Show that if $E(X|G)\le X$ a.s. then $X=E(X|G)$ a.s. 
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.
 A: 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. $\omega \in \Omega$.
A: Ok, with the help of @Michael and @copper.hat, I believe I have a solution:
Note by assumption the two facts:
\begin{align}
&(1)\,\, E(X|\mathcal{G})\le X \,\,\text{a.s} \,\,\implies \,\, 0\le X-E(X|\mathcal{G})\,\,\text{a.s.}\\
&(2)\,\,\text{as $\Omega\in\mathcal{G}$}: \int_{\Omega}XdP=\int_{\Omega}E(X|\mathcal{G})dP\,\,\implies\,\,\int_{\Omega}(X-E(X|\mathcal{G}))dP=0
\end{align}
Now for each $n\in\mathbb{N}$ define: $A_n=\{X-E(X|\mathcal{G})>\frac{1}{n}\}$, then:
\begin{align}
\frac{1}{n}\mathbb{1}_{A_n}\le(X-E(X|\mathcal{G}))\mathbb{1}_{A_n}\le X-E(X|\mathcal{G})
\end{align}
and thus running expectation through, we have:
\begin{align}
\frac{1}{n}P(A_n)\le E(X-E(X|\mathcal{G}))=\int_{\Omega}(X-E(X|\mathcal{G}))dP=0
\end{align}
This implies: $P(A_n)=0$ for all $n\ge1$ and thus we have:
\begin{align}
0=\lim_{n\to\infty}P(A_n)&=\lim_{n\to\infty}E\mathbb{1}_{A_n}\\
&=E(\lim_{n\to\infty}\mathbb{1}_{A_n}) \quad\text{by the DCT}\\
&=E\mathbb{1}_{\{X-E(X|\mathcal{G})>0\}}\\
&=P(X-E(X|\mathcal{G})>0)\\
&\iff P(X-E(X|\mathcal{G})=0)=1\\
\end{align}
Thus, $X-E(X|\mathcal{G})=0$ a.s. and so $X=E(X|\mathcal{G})$ a.s., as we wished to show.
