Suppose that we have a probability space $(\Omega, \mathscr{F},P)$ and a sub-$\sigma-$algebra $\mathscr{G}$ and we want to show a mapping $X$ is $\mathscr{G}-$measurable. Why does it suffice to show that $E[XH]=0$ for all bounded random variable $H$ such that $E[H|\mathscr{G}]=0$?
This result is implicitly used in the proof below from Protter's Stochastic Integration and Differential Equations but I am having difficulty showing it.
Answer assuming that $EX^{2} <\infty$: Let $H=XI_{|X| \leq N} -E(XI_{|X| \leq N}|\mathcal G)$. Then $H$ is bounded and $E(H|\mathcal G)=0$. Hence $E(HX)=0$. This gives $EX^{2}I_{|X| \leq N}=E[XE(XI_{|X| \leq N}|\mathcal G)] $. Leting $N \to \infty$ we get $EX^{2}=E[XE(X|\mathcal G)] $. Now expand $E[X-E(X|\mathcal G)]^{2}$ and use Jensen's inequality to see that this is $ \leq 0$. Hence $X=E(X|\mathcal G)$ a.s. and $X$ is a.s. and so $X$ is $\mathcal G$ measurable.
Jensen's inequality is used as follows: $E[E(X|\mathcal G)]^{2}\leq E [E(X^{2}|\mathcal G)]=EX^{2}$