Does $E(X|\mathcal{F})\ge Y$ imply $X\ge Y$ a.e., if $Y\in\mathcal{F}$

Question

I'd like to show the claim in the title.

I was thinking we could consider the set $A=\{X<Y\}$; if $A\in\mathcal{F}$, letting $Z=E(X|\mathcal{F})$, we'd get

$$\int_A Z-Y=\int_A X-Y\le0,$$ so noting additionally that $\int_A (X-Y)\ge 0$ (by hypothesis), we get $P(A)=0$. But since $X$ is not assumed to be $\mathcal{F}$ measurable, this argument doesn't work.

Does conditional expectation (as a random variable) immediately tell us about the unconditional random variable (something like including the notion of conditioning on nothing)? That is, does $E(X|\mathcal{F})\ge Y$ also tell us $X\ge Y$ a.s.?

Answer 1

no

let $F$ be trivial. Let $X$ be $1,-1$ equal probabilities. Let $Y$ be zero.