Does the conditional expectation converges when the sigma-algebre converges ? e.g: $$Y \in L^{1}, (X_{n})_{n \ge 1} \xrightarrow[n \to \infty]{L^{1}} X_{\infty} \Rightarrow \mathbb{E}[Y | X_{n}] \xrightarrow[n \to \infty]{L^{1}} E[Y | X_{\infty}]$$ ?
My intuition is to say Yes: but I don't know how to properly prove it.
There are statements of the type "the conditional expectation converges if the $\sigma$-algebras converge" (e.g. Lévy's upward theorem). The implication which you stated is, however, in general false.
Consider $(0,1)$ with Lebesgue measure and $Y:=1_{(0,1/2)}$. Set $X_n=0$ if $n$ is even and $X_n = \frac{1}{n} 1_{(0,1/2)}$ if $n$ is even. Clearly, $X_n \to 0$ in $L^1$. Moreover, $\mathbb{E}(Y \mid X_n) = \mathbb{E}(Y)=\frac{1}{2}$ if $n$ is even. If $n$ is odd, then
$$\mathbb{E}(Y \mid X_n) = 1_{(0,1/2)} c_1 + 1_{[1/2,1]} c_2$$
where $$c_1 := 2\mathbb{E}(Y\cdot 1_{(0,1/2)}) \qquad c_2 = 2 \mathbb{E}(Y\cdot 1_{[1/2,1)}).$$ By definition of $Y$, we have $c_1 = 1$ and $c_2=0$; thus, $\mathbb{E}(Y \mid X_n) = 1_{(0,1/2)}$. Consequently, we have shown that
$$\mathbb{E}(Y \mid X_n) = \begin{cases} \frac{1}{2}, & \text{$n$ is even} \\ 1_{(0,1/2)}, & \text{$n$ is odd}, \end{cases}$$
and this sequence clearly fails to converge as $n \to \infty$.