Suppose $\{X_t\}_{t\geq0}$ is a nonnegative discrete-time martingale with $X_0=1$. Then we know by martingale convergence theorem that $X_\infty=\lim_{n\to\infty}X_n$ exists.
Let $\mathcal{F}_n$ be the $\sigma$-algebra generated by $\{X_0,\cdots,X_n\}$. Define new measure $Q$ on $\mathcal{F}_n$ by $Q(V)=E_P[X_n\mathbb{1}_V]$ for $V\in\mathcal{F}_n$. Then by dominated convergence theorem we know that if $E_P[X_\infty]=1$, then $Q\{\sup_nX_n=\infty\}=0$ and $Q\{\sup_nX_n=\infty\}=0$. Now I wonder if the converse can also be shown. Namely, if $Q\{\sup_nX_n=\infty\}=0$ or $Q\{X_\infty=\infty\}=0$, does it hold that $E_P[X_\infty]=1$? Why or why not?
Any help is appreciated.
There is a standard example of a non-negative martingale $(X_n)$ with $EX_n=1$ for all $n$ converging almost surely to $X_\infty=0$. Here $Q\{X_\infty=\infty\}=0$, $Q[\sup_n X_n=\infty]=0$ and $E_P[X_\infty]=0$.