Let $X$ be an integrable real random variable on the probability space $(\Omega,\mathcal A,P)$. If $A\in \mathcal A$ is an event with probability $0<P[A]<1$ then we have that
$$E[X|\sigma(A)]=1_A \frac{E[1_AX]}{P[A]}+1_{A^c} \frac{E[1_{A^c}X]}{P[A^c]} \quad P\text{-a.s.} $$
Now suppose $\mathcal F$ a sub-$\sigma$-algebra of $\mathcal A$. Am wondering if the following holds:
$$E[X|\sigma(A,\mathcal F)]=1_A \frac{E[1_AX|\mathcal F]}{P[A|\mathcal F]}+1_{A^c} \frac{E[1_{A^c}X|\mathcal F]}{P[A^c|\mathcal F]} \quad P\text{-a.s.} \quad \quad (1)$$
assuming that $0<P[A|\mathcal F]<1$ $P$-a.s. (I don't see how to dispense with this assumption).
It seems to me that the answer is yes based on the following argument:
Let $Y$ denote the RHS of $(1)$.Then $Y$ is $\sigma(A,\mathcal F)$ measurable. We check that $Y$ is integrable using the conditional Jensen's inequality,the law of total expectation, and the pull-out property:
$$ E\Bigg[\Bigg|1_A \frac{E[1_AX|\mathcal F]}{P[A|\mathcal F]}\Bigg|\Bigg]\leq E\bigg[1_A \frac{E[1_A |X|\mid\mathcal F]}{P[A|\mathcal F]}\bigg]= E\bigg[E[1_A|\mathcal F] \frac{E[1_A |X|\mid\mathcal F]}{P[A|\mathcal F]}\bigg]=E[1_A|X|]< \infty$$
and similarly $E\Bigg[\Bigg|1_{A^c} \frac{E[1_{A^c}X|\mathcal F]}{P[A^c|\mathcal F]}\Bigg|\Bigg]<\infty$.
Now, we verify that $\sigma(A,\mathcal F)=\Big\{(A\cap B_1) \cup (A^c\cap B_2) : B_1,B_2\in\mathcal F \Big\}$, and therefore by additivity it is is sufficient to check that $E[1_{A\cap B}Y]=E[1_{A\cap B}X]$ and $E[1_{A^c\cap B}Y]=E[1_{A^c\cap B}X]$ for all $B\in\mathcal F$. This we verify by direct computation, using again the law of total expectation and the pull-out property:
$$E[1_{A\cap B}Y]=E\bigg[1_A\frac{E[1_{A\cap B}X|\mathcal F]}{P[A|\mathcal F]}\bigg]=E\bigg[E[1_A|\mathcal F]\frac{E[1_{A\cap B}X|\mathcal F]}{P[A|\mathcal F]}\bigg]=E[1_{A\cap B}X]$$
and similarly $E[1_{A^c\cap B}Y]=E[1_{A^c\cap B}X]$.
Am I missing something? Is there a way to ensure that $0<P[A|\mathcal F]<1$ $P$-a.s.?
Thanks a lot for your help.
Proposition. A necessary and sufficient condition for $0<P[A|\mathcal F]<1$ a.s. is that $P[A\cap B]>0$ and $P[A^c\cap B]>0$ for all $B\in \mathcal F$ with $P[B]>0$. In other words, the partition $\{A,A^c\}$ of $\Omega$ must split each non-null set of $\mathcal F$ into two non-null parts.
Necessity. Suppose $0<P[A|\mathcal F]<1$ a.s.. Let $B\in \mathcal F$ with $P[B]>0$. Then $$P[A\cap B]=E[1_{A\cap B}]=E[P[A|\mathcal F] 1_B]>0$$
for otherwise $P[A|\mathcal F] 1_B=0$ a.s. and so $P[A|\mathcal F]=0$ on an $\mathcal F$-measurable set of positive measure, contradiction. Similarly, $P[A^c\cap B]>0$, for otherwise $P[A|\mathcal F]=1$ on an $\mathcal F$-measurable set of positive measure.
Sufficiency. Suppose $P[A\cap B]>0$ and $P[A^c\cap B]>0$ for all $B\in \mathcal F$ with $P[B]>0$.
Suppose first by contradiction that $P[A|\mathcal F]=0$ on a set $B\in \mathcal F$ with $P[B]>0$. Note that $P[B]<1$, for otherwise $P[A|\mathcal F]=0$ a.s. which implies $P[A]=0$, contradiction. Then
$$E[1_A|\mathcal F]=1_BE[1_A|\mathcal F]=E[1_{A\cap B}|\mathcal F]$$
and so $E[1_{A\cap B^c}|\mathcal F]=E[1_A|\mathcal F]-E[1_{A\cap B}|\mathcal F]=0$ a.s.. This implies $P[A\cap B^c]=0$, contrary to hypothesis.
Next suppose by contradiction that $P[A|\mathcal F]=1$ on a set $B\in \mathcal F$ with $P[B]>0$. Note that $P[B]<1$, for otherwise $P[A|\mathcal F]=1$ a.s. which implies $P[A]=1$, contradiction. Then $E[1_A|\mathcal F]\geq 1_B$ a.s. and so
$$E[1_{A\cap B}|\mathcal F]=1_B E[1_A|\mathcal F]\geq 1_B=E[1_B|\mathcal F]$$
which implies $0\geq E[1_{A\cap B}-1_B|\mathcal F]\geq 0$ or $E[1_{A^c\cap B}|\mathcal F]=0$ a.s.. Hence $P[A^c\cap B]=0$, contrary to hypothesis.