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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.

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    $\begingroup$ This seems similar to the fact that if $Z$ is a nonnegative random variable with $E[Z]=0$, then $P[Z=0]=1$. $\endgroup$
    – Michael
    Jun 21, 2020 at 15:43
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    $\begingroup$ You are overthinking it :-). This is essentially the same as if $EX \le X(\omega)$ ae. then $X(\omega) = EX$ ae. $\endgroup$
    – copper.hat
    Jun 21, 2020 at 16:15
  • $\begingroup$ Bonus question: what happens if we have instead $E[X|G] \ge X$ ? $\endgroup$
    – Thomas
    Jun 21, 2020 at 17:08

2 Answers 2

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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. $\omega \in \Omega$.

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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.

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