Given two stochastic processes on a probability space, will their compound process be a valid stochastic process on the same probability space? Let the stochastic process $M=(M_t, t\ge 0)$ and the stochastic pathwise continuous increasing process $Y=(Y_t,t\ge 0)$ be defined on the probability space $(\Omega, \mathcal F, P)$. Will the compound process $M_Y=(M_{Y_t},t\ge 0)$ also be valid (measureable on the same sigma algebra which $M$ and $Y$ maps from) on this probability space? 
If it is not valid in general, what if $M$ and $v$ are independent of each other? Will it then be 'valid'?
Clarification:
Does 
$\{\omega\in\Omega\colon M(\omega)\in B\}\in\mathcal{F}$, $\forall B\in \mathcal{B}(\mathbb{R})$ and $\{\omega\in\Omega\colon Y(\omega)\in B\}\in\mathcal{F}$, $\forall B\in \mathcal{B}(\mathbb{R})$ $\implies$ $\{\omega\in\Omega\colon M_Y(\omega)\in B\}\in\mathcal{F}$, $\forall B\in \mathcal{B}(\mathbb{R})$
hold? $\mathcal{B}(\mathbb{R})$ being the generated Borel $\sigma$-algebra.
 A: Your "compound" process $M_Y$ at a fixed $t\ge 0$ is the composition of two mappings $\psi_t(\omega):=(\omega,Y_t(\omega))$ and $\varphi(\omega,u):=M_u(\omega)$. The former is an $\mathcal F / \mathcal F\otimes\mathcal B$ measurable mapping of $\Omega$ to $\Omega\times[0,\infty)$. (Where $\mathcal B$ denotes the Borel subsets of $[0,\infty)$.) This is because $\mathcal F\otimes\mathcal B$ is generated by rectangles of the form $F\times B$, $F\in\mathcal F, B\in\mathcal B$, and $\psi_t^{-1}(F\times B)=F\cap Y_t^{-1}(B)$.
To finish you need to know that the latter mapping $\varphi$ is $\mathcal F\otimes\mathcal B / \mathcal R$ measurable. (Here I use $\mathcal R$ to denote the Borel subsets of $\Bbb R$.) This situation is referred to as $M$ being a "measurable process" and is an additional hypothesis. For example, if each random variable $M_u$ is $\mathcal F$ measurable and $u\mapsto M_u(\omega)$ is right continuous for each $\omega$, then $M$ is a measurable process.
If both $\psi_t$ and $\varphi$ are measurable as indicated, then the composite function $M_{Y_t}=\varphi\circ\psi$ is $\mathcal F/\mathcal R$ measurable.
A: I guess that by "same sigma algebra which $M$ and $Y$ maps onto" you mean the sigma algebra generated by the two processes.
However, this is not the case. For instance take $Y$ to be the deterministic process defined by $Y_t=2t$ and $M_t$ any stochastic process such that its filtration satisfies that $\mathcal{F}_t\subsetneq\mathcal{F}_s$ for $t,s$.
At time $t$ the sigma algebra generated by $M$ and $Y$ is the same one as the one generated by $M$.
However, even $M_{Y_t}=M_{2t}$ is not measurable at time $t$.
