Let $(\Omega, \mathcal{F},P)$ be a probability space and $\mathcal{G}$ a sub-$\sigma$-algebra of $\mathcal{F}$. Let $X$ be an integrable random variable such that $E(X|\mathcal{G})\le X$ a.s. Show that $X=E(X|\mathcal{G})$ a.s.
I think that I have overthought myself into a circle on this one. I know that:
\begin{equation*}
\text{If $X$ and $Y$ are integrable r.v.'s then:}\,\, X=Y \,\,\text{a.s.} \iff \int_{A}XdP=\int_AYdP \,\,\text{for all} \,\,A\in\mathcal{F}
\end{equation*}
and I know that:
\begin{align*}
\int_GXdP=\int_GE(X|\mathcal{G})dP \,\,\text{for all}\,\,G\in\mathcal{G}
\end{align*}
I know we need to be careful using these facts as $X\in\mathcal{F}$ whereas $E(X|\mathcal{G})\in\mathcal{G}$ but I don't see any way to connect these facts with the assumption that $E(X|\mathcal{G})\le X$ a.s., so any help here would be greatly appreciated.
Best Answer
Hint:
Note that $\int_\Omega (X-E[X|{\cal G}]) d \mu = 0$ and $X(\omega)-E[X|{\cal G}](\omega) \ge 0$ for ae. $\omega \in \Omega$.