Background. (TL:DR you can skip to Question. below.) This is a followup question to one of my previous questions (linked here) on this website.
In short, the other question was about how to express the conditional expectation with respect to a simple random variable $Y$ in terms of a polynomial in $Y$. More specifically, suppose that $Y$ can be written as \begin{align}\tag{1} Y=\sum_{i=1}^na_i\chi_{A_i}, \end{align} where the $A_i$ partition the sample space into sets of positive measure. Then, given any random variable $X$, one has \begin{align}\tag{2}\mathbb E\big[X \, \big| \, Y\big]=\sum_{i=1}^n\left(\mathbb E\big[X \, \big| \, Y=a_i\big]\prod_{j\neq i}\frac{(Y-a_j)}{(a_i-a_j)}\right).\end{align}
After providing an answer to my question, angryavian questioned the usefulness of writing the conditional expectation in those terms.
After thinking about it for some time, I had an idea: Given that, in any $L^p$ space, a random variable $Y$ may be approximated in the $\|\cdot\|_{p}$ by simple random variables (i.e., random variables that can be written as a sum like equation $(1)$), maybe one could approximate the random variable $\mathbb E\big[X \, \big| \, Y\big]$ using the polynomials $$\mathbb E\big[X \, \big| \, Y_n\big]=\sum_{i=1}^n\left(\mathbb E\big[X \, \big| \, Y_n=a_i\big]\prod_{j\neq i}\frac{(Y_n-a_j)}{(a_i-a_j)}\right),$$ where $\{Y_n\}_n$ is a sequence of simple random variables that converges to $Y$ in a suitably chosen $L^p$-norm.
This then leads me to the following question:
Question. If $Y_n$ converges to $Y$ in distribution, in probability, almost surely, or with respect to some $L^p$ norm, then does $\mathbb E\big[X \, \big| \, Y_n\big]$ also converge to $\mathbb E\big[X \, \big| \, Y\big]$ in distribution, in probability, almost surely, or with respect to some $L^p$ norm?
Every convergence result I have seen with conditional expectation is about $\mathbb E[X_n \mid Y]\to\mathbb E[X_n \mid Y]$ where $X_n\to X$, not the other way around.
Furthermore, given that $\mathbb E[X \mid Y]$ is actually defined as $\mathbb E[X \mid \mathcal A]$, where $\mathcal A$ is the $\sigma$-algebra generated by $Y$, I'm having trouble imagining what a statement like
$\mathbb E[X \mid \mathcal A_n]\to\mathbb E[X \mid \mathcal A]$ when $\mathcal A_n\to\mathcal A$
should mean, given that I'm not aware of any convergence concepts for $\sigma$-algebras.
$\mathcal{A}:=\sigma\left(\bigcup_{n\in\mathbb{N}} \mathcal{A}_n\right)$ is the $\sigma$-algebra that contains all the $\sigma$-algebras: $\mathcal{A}_1,\mathcal{A}_2,\dots$ so there is really no general "convergence" of $\sigma$-algebras.