Martingale in probability and Statistics Let $(X_n)$ be a supermartingale such that $EX_n$ is constant. show that $X_n$ is a martingale. My problem here in this question is how can I use the constant expectation concept given to prove the statement.
 A: We have $\mathbb{E}(X_{n+1}| \mathcal{F}_n) \le X_n $, because $(X_n)_{n \in \mathbb{N}}$ is a supermartingale. Let $$A_\varepsilon = \{ \mathbb{E}(X_{n+1}| \mathcal{F}_n)+\varepsilon \le X_n\},$$
then
$$ \int_{A_\varepsilon} X_{n+1} \, d \mathbb{P}+ \varepsilon \mathbb{P}(A_\varepsilon) = \int_{A_\varepsilon} \mathbb{E}(X_{n+1}| \mathcal{F}_n) \, d \mathbb{P} + \varepsilon \mathbb{P}(A_\varepsilon) \le  \int_{A_\varepsilon} X_n \, d \mathbb{P}.$$
On the other hand
$$\int_{A_\varepsilon^c} X_{n+1} \, d \mathbb{P} = \int_{A_\varepsilon^c} \mathbb{E}(X_{n+1}| \mathcal{F}_n) \, d \mathbb{P} \le \int_{A_\varepsilon} X_n \, d \mathbb{P}.$$
Adding both, we get
$$\mathbb{E}[X_n]+\varepsilon \mathbb{P}(A_\varepsilon)  =\mathbb{E}[X_{n+1}]+\varepsilon \mathbb{P}(A_\varepsilon)  \le \mathbb{E}[X_n].$$
Thus $\mathbb{P}(A_\varepsilon) =0$ for any $\varepsilon >0$. Finally, by $\sigma$-additivity we find that
$$\mathbb{P}(\mathbb{E}(X_{n+1}| \mathcal{F}_n) \neq X_n) = \mathbb{P}(\bigcup_{n=1}^\infty A_{1/n}) \le \sum_{n=1}^\infty \mathbb{P}(A_{1/n}) =0.$$ 
