# 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?

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. $$\mathbb{E}(M_t \mid \mathcal{F}_s) =M_s$$.

$$EM_t$$ 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.