Given the product is measurable, is each factor measurable? Given a random variable $M$ on $(\Omega,\mathscr F, \Bbb P)$ and $M=X\cdot Y$, can we proof that $X$ and $Y$ are also measurable?
To be more specific, I was thinking about if a process $M_t=X_t\cdot Y_t$ is adapted to its natural filtration $\mathscr F_t^M$. Is $X_t$ and $Y_t$ also adapted to the natural filtration?
Thanks for any thought.

No; not necessarily.
Witness a sample space, $(\Omega,\mathcal F) =\big( \{(1,1),(1,2),(2,1)\},\{\emptyset, \{(1,1)\}, \{(1,2),(2,1)\}, \Omega\}\big)$.
We define the random variables $X:(x,y)\mapsto x$, $Y:(x,y)\mapsto y, M:(x,y)\mapsto xy$.
Then $M(\omega)=[X\cdot Y] (\omega)$, and $M$ is $\mathcal F$-measureable, but neither $X$ nor $Y$ are.


*

*$M^{-1}\{1\}= \{(1,1)\}\in \mathcal F\\M^{-1}\{2\}=\{(1,2),(2,1)\}\in \mathcal F$

*$X^{-1}\{1\}= \{(1,1),(1,2)\}\notin \mathcal F\\X^{-1}\{2\}=\{(2,1)\}\notin \mathcal F$

*$Y^{-1}\{1\}= \{(1,1),(2,1)\}\notin \mathcal F\\Y^{-1}\{2\}=\{(1,2)\}\notin \mathcal F$


So $M$ being $\mathcal F$-measurable and $M=X\cdot Y$ is insufficient to prove that either $X$ or $Y$ are $\mathcal F$-measurable.
