If $X$ and $Y$ are random variables on a probability space $(\Omega,\cal A,P)$ and $\beta\subset\cal A$ sub $\sigma$-algebra and $\cal C$ the smallest $\sigma$-algebra containing $\beta$ s.t. $Y$ is $\cal C$ measurable. Assume $X$ is independent of $\cal C$ then show,
$$E(XY|\beta)=E(X)E(Y|\beta)$$
I tried something but I don't know whether it is true, I think certain steps need more constraints (for example 4th equality, $Y$ needs to be non-negative or $L^1$, but in the question nothing is mentioned about that). I cannot justify some steps. Can you help ?
\begin{equation} \label{eq1} \begin{split} E(XY|\beta) & = E\left(E(XY|\cal C)\big|\beta\right) \\ & = E\left(YE(X|\cal C)\right|\beta)\\ & = E\left(YE(X|\beta)\right|\beta) \\ & = E\left(E(Y|\beta)E(X|\beta )|\beta\right) \\ & = E\left(E(Y|\beta)E(X)|\beta\right) \\ & = E(X)E\left(E(Y| \beta)|\beta\right) \\ & = E(X)E(Y|\beta) \\ \end{split} \end{equation}
You're on the right track, but as you notice the fourth equality in your attempt is suspect but not just because $Y$ need not be $L^1$ or non-negative but because $Y$ need not be $\beta$-measurable.
To salvage your work, notice that $X$ is independent of $\mathcal{C}$ so that $E[X \mid \mathcal{C}] = E[X]$. Then by the first couple of lines of your work $$E[XY \mid \beta] = E[ Y E[X \mid C] \mid \beta] = E[Y E[X] \mid \beta] = E[X] E[Y \mid \beta]$$ as desired.