# Does $X^p_n \longrightarrow X^p$ in $L^1$ imply $X_n \longrightarrow X$ in $L^p$?

Let $\{X_n\}$ be a sequence of random variables s.t $X^p_n \longrightarrow X^p$ in $L^1$, do I have that $X_n \longrightarrow X$ in $L^p$?

What if I have that $X_n \longrightarrow X$ almost surely?

I am aware that $X_n \longrightarrow X$ in $L^p$ implies $X^p_n \longrightarrow X^p$ in $L^1$ (as one can find here). But I struggle to either prove or find a counter example.

The only thing that I observed is that the hypothesis implies that both$\{X_n\}$ and $\{|X_n|^p\}$ are U.I.

• Why implies $X_n \longrightarrow X$ in $L^p$ that $X^p_n \longrightarrow X^p$ in $L^1$? The source you are referring to is not showing this, right? Commented Aug 2, 2021 at 9:11

Yes $X_n^p \to X^p$ in $L^1$ implies $X_n \to X$ in $L^p$. This can be proven by the following argument: For any subsequence of $\{X_n\}$ we may pass to a subsequence such that $X_n^p$ converges pointwise almost everywhere to $X^p$. Furthermore, we have $$\limsup_{n\to\infty}\int_E |X_n - X|^p \leq 2^{p-1} \limsup_{n\to\infty} \int_E |X_n|^p + |X|^p = 2^p \int_E|X|^p$$ for any measurable set $E$ and hence by Vitali's Theorem we conclude $\int |X_n - X|^p \to 0$.