Show $E[Y_{n}\mid \mathcal{F}_{n}] \xrightarrow{L^{1}} E[Y\mid \mathcal{F}_{\infty}]$

Question

Let $Y_{n} \xrightarrow{L^{1}} Y$ and let $\mathbb F$ be the corresponding filtration where $\mathcal{F}_{n} \to \mathcal{F}_{\infty}$. Show $$E[Y_{n}\mid \mathcal{F}_{n}] \xrightarrow{L^{1}} E[Y\mid \mathcal{F}_{\infty}]$$

My idea:

$E[\vert E[Y_{n}\mid \mathcal{F}_{n}] - E[Y\mid \mathcal{F}_{\infty}]\vert ]$ and we know that $E[Y\mid \mathcal{F}_{\infty}]=\lim\limits_{n \to \infty} E[Y\mid \mathcal{F}_{n}]$.

Hence, \begin{align} E[\vert E[Y_{n}\mid \mathcal{F}_{n}] - E[Y\mid \mathcal{F}_{\infty}]\vert ] &=E[\vert E[Y_{n}\mid \mathcal{F}_{n}] - \lim_{n \to \infty}E[Y\mid \mathcal{F}_{n}]\vert ] \\&= E[\liminf\limits_{m \to \infty}\vert E[Y_{n}\mid \mathcal{F}_{n}] - E[Y\mid \mathcal{F}_{m}]\vert ] \\&\leq \liminf\limits_{m \to \infty}E[\vert E[Y_{n}\mid \mathcal{F}_{n}] - E[Y\mid \mathcal{F}_{m}]\vert ] \\&=\liminf\limits_{m \to \infty}E[\vert E[\vert E[Y_{n}\mid \mathcal{F}_{n}]-Y\mid \mathcal{F}_{m}]\vert ] \end{align}

Am I on the right track?

Answer 1

Your argument does not seem to work.

$E|E(Y_n|\mathcal F_n)-E(Y|\mathcal F_n)|\leq E(E|Y_n-Y||\mathcal F_n) =E|Y_n-Y| \to 0$ and $E|E(Y|\mathcal F_n)-E(Y|\mathcal F)| \to 0$. Hence $E|E(Y_n|\mathcal F_n)-E(Y|\mathcal F)| \to 0$.