Convergence of a conditional expectation Consider a filtered probability space ($\Omega,\mathcal{F}_t,\mathcal{F},\mathbb{P}$) and a random variable $X$ defined on $\Omega$ with values in a set $E$.
We consider the process $X_t = \mathbb{E}[ X | \mathcal{F}_t]$ and the question is the following:
What conditions should be put on $\mathcal{F_t}$ so that $X_t$ has continuous paths $\mathbb{P}$-a.s. That is to say 
$X_t = \lim_{s\rightarrow t} X_s$ a.s
Thank you! 
 A: Obviously, $(X_t,\mathcal{F}_t)_{t \geq 0}$ is a uniformly integrable martingale. Fix $t>0$. By the martingale convergence theorem, there exists $Y \in L^1$ such that
$$X_s = \mathbb{E}(X \mid \mathcal{F}_s) \to Y, \qquad s \uparrow t$$
in $L^1$ and almost surely. Using the tower property, we find
$$\mathbb{E}(Y \mid \mathcal{F}_s) = X_s = \mathbb{E}(X_t \mid \mathcal{F}_s), \qquad s<t$$
and therefore
$$\int_F Y \, d\mathbb{P} = \int_F X_t \, d\mathbb{P}$$
for any $F \in \mathcal{F}_s$. This implies
$$\int_F Y \, d\mathbb{P} = \int_F X_t \, d\mathbb{P}$$
for any $F \in \sigma(F_s; s <t)=: \mathcal{F}_{t-}$. As $Y$ is $\mathcal{F}_{t-}$-measurable, we conclude
$$Y = \mathbb{E}(X_t \mid \mathcal{F}_{t-})$$
This yields the (necessary and sufficient) condition for left-continuity of the process $(X_t)_{t \geq 0}$:
$$X_t \stackrel{!}{=} Y=\mathbb{E}(X_t \mid \mathcal{F}_{t-}) = \mathbb{E}(X \mid \mathcal{F}_{t-}) \tag{1}$$
A similar reasoning shows that
$$X_t \stackrel{!}{=} \mathbb{E}(X \mid \mathcal{F}_{t+}) \tag{2}$$
with $\mathcal{F}_{t+} := \bigcap_{s>t} \mathcal{F}_s$ ensures the right-continuity of $(X_t)_{t \geq 0}$.
Consequently, we conclude that $(X_t)_{t \geq 0}$ is (a.s.) continuous if and only if $(1)$ and $(2)$ are satisfied. This is in particular the case if
$$\mathcal{F}_{t-} = \mathcal{F}_t = \mathcal{F}_{t+}$$
for any $t \geq 0$.
