Is it possible to obtain the Radon-Nikodym derivative $\left. \frac{dQ}{dP} \right|_{\mathscr{F}}$ from a UI martigale?

Question

I have the following theorem:

Theorem. Let P be a probability measure on $(\Omega, \{ \mathscr{F} \}, \mathscr{F}).$ Given an a.s. strictly positive UI $(\{ \mathscr{F} \},P)$-martingale $\rho$ with $E_P[\rho_{\infty}]=1$, $$ \left. \frac{dQ}{dP} \right|_{\mathscr{F}_{\infty}}:= \rho_{\infty}$$ defines via $$Q(A) = \int_A \rho_{\infty} dP \quad \text{for every } A \in \mathscr{F}_{\infty}$$ a measure Q equivalent to P with respect to $\mathscr{F}_{\infty}.$

Remark. Note that if $\mathscr{F}$ is richer than $\mathscr{F}_{\infty},$ we can not recover $\left. \frac{dQ}{dP} \right|_{\mathscr{F}}$ from the martingale $\rho.$

My question: I don't understand the remark. Why can't we get Q equivalent to P with respect to the "richer" $\mathscr{F}$? Why the "best" we can do is to obtain $ \left. \frac{dQ}{dP} \right|_{\mathscr{F}_{\infty}}$?

We know that $\rho_{\infty} \in L^1(\Omega, \mathscr{F}_{\infty}, P)$ but I think it is also true that $\rho_{\infty} \in L^1(\Omega, \mathscr{F}, P).$ Hence, I could have defined $Q(A) = \int_A \rho_{\infty} dP \quad \text{for every } A \in \mathscr{F},$ and then we would have $Q \ll P$ with respect $\mathscr{F}.$ Also, since $\rho$ is an a.s. strictly positive UI martingale, we have $P \ll Q.$ Is there anything wrong with this argument?

Answer 1

If we define $$Q(A) = \int_A \rho_{\infty} dP \quad \text{for all } A \in \mathscr{F},$$ then $\rho_{\infty}$ is the Radon-Nikodym derivative of Q with respect to P on $\mathscr{F}.$ Since $\rho_{\infty}$ is $\mathscr{F}$-measurable, $$E[\rho_{\infty} | \mathscr{F}] = \rho_{\infty}.$$ But if $\mathscr{F}$ is richer than $\mathscr{F}_{\infty},$ it could happen that $$E[\rho_{\infty} | \mathscr{F}] = \rho_{\infty} \neq E[\rho_{\infty} | \mathscr{F}_{\infty}],$$ so the UI martingale wouldn't be $\{ \mathscr{F}_t \}$-adapted, which would be a contradiction. Thus, we are forced to define $Q$ only for all $A \in \mathscr{F}_{\infty}.$