Positive continuous supermartingale is a proper martingale Let $M$ be continuous positive supermartingale with $\mathbb{E}[M_0]< \infty$. By the supermartingale convergence theorem $M_\infty = \lim M_t$ exists almost surely. How do I show that, if $\mathbb{E}[M_0]=\mathbb{E}[M_\infty]$, $M$ is a proper martingale?
 A: By Fatou's lemma,
$$\mathbb{E}(M_0) = \mathbb{E}(M_{\infty}) \leq \liminf_{t \to \infty} \mathbb{E}(M_t).$$
Since $(M_t)_{t \geq 0}$ is a supermartingale, we have $\mathbb{E}(M_t) \leq \mathbb{E}(M_T) \leq \mathbb{E}(M_0)$ for all $t \geq T$, and so
$$\mathbb{E}(M_0) = \mathbb{E}(M_{\infty}) \leq \mathbb{E}(M_T) \leq \mathbb{E}(M_0).$$
Thus,
$$\mathbb{E}(M_0) = \mathbb{E}(M_{\infty}) = \mathbb{E}(M_T)$$
for all $T \geq 0$, i.e. the supermartingale has constant expectation.
By the supermartingale property,
$$\int_F M_t \, d\mathbb{P} \leq \int_F M_s \, d\mathbb{P} \tag{1}$$
for all $s \leq t$ and $F \in \mathcal{F}_s$. Replacing $F$ by $F^c$ we get
$$\underbrace{\mathbb{E}(M_t)}_{=\mathbb{E}M_0} - \int_F M_t \, d\mathbb{P} \leq \underbrace{\mathbb{E}(M_s)}_{\mathbb{E}(M_0)} - \int_F M_s \, d\mathbb{P},$$
i.e.
$$\int_F M_t \, d\mathbb{P} \geq \int_F M_s \, d\mathbb{P} \tag{2}$$
Combining $(1)$ and $(2)$ gives
$$\int_F M_t \, d\mathbb{P} = \int_F M_s \, d\mathbb{P}, \qquad F \in \mathcal{F}_s, s \leq t,$$
i.e. $\mathbb{E}(M_t \mid \mathcal{F}_s) =M_s$.
A: $EM_t$ is decreasing by super-martingale property. (Just take expectation in the definition). 
Thus $EM_{\infty}=E\lim M_t \leq \lim \inf EM_t\leq EM_t \leq EM_)$ for all $t$ where I have used Fatou's lemma for the inequality. It follows now that $EM_0=EM_{\infty} \leq EM_t \leq EM_0$ for al $t$ which implies that $EM_0=EM_{\infty}=EM_t$ for all $t$. 
Finally $E(M_{t+s}|F_s) \leq M_t$ and the non-negative random variable $M_t-E(M_{t+s}|F_s) $ has mean $0$ by what we just proved. Hence $M_t=E(M_{t+s}|F_s) $ almost surely. Thus $(M_t)$ is  a martingale. 
