Can a filtration on a measure space have real-number indexing (giving a continuum of $\sigma$-algebras? My book "Measures, Integrals and martingales" defines a filtration as a sequence $F_t$ of $\sigma$-algebras. That means that there are at most countable $\sigma$ algebras in the filtration.
But in stochastic processes over a space of continuous functions (such as in Brownian motion), it is reasonable to ask for a $\sigma$ algebra at every point $t\in \mathbb R$. This would give us a "filtration" that is not a sequence (i.e. a map $\mathbb N \to \mathcal P(\Omega)$), but a function $\mathbb R \to \mathcal P(\Omega)$. 
Would such a "Continuum of $\sigma$ algebras" still be called a "filtration"?
 Is it valid? 
 A: It is not only possible, but also the right definition. In general, given a probability space $(\Omega,\mathcal E,P)$ in which it is defined a stochastic process $(X_t)_{t\in \mathcal I}$ with time domain on a totally ordered set $\mathcal I$, a filtration of $X_t$ is a collection $(\mathcal E_t)_{t\in\mathcal I}$ of increasing (i.e., non-decreasing) sub-$\sigma$-algebras of $\mathcal E$, that is:


*

*$\mathcal E_t \subseteq \mathcal E$ for every $t\in \mathcal I$,

*$\mathcal E_t$ is a $\sigma$-algebra for every $t\in \mathcal I$, and

*$t_0\le t_1 \implies \mathcal E_{t_0} \subseteq \mathcal E_{t_1}$.


Just note that in your case it won't be a mere function
$$\mathbb R \to \mathcal P(\Omega)$$
but a function
$$\mathbb R \to \mathcal P\big(\mathcal P(\Omega)\big).$$
And even so, it could well be the case that a filtration could not be a function with image $\mathcal P\big(\mathcal P(\Omega)\big)$. All this is true because:


*

*the elements of $\mathcal P(\Omega)$ are not $\sigma$-algebras over $\Omega$ but subsets of it;

*it might well be the case that $\mathcal P(\Omega)$ not be a suitable $\sigma$-algebra over whose events a valid probability function can be defined (perhaps you are working with the probability space $(\mathbb R, \mathcal B (\mathbb R),P)$, and so your filtration gives [not subsets but] sub-$\sigma$-algebras of $\mathcal B(\mathbb R)$ or whatever $\sigma$-algebra over $\Omega$ is defined on your space.)


But, short answer, yes: such a function would still be called a filtration provided it satisfies the conditions mentioned above.
