Ytukyg

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.

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
$$mathbb{P}(mathbb{E}(X_{n+1}| mathcal{F}_n) neq X_n) = mathbb{P}(bigcup_{n=1}^infty A_{1/n}) le sum_{n=1}^infty mathbb{P}(A_{1/n}) =0.$$