Shouldn't conditional expectation be defined $P\lvert_{\mathcal{F}}$ almost surely rather than $P$ almost surely Consider some integrable random variable $X \in \mathcal{L}^{1}(\Omega,\mathcal{A},P)$ and let $\mathcal{F}$ be some sub-$\sigma$-field. 
In all of the literature I have seen, all results surrounding $E[X\lvert \mathcal{F}]$ are given $P$-a.s. But the way I see it, is that $E[X\lvert \mathcal{F}]$ is by definition $\mathcal{F}$-measurable such that all results can only be defined $P\lvert_{\mathcal{F}}$ almost surely. This may be a minor detail but I would still like to know whether my thinking that the above is an "abuse of notation" is correct or not. 
Clearly $P << P\lvert _{\mathcal{F}}$, such that $P\lvert _{\mathcal{F}} \text{ a.s. } \implies P \text{ a.s. }$, so that 
$E[X\lvert \mathcal{F}]=Y \; P\lvert_{\mathcal{F}}\text{ a.s. }\implies E[X\lvert \mathcal{F}]=Y \; P\text{ a.s. }.$
Any clarification would be of great help. 
 A: No, it's not an abuse of notation. Conditional expectation is defined using the Radon-Nikodym derivative, and the Radon-Nikodym Theorem guarantees that, if $\nu \ll \mu$, and $f$ and $g$ are two Radon-Nikodym derivatives of $\nu$ in respect to $g$, then $f=g$, $\mu$-almost everywhere, a stronger condition than saying $\nu$-almost everywhere.
It's not difficult to see why this should happen. Note that we want, for every $A \in \mathcal{F}$,
$$\int_A E[X \mid \mathcal F] \ dP = \int_AX \ dP.$$
Note that, if we we modify the conditional expectation in a set of $P$-measure $0$, the equality is mantained. This is stronger that the result you claim. 
A: The expression $P \ll P|_\mathcal F$ doesn't make sense (according to the standard definition of $\ll$) because the two measures are defined on different domains. 
Also, expressions like "$E[X \mid \mathcal F] = Y \ P|_\mathcal F$ a.s." make sense only if $Y$ is also $\mathcal F$-measurable.
I think you are overlooking a very simple point, namely that any random variable that is $\mathcal F$-measurable is $\mathcal A$-measurable as well. So any statement we can make about it using $P|_\mathcal F$ we can also make about it using $P$. This isn't an abuse of notation, it's just true. 
