Let $M_t$ be a non-negative supermartingale, $EM_0 < \infty$ and $M_t \to M_\infty$ a.s. I want to show that if $E M_\infty = EM_0$, then $M$ is a uniformly integrable martingale. I thought about doing so by showing that $M_t = E ( M_\infty | \mathcal{F}_t )$ but couldn't figure out the necessary steps. Any hints? Just somewhere to start would be great.
Edit: An additional question: Can I follow from the above that $E (sup_{t\geq0} M_t) < \infty$? That would also give uniform integrability, right?
Best Answer
In a supermartingale $EM_t$ is decreasing. If $EM_{\infty} =EM_0$ then $EM_t$ is independent of $t$. For $s >t$ the non-negative random variable $M_t-E(M_s|\mathcal F_t)$ has mean $0$ and hence it vanishes a.s. This proves that $(M_t)$ is a martingale. Also, $M_t \to M_{\infty}$ a.s and $EM_t \to EM_{\infty}$ together imply convergence in $L^{1}$ because of non-negativity. [See https://en.wikipedia.org/wiki/Scheff%C3%A9%27s_lemma ] This also implies uniform integrability.