Every martingale that converges in $L^p$ converges in $L^1$? If not, what is a conterexample?

Question

We know that, for a martingale $\{X_n\}_n$ converge in $L^1$ it is necessary that $$\lim_{M \to \infty}\left(\sup_iE[|X_i|;|X_i|>M]\right)=0$$ that is, $\{X_n\}_n$ is uniformly integrable.

On the other hand, to ensure that $\{X_n\}_n$ converges in $L^p$ for $p>1$ it is enough that $\sup_n E[|X^n|^p]<\infty$. (Result follows using Doob's $L^p$ maximal inequality, which fails for $p=1$, for example in a simple random walk starting at $1$ and stopping whenever it reaches $0$).

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Now, if $\sup_n E[|X_n|^p]<\infty$ then by holder inequality $$\sup_i E[|X_i|;|X_i|>M]\leq \sup_i \|X_i\|_p\|\chi_{\{|X_i|>M\}}\|_q,$$ and $\chi_{\{|X_i|>M\}} \to 0 $ as $M \to \infty$, so $\{X_n\}_n$ is uniformly integrable.

So every martingale bounded in $L^p$ converges in $L^p$ AND in $L^1$.

In other words, the only way I have to ensure a martingale converges in $L^p$ forces convergence in $L^1$. How do I find a martingale that converges only in $L^p$ (if it exists)?

Answer 1

This is probability, so the total measure of the simple space is $1$. For $p \ge 1$, $$ \|X_n-X\|_1 \le \|X_n-X\|_p $$ So convergence in $L^p$ implies convergence in $L^1$. No need for it to be a martingale.