# Conditional expectation characterization of sub-$\sigma$-algebra measurability

Suppose $$(\Omega,\mathscr{F},\mathbb{P})$$ is a complete probability space and $$\mathscr{G}$$ is a sub-$$\sigma$$-algebra of $$\mathscr{F}$$. I with to show that a random variable $$X\in\mathcal{L}^1$$ is $$\mathscr{G}-$$measurable if and only if $$E[X\eta]=E[XE[\eta|\mathscr{G}]],\forall\eta\in\mathcal{L}^\infty.$$ The "only if" statement follows from the definition of conditional expectation and the use of standard method (simple function approximation), and I tried to show the "if" statement by showing $$\sigma(X)\subset\mathscr{G}$$. However, I got stuck in doing so, and wish to know how I may show $$\sigma(X)\subset\mathscr{G}$$?

• I have an incomplete idea, which might help you. Imagine that $X \in \mathcal{L}^2$, so that $E[-|\mathscr{G}]=p$ is the orthogonal projection $p$ onto the subspace $\mathcal{L}^2(X, \mathscr{G}) \subset \mathcal{L}^2(X, \mathscr{F})$. Then you condition says that $(X, \eta-p(\eta))=0$ for all $\eta$, so $X \in \mathcal{L}^2(X, \mathscr{G})^{\perp \perp}$. Commented Apr 16, 2022 at 3:49
• @NicolásVilches Your argument is almost complete. All you need is the fact that $L^{\infty}$ is dense in $L^{2}$. Commented Apr 16, 2022 at 4:49

For any $$\eta \in L^{\infty}$$, $$E(X\eta) = E(XE(\eta \mid \mathscr{G})) = E(E(X \mid \mathscr{G})E(\eta \mid \mathscr{G})) = E(E(X \mid \mathscr{G})\eta).$$ Taking $$\eta$$ to be suitable indicator functions, this implies that $$X = E(X \mid \mathscr{G})$$ a.s.. So $$X$$ is a.e. equal to a $$\mathscr{G}$$-measurable function.
• Is it to choose $\eta_1=1_{X\geq E(X|\mathscr{G})},\eta_2=-1_{X<E(X|\mathscr{G})}$?Since $E((X-E(X|\mathscr{G}))\eta)=0$ and $(X-E(X|\mathscr{G}))\eta\in\mathcal{L}_0^+$, we have $X=E(X|\mathscr{G})$ a.e. on both $\{X\geq E(X|\mathscr{G})\}$ and $\{X<E(X|\mathscr{G})\}$? Commented Apr 17, 2022 at 9:59