Suppose $Y_n = \mathbb{E}[Y|X_1 \dots X_n]$ is a Doob's Martingale sequence . let $H_n$ be $\sigma$-algebra generate by $X_1 \dots X_n$,then if for any random variable $Z \in H_n$ we have :$\mathbb{E}[(Y_{n+1} - {Y_n})Z] =0.$ then we have $Y_n = \mathbb{E}[Y_{n+1}|H_n]$
how to see that ?
On
If $Z = \mathbf{1}_\mathrm{X}$ for $\mathrm{X} \in \sigma(X_1, \ldots, X_n)$ then $\mathbf{E}((Y_{n+1}-Y_n)Z) = 0$ signifies $\int\limits_\mathrm{X} Y_{n+1} d\mathbf{P} = \int\limits_\mathrm{X} Y_n d\mathbf{P},$ and since $Y_n$ is $\mathscr{H}_n$-measurable, the result follows upon remembering what the conditional expectation means.
By the hypothesis, $$0 = E[(Y_{n+1} - Y_n) Z] = E[E[Y_{n+1} \mid H_n]Z] - E[Y_n Z].$$ Now recall the measure-theoretic definition of the conditional expectation $E[Y_{n+1} \mid H_n]$.