Why are these two definitions of Martingales equivalent? I am recently reading books on Probability theory, in Durrett's book Probability: Theory and Examples, the definition is following:

If $X_{n}$ is sequence with
(1) $\mathbb{E}\left|X_{n}\right|<\infty$
(2) $X_{n} \text { is adapted to } \mathcal{F}_{n}$
(3) $\mathbb{E}\left(X_{n+1} | \mathcal{F}_{n}\right)=X_{n} \text { for all } n$
Then $X$ is said to be a martingale (with respect to $\mathcal{F}_n$ )

And, in Erhan Cinlar's book Probability and Stochastics, the definition is

A real-valued stochastic process $X=\left(X_{t}\right)_{t \in \mathbb{T}}$ ($\mathbb{T}$ is a subset of $\overline{R}$.) is called $\text {an } \mathcal{F} \text { -submartingale if } X \text { is adapted to } \mathcal{F}, \text { each } X_{t} \text { is integrable, and}$ $$\mathbb{E}_{s}\left(X_{t}-X_{s}\right) \geq 0$$
$\text { whenever } s<t$, $\text {It is called an } \mathcal{F} \text { -supermartingale if }-X \text { is an } \mathcal{F} \text {-submartingale}$ $\text { and an } \mathcal{F}\text{-martingale} $ is it's both $\mathcal{F} \text { -supermartingale}$ and $\mathcal{F} \text { -submartingale}$.

From first looks I don't see the connection immediately, can you please illuminate me on this?
 A: The second definition is slightly more general, but if you take $\mathbb{T}=\mathbb{N}$ then they are the same. Note that their definitions of adapted are the same ($X_t$ is measurable with respect to $\mathcal{F}_t$ for all $t$). And saying $X_t$ is integrable is the same a saying $\mathbb{E}|X_t|<\infty$. So the only part that really needs to be checked for equivalence is Durrett's condition 3, and the centered equation in Çinlar's definition. Also note that Çinlar's notation $\mathbb{E}_t[\cdot]$ is defined to mean $\mathbb{E} [\cdot \mid \mathcal{F}_t]$.
A consequence of Durrett's definition is that for all $s<t$, $\mathbb{E}[X_t \mid \mathcal{F}_s] = X_s$. Also note that $\mathbb{E}[X_s \mid \mathcal{F}_s] = X_s$, so by linearity of conditional expectation $\mathbb{E}[X_t-X_s \mid \mathcal{F}_s]=0$. Therefore $\{X_t\}$ is a submartingale by Çinlar's definition. But we also have $\mathbb{E}[(-X_t)-(-X_s) \mid \mathcal{F}_s] = 0$, so $\{-X_t\}$ is a submartingale by Çinlar's definition, i.e., $\{X_t\}$ is a supermartingale. So $\{X_t\}$ is both a sub- and supermartingale, so it is a martingale.
Going the other direction, suppose $\{X_t\}$ is a martingale. Then take $s=n$, $t=n+1$. Since our sequence is a submartingale, $\mathbb{E}[X_{n+1}-X_n \mid \mathcal{F}_n] \geq 0$, i.e., $\mathbb{E}[X_{n+1} \mid \mathcal{F}_n] \geq \mathbb{E}[X_n\mid \mathcal{F}_n]=X_n$. Since the negative of our sequence is a supermartingale, $\mathbb{E}[(-X_{n+1})-(-X_n) \mid \mathcal{F}_n] \geq 0$, i.e., $\mathbb{E}[X_{n+1}\mid \mathcal{F}_n]\leq \mathbb{E}[X_n\mid \mathcal{F}_n]=X_n$. So $\mathbb{E}[X_{n+1}\mid \mathcal{F}_n]=X_n$ for all $n$, and so we have a martingale by Durrett's definition.
A: From Cinlar's definition, if $X_t$ is a martingale, it is both a supermartingale and a submartingale.
The submartingale property implies $E_s[X_t - X_s] \ge 0$, and the supermartingale property implies $E_s[-(X_t - X_s)] \ge 0$, and thus $E_s[X_t - X_s] = 0$. Since $X_s \in \mathcal{F}_s$, this can be rewritten as $E_s[X_t] = X_s$. (Presumably, $E_s[\cdot]$ in Cinlar's notation is $E[\cdot \mid \mathcal{F}_s]$ in Durrett's notation?)
