Prove/Disprove probability of 0 or 1 (almost surely) will never change and has never been different Prove/Disprove $E[1_A | \mathscr{F_t}] = 0 \ \text{or} \ 1 \ \text{a.s.} \ \Rightarrow E[1_A | \mathscr{F_{s}}] = E[1_A | \mathscr{F_t}] \ \text{a.s.}$

Given a filtered probability space $(\Omega, \mathscr{F}, \{\mathscr{F}_n\}_{n \in \mathbb{N}}, \mathbb{P})$, let $A \in \mathscr{F}$.
Suppose $$\exists t \in \mathbb{N} \ \text{s.t.} \ E[1_A | \mathscr{F_t}] = 1 \ \text{a.s.}$$ Does it follow that $$E[1_A | \mathscr{F_{s}}] = E[1_A | \mathscr{F_t}] \ \text{a.s.} \ \forall s > t \ ?$$ What about $\forall s < t$?
What if instead $$\exists t \in \mathbb{N} \ \text{s.t.} \ E[1_A | \mathscr{F_t}] = 0 \ \text{a.s.} \ ?$$ Or what if $$E[1_A | \mathscr{F_t}] = p \ \text{a.s.} \ \text{for some} \ p \in (0,1) \ ?$$

What I tried:

If $\Bbb E[1_A|\mathscr F_t]=1$, then $\Bbb E[1_A]=1$, which is the same as $1_A=1$ (almost surely).  In this case $\Bbb E[1_A|\mathscr F_s]=1$ (almost surely) for each $s$.
Likewise, if $\Bbb E[1_A|\mathscr F_t]=0$, then $\Bbb E[1_A]=0$, which is the same as $1_A=0$ (almost surely).  In this case $\Bbb E[1_A|\mathscr F_s]=0$ (almost surely) for each $s$.
If $\Bbb E[1_A|\mathscr F_t]=p$, for a constant $p\in(0,1)$, then we have
$\Bbb E[1_A|\mathscr F_s]=E[E[1_A|\mathscr F_t]|\mathscr F_s] = E[p|\mathscr F_s] = p$. This may fail if $s>t$.
Alternatively for $= p$ case:
Let $F$ be a bounded $\mathscr F_t$-measurable random variable.
$$\Bbb E[1_A\cdot F]=\Bbb E[E[1_A\cdot F|\mathscr F_t]]=\Bbb E[F\cdot E[1_A|\mathscr F_t]]$$
$$=\Bbb E[p\cdot F]=p\Bbb E[F]=\Bbb E[1_A]\cdot\Bbb E[F]$$
meaning that $1_A$ and $F$ are independent. In other words, $\sigma(A)$ and $\mathscr F_t$ are independent. So $\sigma(A)$ and $\mathscr F_s$ are also independent if $s<t$ and hence $E[1_A|\mathscr F_s] = E[1_A] = p$ . This may fail if $s>t$.
I guess the idea is that a constant is both independent of $\mathscr F_s$ and $\mathscr F_s$-measurable.
 A: Your argument seems to be valid, but you start off by assuming that $E[1_A | \mathscr{F_t}] = 1$. However, the question states that $E[1_A | \mathscr{F_t}] \in \{0, 1\}$, which I would take to mean that the random variable $E[1_A | \mathscr{F_t}]$ takes values in the set $\{0, 1\}$ i.e. $E[1_A | \mathscr{F_t}]=1_B$ where $B\in\mathscr{F_t}$. The defining property of this conditional expectation is that $\int_F 1_B d\mathbb{P}=\int_F 1_A d\mathbb{P}$ for all $F\in\mathscr{F_t}$. In particular, taking $F=B$ leads to $P(B)=P(A\cap B)$, from which we can conclude that $B\subset A$ (except possibly on a set of probability zero). However, we also know (as in the argument you have written) that $E[E[1_A | \mathscr{F_t}]] = E[1_B]$ i.e. $P(A)=P(B)$, so the only possible conclusion is that $A=B$ (except possibly for a set of probability zero).
For $s\gt t$, $\mathscr{F_t}\subset\mathscr{F_s}$, so the tower law for conditional expectations implies that $E[1_A | \mathscr{F_t}]=E[E[1_A | \mathscr{F_t}] | \mathscr{F_s}]$. But $E[1_A | \mathscr{F_t}]=1_A$, so $E[1_A | \mathscr{F_s}]=1_A$. So all the conditional expectations for $s>t$ are equal (to $1_A$). For $s<t$, if $A\in\mathscr{F_s}$ then we will still have $E[1_A | \mathscr{F_s}]=1_A$. On the other hand, if we go back to a time where $A$ is not in $\mathscr{F_s}$, then I don't think anything can be said about $E[1_A | \mathscr{F_s}]$ in general. For a concrete example, see this paper, Figure 1. Taking $A=\{\omega_2\}\in\mathscr{F_2}\setminus\mathscr{F_1}$, for example, gives the sequence of conditional expectations $E[1_A | \mathscr{F_0}]=\frac{1}{8} 1_\Omega$, $E[1_A | \mathscr{F_1}]=\frac{1}{2}1_{\{\omega_1,\omega_2\}}$, $E[1_A | \mathscr{F_2}]=1_{\{\omega_2\}}$, $E[1_A | \mathscr{F_3}]=1_{\{\omega_2\}}$.
