Law of Total (Conditional) Probability and Filtration The law of total probability establishes that 
$$
\mathbb{P}[A\mid C] = \sum_{n}\mathbb{P}[A\mid C\cap B_n]\,\mathbb{P}.[B_n\mid C] 
$$
Suppose that I have a filtration $\mathcal{F}_t$ and $A_t$ and $B_t$ are  $\mathcal{F}_t$-adapted stochastic processes. In particular $B_t$ can be either $1$ or $0$. Let $I\subset\mathbb{R}$ be an interval.
I am wondering if I can apply the law of total probability and write
$$
\begin{eqnarray}
\mathbb{P}[A_t\in I\mid\mathcal{F}_{t-1}]&=&\mathbb{P}[A_t\in I\mid \mathcal{F}_{t-1}\cap \{B_t=1\}]\,\mathbb{P}[B_n=1\mid\mathcal{F}_{t-1}]+\\
& & +\mathbb{P}[A_t\in I\mid \mathcal{F}_{t-1}\cap \{B_t=0\}]\,\mathbb{P}[B_n=0\mid\mathcal{F}_{t-1}] 
\end{eqnarray}.
$$ 
Is the writing above formally correct? 
 A: No.


*

*I think you mean $\mathbb E$ instead of $\mathbb P$

*It does not make sense to take the intersection of a collection of events (e.g. $\mathcal F_{t-1}$) and a collection of sample points (e.g. $(B_t = 1)$). Perhaps you meant $\sigma((B_t = 1))$.

*Careful about the extension you're trying to make here. You seem to be thinking we can do something like:
$$E[X] = E[Y] \to E[X|\mathscr F] = E[Y|\mathscr F]$$
And is that really possible? Go back to the definition of conditional expectation.


*If you want specifically to condition on $(B_t = 1)$, try (note the correction for the indices):


$$
\begin{eqnarray}
\mathbb{E}[\mathbb{1}_{\left\{A_t\in I\right\}}\mid\mathcal{F}_{t-1}]&=&\mathbb{E}[\mathbb{1}_{\left\{A_t\in I\right\}| \{B_t=1\}}\mid \mathcal{F}_{t-1}]\,\mathbb{P}[B_t=1\mid\mathcal{F}_{t-1}]+\\
& & +\mathbb{E}[\mathbb{1}_{\left\{A_t\in I\right\}| \{B_t=0\}}\mid \mathcal{F}_{t-1}]\,\mathbb{P}[B_t=0\mid\mathcal{F}_{t-1}] 
\end{eqnarray}.
$$ 
where we seem to have $E[1_{A|B}] := P(A|B)$, where in our case $A = \{A_t \in I \}$ and $B = \{B_t=1\}$, not that the above should be correct or sensible:
For example, what is $1_{A|B}$?
$1_{A|B}(\omega) = 1 \times 1_{\omega \in A} + 0 \times 1_{\omega \in A^C}$ but $\omega \in B$?
So what if $\omega \notin B$? What is the value of $1_{A|B}$ (This is similar to asking what is the $P(A|B)$ if $P(B) = 0$)? What kind of event or object is $A|B$ anyway?
Well assuming $1_{A|B}(\omega)$ is a well-defined object and a well-defined random variable, we seem to have:
$$E[1_{A|B}(\omega)|\mathscr F_{t-1}] E[1_{B}(\omega)|\mathscr F_{t-1}]$$
$$ = E[ E[1_{A|B}(\omega)|\mathscr F_{t-1}] 1_{B}(\omega)|\mathscr F_{t-1}]$$
Now if $B_t$ is previsible, we have:
$$ = E[1_{A|B}(\omega) 1_{B}(\omega) |\mathscr F_{t-1}]$$
///ly, we have
$$E[1_{A|B^C}(\omega)|\mathscr F_{t-1}] E[1_{B^C}(\omega)|\mathscr F_{t-1}]$$
$$ = E[1_{A|B^C}(\omega) 1_{B^C}(\omega) |\mathscr F_{t-1}]$$
It seems that
$$E[1_{A|B}(\omega) 1_{B}(\omega) |\mathscr F_{t-1}] + E[1_{A|B^C}(\omega) 1_{B^C}(\omega) |\mathscr F_{t-1}] = E[1_A | \mathscr F_{t-1}]$$

Possibly related section from Probability w/ Martingales:

