How to determine if conditional expectations with respect to different measures are equal a.s.? Let $(\Omega, \mathcal{F}, P)$ be a probability space and let $\mathcal{A}$ be a sub-$\sigma$-algebra of $\mathcal{F}$. Let $Q_{\mathcal{A}}$ be a probability measure on $(\Omega, \mathcal{A})$ and define, for all $P$-integrable $f$,
$$Q(f) := \int E_P(f \mid \mathcal{A})dQ_{\mathcal{A}}.$$
Note that $E_P(f \mid \mathcal{A})$ is the conditional expectation with respect to $P$. Also note that $Q$ defines a measure on $(\Omega, \mathcal{F})$ by taking $f$ to be an indicator function (we abuse notation by writing $Q(A)$ for $A \in \mathcal{F}$).
Motivation. The idea is that we start with a "prior" probability space $(\Omega, \mathcal{F}, P)$. This extends to a linear functional (expectation) on the space $L^1$ of $P$-integrable functions $f$. Then we "learn" something about the sub-algebra $\mathcal{A}$ and adopt the new probability $Q_{\mathcal{A}}$ defined on $\mathcal{A}$. The question arises: How to extend this new probability $Q_{\mathcal{A}}$ to all of $\mathcal{F}$ (and thereby $L^1$)? We consider the extension $Q$ defined above.
Question. Does $E_P(f \mid \mathcal{A}) = E_Q(f \mid \mathcal{A})$ a.s. ($P$)?
Added Question. Is $Q \ll P$?
The a.s. equality would follow if I could show that $$\int_A E_Q(f \mid \mathcal{A}) dP = \int_A f dP$$
for all $A \in \mathcal{A}.$ But I'm not sure what can be said when integrating a $Q$-conditional expectation against the measure $P$.

I believe I can show the result for the case where $\mathcal{A}$ is generated by a countable partition $\{A_i \}_{i \in I}$ with $P(A_i)>0$ and $Q_{\mathcal{A}}(A_i) > 0$ for all $i \in I$.
In that case, we have
$$E_Q(f \mid \mathcal{A}) = \sum_i E_Q(f \mid A_i) \mathbf{1}_{A_i},$$
so it suffices to show that $E_Q(f \mid A_i) = E_P(f \mid A_i)$ for all $i \in I$.
To that end we calculate (I abuse notation, writing $A_i = \mathbf{1}_{A_i}$)
$$\begin{align} E_Q(f \mid A_i) &= \frac{1}{Q(A_i)} \int_{A_i}fdQ \\
 &= \frac{1}{Q(A_i)} \int E_P(fA_i \mid \mathcal{A}) dQ_{\mathcal{A}} \\
 &= \frac{1}{Q(A_i)} \int \left(\sum_{j \in I} \mathbf{1}_{A_j} \frac{1}{P(A_j)} \int_{A_i}fA_j dP \right)dQ_{\mathcal{A}} \\
&= \frac{Q_{\mathcal{A}}(A_i)}{Q(A_i)} \frac{1}{P(A_i)} \int_{A_i} f dP \\
&= E_P(f \mid A_i).\end{align}$$
The problem with extending this to general $\mathcal{A}$ is that I can't say explicitly what the conditional expectations (almost surely) are.
 A: It's a very interesting question, because it leads to a useful discussion on matters, which usually are overlooked when one speaks of conditional expectations.
The first thing to note is that one should be extremely careful with writing things like 
$$
E_Q(f \mid \mathcal{A}) = E_P(f \mid \mathcal{A}) \tag{1}
$$
for some measures $P$ and $Q$ (with $Q$ not necessarily defined as in the OP). The sad truth is that the conditional expectation $E_Q(f \mid \mathcal{A})$ is defined modulo $Q$-null sets. This makes equality (1) totally meaningless unless the measures $Q$ and $P$ have the same null sets, i.e. they are equivalent. 
Now turning to the particular situation, $E_{Q_{\mathcal A}}(E_P(f\mid\mathcal A))$ cannot be defined if $Q_{\mathcal A}\not\ll P|_{\mathcal A}$. Indeed, in this case case there is some $A\in\mathcal A$ with $Q_{\mathcal A}(A)>0 = P(A)$. Changing the values of $E_P(f\mid\mathcal A)$ on $A$ does not change this conditional expectation. However, it does change the value of $E_{Q_{\mathcal A}}(E_P(f\mid\mathcal A))$. 
Consequently, we must assume that $Q_{\mathcal A}\ll P|_{\mathcal A}$. Moreover, if $P|_{\mathcal A}\not \ll Q_{\mathcal A}$, then (1) cannot hold $\pmod P$. Indeed, in this case there is some set $B\in\mathcal A$ such that $P(B)>0 = Q(B)$. Then we can change the left-hand side of (1) on $B$ arbitrarily. On the bright side, (1) does hold $\pmod Q$ even in this case. To show this, note that by definition of $E_Q(\cdot\mid\mathcal A)$, (1) holds iff for any $A\in \mathcal A$ 
$$
E_Q(A\,f ) = E_Q(A\,E_P(f \mid \mathcal{A})).
$$
But 
$$
E_Q(A\,f) = Q(A\, f) = E_{Q_{\mathcal{A}}}(E_P(A\,f \mid \mathcal{A})) \\ = E_{Q}(E_P(A\,f \mid \mathcal{A})) = E_{Q}(A\,E_P(f \mid \mathcal{A})),
$$
as required.
The answer to the second question is positive (but, I repeat, you must assume $Q_{\mathcal A}\ll P|_{\mathcal A}$). Indeed, if $P(A) = 0$, then $E_P(A\mid\mathcal A) = 0 \pmod P$ thanks to the tower property, therefore, $E_P(A\mid\mathcal A) = 0 \pmod Q_\mathcal A$ in view of the absolute continuity, so
$$
Q(A) = E_{Q_{\mathcal{A}}}(E_P(A\mid \mathcal{A})) = 0,
$$
as claimed. (Alternatively, you can just show, by definition, that $dQ/dP = dQ_{\mathcal A}/dP|_{\mathcal A}$.)
