From randomservices.org, a central property in conditional expectation is $$ E[r(X) E[Y|X]] = E[r(X) Y] $$ where $X$ is a random variable taking values in $S \subseteq \mathbb{R}^n$, $Y$ is a random variable taking values in $T \subseteq \mathbb{R}$, and $r:S\to\mathbb{R}$ is any function.
How can we use this to prove the following "taking out what is known" rule: $$ E[s(X) Y | X] = s(X) E[Y|X] $$ for some function $s:S\to\mathbb{R}$.
I see that for any particular value $x \in S$, we have $s(x) E[Y|X]$ by stepping through the integrals, but I'm not sure how to use the central property to show this for random variable $X$. My attempt is below.
Consider $E[s(X) Y | X]$ as a random variable $f(X)$. Then by the central property, we have that for any $r:S\to\mathbb{R}$ \begin{align*} E[f(X) \, r(X) \, E[Y|X] ] = E[r(X) \, s(X) \, Y]. \end{align*} Choose $r(X) = 1 | X$ so we have \begin{align*} E[f(X) \, 1 \, E[Y|X] \mid X] = E[1 \, s(X) \, Y \mid X]. \end{align*} Intuitively I want to pull $s(X)$ outside of the expectation, but I'm not sure of the appropriate way of justifying/doing that mathematically.
Use the definition of conditional expectations. For any $A\in\mathcal{B}(S)$, $$ \mathsf{E}[s(X)\mathsf{E}[Y\mid X]\times 1_A(X)]=\mathsf{E}[s(X)Y\times 1_A(X)], $$ where we take $r(X)=s(X)1_A(X)$ in the "central property".