Why is it enough to check the usual martingale property in the following way?
$X_t$ is a martingale iff $s
Alternative Check for martingale property
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stochastic-processes
stochastic-calculus
martingales
1 Answers
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Let us suppose you have the proper integrability assumptions of $Y$ and $X_t$.
For the if part, given $A\in\mathbb{F}_s$, then $1_A$ is $\mathbb{F}_s$ measurable, hence by assumption, $$E[1_AX_s]=E[1_A X_t],\tag{1}$$ which is equivalent to the requirement of martingale that $$\int_{A}X_t=\int_A E[X_t\mid\mathbb{F}_s]=\int_A X_s,\quad \forall A\in \mathbb{F}_s.$$
For the only if part, suppose $X_t$ is a martingale, then (1) holds. Then similar identities hold for simple functions and integrable functions by passing to limit.