equivalence of $E[X_\infty]=1$ and $X$ is a u.i. martingale on $[0,\infty]$ Let $(X_t)$ be a strictly positive supermartingale on $[0,\infty)$. Hence $X_t$ covnerge to $X_\infty$ a.s. Now how can I show the following: $E[X_\infty]=1$ is equivalent to $(X_t)$ is a uniformly integrable martïngale on $[0,\infty]$.
hulik
 A: This result does not hold, as the following classical example shows. Let $U$ be uniform on $[0,1]$. For every $n\geqslant0$, let $U_n$ denote the integer part of $2^nU$, $\mathcal F_n=\sigma(U_n)$, and $X_n=2^n\cdot[U_n=0]$. 
Then $(\mathcal F_n)$ is a filtration and $(X_n)$ is a nonnegative martingale with respect to $(\mathcal F_n)$, starting from $X_0=1$. Furthermore, $X_n\to X_{\infty}=0$ almost surely, hence $\mathrm E(X_{\infty})=0$, and $(X_n)$ is not uniformly integrable since $\mathrm E(X_n)=1$ does not converge to $\mathrm E(X_\infty)=0$.
For a strictly positive example, add a positive constant to each $X_n$.

About the revised version: Assume first that $(X_n)$ is in fact a martingale and that $\mathrm E(X_0)=1$.
In one direction, if $\mathrm E(X_\infty)=1$, then Scheffé's lemma ensures that $X_n\to X_\infty$ in $L^1$. Any sequence in $L^1$ which converges in $L^1$ is uniformly integrable hence $(X_n)$ is uniformly integrable. In the other direction, if $X_n\to X_\infty$ in $L^1$, $\mathrm E(X_n)\to \mathrm E(X_\infty)$. Since $\mathrm E(X_n)=1$ for every $n$, $\mathrm E(X_\infty)=1$. QED.
Finally, if one only assumes that $(X_n)$ is a supermartingale, what could impose that $(X_n)$ is in fact a martingale? To wit, consider the deterministic example $X_n=1+\frac1{n+1}$.
