Let $(X_t)_{t\ge0}$ be a stochastic process and $f=(f_t)_{t\ge 0}$ the filtration. Is the property $E[X_t-X_s|f_s]=0$ for $t>s$ of a martingale equivalent to
for every $W$ random variable depending only on the sigma-field $f_s$ we have $E[W(X_t-X_s)]=0$.
?
I think one direction could be $E[W(X_t-X_s)]=E[E[W(X_t-X_s)|f_s]]=E[W\underbrace{E[X_t-X_s|f_s]}_{=0}]=0$ where we use the pull-out property of conditional expectation. However, I am not sure how to show the other direction.
Independent of $f_s$ should be changed to measurable w.r.t $f_s$. This is what you used in the direct part.
For an arbitrary $W$ measurable w.r.t $f_s$ there is no reason why $EW(X_t-X_s)$ should exist. The condition can should be rephrased as $EW(X_t-X_s)=0$ for every bounded r.v. $W$ measurable w.r.t. $f_s$.
The converse part is very simple: Take $W=I_E$ where $E \in f_s$. You get $EX_tI_E=EX_sI_E$ for all $E \in f_s$ and the definition of conditional expectation shows that $E(X_t|f_s)=X_s$ which gives $E(X_t-X_s|f_s)=0$.