Verification of result I've recently I've come across this and am questioning it's validity but can't fully convince myself of (or against) it.  Here are my intuitions why it may be true but I can't fully but I can't find a counter-example of prove it....

Claim:
$$
E_{\mathbb{P}}[Y|\mathscr{F}_t]=E_{\mathbb{Q}}[Y\frac{Z_t}{Z_T}|\mathscr{F}_t].
.
$$

Thoughts:
Suppose that $X_t$ is an integrable $\mathscr{F}$-adapted process, $\mathbb{Q}\sim\mathbb{P}$ are equivalent measures and $Y$ is an $\mathscr{F}_T$-measurable random-variable for some $T>t$.  Let $Z_t$ be the corresponding density process
$$
Z_t \triangleq \frac{d\mathbb{Q}}{d\mathbb{P}}|_{\mathscr{F}_t}.
$$
It is known, for example in Theorem 3.8 of Jacod and Shiryaev's book, that
$$
E_{\mathbb{Q}}[Y|\mathscr{F}_t]=E_{\mathbb{P}}[Y\frac{Z_T}{Z_t}|\mathscr{F}_t].
$$
Moreover, in Corollary 15.2.2 of Cohen and Elliot's book, it is known that $(Z_t)^{-1}$ is also a $(\mathbb{Q},\mathscr{F}_t)$-martingale since both measures are equivalent.  
This seems wrong, but the Cohen result makes me believe otherwise....  Can someone confirm this?
So the claim seems to be a mis-reading of the Jacod and Shiryaev result but on second glance maybe the Samuel and Cohen result justifies it...Just don't see why/not..
 A: Here are my two cents. I think the claim is true at least if the condition (c) below holds true (sorry for the first posts where I mixed up the algebra). 
First let's recall things a bit following your post :
$Z \triangleq \frac{d\mathbb{Q}}{d\mathbb{P}}$ and $Z^{-1} \triangleq \frac{d\mathbb{P}}{d\mathbb{Q}}$ 
and we have $Z^{-1}=\frac{1}{Z}$ as $\mathbb{Q}\sim\mathbb{P}$ by hypothesis (this is easy to show there posts on the forum on this I think). 
Now the condition (c)
Let's redefine following your post for any $\forall t>0$ :
$$Z_t = E_{\mathbb{P}}[Z|\mathscr{F}_t] \triangleq \frac{d\mathbb{Q}}{d\mathbb{P}}|_{\mathscr{F}_t} $$ 
and 
$$Z_t^{-1} = E_{\mathbb{Q}}[Z^{-1} |\mathscr{F}_t] \triangleq \frac{d\mathbb{P}}{d\mathbb{Q}}|_{\mathscr{F}_t} $$ 
If we have in this setting : 
$$Z_t^{-1}=\frac{1}{Z_t} ~~~~(c)$$
Your claim is true under condition (c).
Proof :
Starting with the Jacod and Shiryaev theorem but switching $\mathbb{P}$ and $\mathbb{Q}$ we get :
$$E_{\mathbb{P}}[Y|\mathscr{F}_t]=E_{\mathbb{Q}}[Y\frac{Z_T^{-1}}{Z_t^{-1}}|\mathscr{F}_t]$$
then using the property (c) above we get your claim because :
$$E_{\mathbb{Q}}[Y\frac{Z_T^{-1}}{Z_t^{-1}}|\mathscr{F}_t]=E_{\mathbb{Q}}[Y\frac{Z_t}{Z_T}|\mathscr{F}_t]$$
So the condition above is sufficient for your claim to be true and it remains to find suitable conditions for which it holds true, but I think it holds in quite a general framework, the question to know if the condition above is also necessary is unknown me, but note that I didn't use the theorem of Cohen and Elliott's book. 
Note that condition (c) can be rewritten : 
$$E_{\mathbb{Q}}[Z^{-1} |\mathscr{F}_t] =\frac{1}{E_{\mathbb{P}}[Z|\mathscr{F}_t]}$$ or 
$$E_{\mathbb{P}}[Z|\mathscr{F}_t].E_{\mathbb{Q}}[Z^{-1} |\mathscr{F}_t] = 1$$ 
Unless mistaken this is the case when Girsanov's theorem applies because it holds true that $Z_t$ and $Z^{-1}_t$ can be expressed as :
$Z_t=exp(X_t-\frac{1}{2}<X>_t)$ and $Z^{-1}_t=exp(-X_t+\frac{1}{2}<X>_t)$
For some martingale $X_t$, but I don't know if for strict local martingale this is true. 
Here is another one line argument :
Let's take Jacod and Shiryaev theorem (JST) with $Y'=Y.\frac{Z_t}{Z_T}$, note that  $Y'$ is $\mathcal{F}_T$-measurable so that : 
$$E_{\mathbb{Q}}[Y.\frac{Z_t}{Z_T}|\mathscr{F}_t]=\underbrace{E_{\mathbb{Q}}[Y'|\mathscr{F}_t]=E_{\mathbb{P}}[Y'\frac{Z_T}{Z_t}|\mathscr{F}_t]}_\text{by Jacod and Shiryaev's theorem applied to $Y'$}=E_{\mathbb{P}}[Y.\frac{Z_t}{Z_T}\frac{Z_T}{Z_t}|\mathscr{F}_t]=E_{\mathbb{P}}[Y|\mathscr{F}_t]$$
QED
