When does $\mathbf{E}(XY\mid\mathcal{A})=X\mathbf{E}(Y\mid\mathcal{A})$ hold?

Question

Les $(\Omega,\mathcal{B},\mathbb{P})$ be a probability space, and let $\mathcal{A}\subset\mathcal{B}$ be a $\sigma$-subalgebra. One of the basic properties of conditional expectation (see : pulling out known factors) is the following

Theorem 1. If $X\in L^\infty(\mathcal A)$ and $Y\in L^1(\mathcal B)$, then $$\mathbf{E}(XY\mid\mathcal{A})=X\mathbf{E}(Y\mid\mathcal{A})$$

This isn't the most general form of this property : if $(M_i)_{i\geq 0}$ is a discrete time, square integrable martingale relative to some filtration $(\mathcal{A}_i)_{i\geq 0}$, one comes across the following computation : for any $i\geq 1$ $$\begin{array}{rcl} \mathbf{E}(M_iM_{i-1}) & = & \mathbf{E}(\mathbf{E}(M_iM_{i-1}\mid\mathcal{A}_{i-1})) \\ & = & \mathbf{E}(\mathbf{E}(M_i\mid\mathcal{A}_{i-1})M_{i-1})\\ & = & \mathbf{E}(M_{i-1}^2). \end{array}$$ The previous theorem may not be sufficient to pull out $M_{i-1}$ (as $M_{i-1}$ may be unbounded), but a straightforward argument involving the density of $L^\infty(\mathcal A)\subset L^p(\mathcal A)$ and continuity of $L^p(\mathcal A)\to\mathcal{L}(L^q(\mathcal B),L^1(\mathcal B)),\quad X\mapsto (X\cdot -)$ shows that

Theorem 2. If $X\in L^p(\mathcal A)$ and $Y\in L^q(\mathcal B)$, then $$\mathbf{E}(XY\mid\mathcal{A})=X\mathbf{E}(Y\mid\mathcal{A})$$

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Question : Suppose $(\Omega,\mathcal{B},\mathbb{P})$ is a probability space, and $\mathcal{A}\subset\mathcal{B}$ is a $\sigma$-subalgebra of $\mathcal{B}$. Let $X:(\Omega,\mathcal{A})\to(\mathbb{R},\mathcal{B}(\mathbb{R}))$ and $Y:(\Omega,\mathcal{B})\to(\mathbb{R},\mathcal{B}(\mathbb{R}))$ be measurable functions

1. When does $\mathbf{E}(Y\mid\mathcal{A})$ make sense ?
2. When is the equation $\mathbf{E}(XY\mid\mathcal{A})=X\mathbf{E}(Y\mid\mathcal{A})$ true ?

Regarding the first point : I know conditional expectation is defined when either $Y\geq 0$ or $Y\in L^1$. It should thus make sense more generally if either $Y^+$ or $Y^-$ has finite expectation. Is this the broadest context wherein it is defined ?

Answer 1

Regarding 1.: Recall the definition of conditional expectation. One point was $$\Bbb{E}[\Bbb{1}_AY] = \Bbb{E}[\Bbb{1}_A\Bbb{E}[Y|\mathcal{A}]] $$ for all $A\in \mathcal{A}$. If neither $Y \geq 0$ nor $Y\in L^1$ holds, what is - for example - $\Bbb{E}[\Bbb{1}_\Omega Y] = \Bbb{E}[ Y]$ supposed to be? This integral does not have to be defined then.

Regarding 2.: The equation holds, when $X$ ist $\mathcal{A}$-measurable. However, conversely $X$ can be not-$\mathcal{A}$-measurable, but is fulfilling the equation, for example with $Y=0$.