Monotonical Lévy's Upward Theorem By Lévy's Upward Theorem, we know that if $f \in \mathcal{L}^1(\Omega, \mathcal{F}, P)$ and  $(\mathcal{F}_n)_{n\in \mathbb{N}}$ is a filtration of $\mathcal{F}$ and $\mathcal{F}_{\infty} = \sigma(\bigcup_n \mathcal{F}_n)$, then
$$ \mathbb{E}[f\mid \mathcal{F}_n] \to \mathbb{E}[f\mid \mathcal{F}_\infty] \qquad\text{ in } \mathcal{L}^1$$
Now, my question is, does this sequence converge $\mathcal{L}^1$-monotonically to its limit? Or better put, is
$$\big\lVert \mathbb{E}[f\mid \mathcal{F}_n] - \mathbb{E}[f\mid \mathcal{F}_\infty] \big\rVert$$
a decreasing sequence?
My intuition is that $\mathbb{E}[f\mid \mathcal{F}_n]$ gives a better and better approximation of $f$ as $\mathcal{F}_n$ gets larger.
 A: The following example might be instructive.
Suppose that $\mathcal F=\mathcal F_\infty$ is generated by a partition $\{B_n\}_{n=1}^\infty$ of $\Omega$ with $\Bbb P(B_n)>0$ for each $n$.
Let $\mathcal F_n:=\sigma(B_1,B_2,\ldots,B_n)$.
A function $X:\Omega\to\Bbb R$ is $\mathcal F$-measurable iff there are reals $b_k$ such that
$$
X(\omega) =\sum_{k=1}^\infty b_k 1_{B_k}(\omega)\qquad \forall \omega\in\Omega.
$$
If such an $X$ is to be integrable we must have
$$\sum_{k=1}^\infty |b_k|\cdot\Bbb P(B_k)<\infty,
$$
and in this case we have
$$
\Bbb E[X\mid\mathcal F_n] =\sum_{k=1}^n b_k1_{B_k}+1_{G_n}\cdot\Bbb E[X|G_n],
$$
where $G_n:=\cup_{k=n+1}^\infty B_k$ and $\Bbb E[X|G_n]$ is the "elementary" conditional expectation
$$
\Bbb E[X|G_n]=\Bbb E[X\cdot 1_{G_n}]/\Bbb P(G_n)=\sum_{k=n+1}^\infty b_k P(B_k)/P(G_n).
$$
Therefore
$$
X-\Bbb E[X\mid\mathcal F_n]=\sum_{k=n+1}^\infty b_k1_{B_k}-\sum_{k=n+1}^\infty b_k P(B_k)/P(G_n).
$$
The disjointness of the $B_k$ makes it easy to calculate the mean of the magnitude of this last expression; it is
$$
\|X-\Bbb E[X\mid \mathcal F_n]\|_1= \sum_{k=n+1}^\infty|c_k|+\Big|\sum_{k=n+1}^\infty c_k\Big|,
$$
where $c_k:=b_k\cdot\Bbb P(B_k)$. By the triangle inequality
$$
\Big|\sum_{k=n+1}^\infty c_k\Big|\le|c_n|+\Big|\sum_{k=n}^\infty c_k\Big|,
$$
so
$$
\|X-\Bbb E[X\mid \mathcal F_{n-1}]\|_1= \sum_{k=n}^\infty|c_k|+\Big|\sum_{k=n}^\infty c_k\Big|\ge \sum_{k=n}^\infty|c_k|+\Big|\sum_{k=n+1}^\infty c_k\Big|-|c_n| =\|X-\Bbb E[X\mid \mathcal F_{n}]\|_1.
$$
That is, $\|X-\Bbb E[X\mid \mathcal F_{n}]\|_1$ is indeed decreasing in $n$.
It's not clear to me whether the conclusion is true in general.
