Let $(\Omega,\left(\mathcal F_t\right)_{t\geq 0},\mathcal F,\mathbb P)$ a filterated probability space fulfiling the usual measurability conditions (see "Diffusions, Markov processes and martingales - Roger and Williams" Definition II.67.1). If needed it can be assumed that:
- $\Omega := \mathcal C_0^0(\mathbb R_+,\mathbb R)=\{\omega:\mathbb R_+\to \mathbb R;\ \omega\ \text{is continuous and }\omega(0) =0\}.$
- $\mathcal F_{t} = \sigma\left(W_s; 0\leq s\leq t\right),$ where $W_s$ is the standard Brownian motion.
- $\mathcal F = \sigma(W_s; s\geq 0),$
- $\mathbb P$ the Wiener measure.
Let $\{M_s:\Omega \to \mathbb R_+\}_{s\geq 0},$ be a martingale with respect to the filtration $\{\mathcal F_s\}_{s\geq 0},$ satisfying $\mathbb E[M_s] = 1.$
Consider the measure $\mathbb Q_s$ on $\mathcal F_s$ defined as $$\mathbb Q_s[A] = \mathbb E[\mathbb 1_A M_s] \ \text{for every }s\geq 0\ \text{and }A\in\mathcal F_s.$$
My question: Is it possible to guarantee that there exists a measure $\mathbb Q_\infty$ on $(\Omega,\mathcal F),$ such that $$ \mathbb Q_\infty (A_s) = \mathbb Q_s [A_s], \ \text{for every }A_s\in\mathcal F_s?$$
I do not know if this is true. Can anyone help me?