Can we show that $1_A\text E[X\mid\mathcal F]=1_AY\Leftrightarrow\forall F\in\mathcal F:\text E[1_A1_FX]=\text E[1_A1_FY]?$

Question

Let

• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
• $\mathcal F\subseteq\mathcal A$ be a $\sigma$-algebra on $\Omega$
• $X\in\mathcal L^1(\operatorname P)$
• $A\in\mathcal A$
• $Y:\Omega\to\mathbb R$ be $\mathcal A$-measurable with $\operatorname E\left[1_A|Y|\right]<\infty$

Are we able to show that $$1_A\operatorname E\left[X\mid\mathcal F\right]=1_AY\text{ almost surely}\Leftrightarrow\forall F\in\mathcal F:\operatorname E\left[1_A1_FX\right]=\operatorname E\left[1_A1_FY\right]\tag1?$$

I'm especially unsure whether we need to impose any stronger measurability assumption on $Y$ and how $(1)$ is related to $$\operatorname E\left[\left.X\right|_A\mid\left.\mathcal F\right|_A\right]=\left.Y\right|_A\;\left.\operatorname P\right|_A\text{-almost surely}\Leftrightarrow\left.Y\right|_A=\left.\operatorname E\left[X\mid\mathcal F\right]\right|_A\;\left.\operatorname P\right|_A\text{-almost surely}\tag2.$$

In order to prove "$\Rightarrow$" in $(1)$ we obviously need to use that, by definition, $\forall F\in\mathcal F:\operatorname E\left[1_FX\right]=\operatorname E\left[1_A\operatorname E\left[X\mid\mathcal F\right]\right]$ (it would be clear, if $A\in\mathcal F$; do we need this?). And for "$\Leftarrow$" we somehow to argue with measure uniqueness. But I struggle to fill out the details in both directions.

Answer 1

In general, (1) fails: let $\mathcal F$ be the $\sigma$-algebra containing $\emptyset$ and $\Omega$. Then (1) reads $$ 1_A\operatorname E\left[X \right]=1_AY\text{ almost surely}\Leftrightarrow \operatorname E\left[1_A X\right]=\operatorname E\left[1_A Y\right], $$ which is not true if $A=\Omega$, $X$ and $Y$ have the same expectation but $Y$ is not almost surely constant.