1

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, $\limsup_{n\to +\infty}\left\lvert X_n-X\right\rvert=1$.

Answer 2

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.