Stochastic processes: understanding a complete filtration

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My understanding is that the filtration for a stochastic process represents the information known, meaning that at time $t$ I know which sets in $\mathscr{F}_t$ are true or false. A filtration is complete if the sigma algebra for the probability space is complete and $\mathscr{F}_0$ contains all the null sets. Suppose my stochastic process consists of a sequence of normal random variables. Then the set {(0,0,0,...)}, the event that every variable equals zero, is a null set and is therefore an element of $\mathscr{F}_0$. The same is true for every sequence of real numbers. Since all these sets are in $\mathscr{F}_0$, doesn't this mean that at time 0 I know the whole future of the process, because I know which of these sets are true and which are false? Why is it okay to add this information to the filtration?

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Let $(\Omega,\mathcal A,\mathbb P)$ be a probability space and $\{X(t):t\geqslant 0\}$ a real-valued stochastic process - that is, for each nonnegative $t$, $X(t):\Omega\to\mathbb R$ is $(\mathcal A,\mathcal B(\mathbb R))$ a measurable function. The natural filtration of $X$ with respect to $\mathcal A$ is $\mathbb F=(\mathcal F_t)_{t\geqslant 0}$ where $$ \mathcal F_t = \sigma\left\{X_s^{-1}(A) : 0\leqslant s\leqslant t, A\in\mathcal A \right\}. $$ $\mathbb F$ is a complete filtration if $(\Omega, \mathcal F_t, \mathbb P)$ is a complete measure space for all $t$ - that is, if every $\mathcal F_t$ contains the set of all sets that are contained within a $\mathbb P$-null set, $$ \mathcal N_{\mathbb P} := \{A\subset\Omega: A\subset B \text{ for some } B\in\mathcal F \text{ with } \mathbb P(B)=0 \}. $$ This is mostly a technical matter of measure theory that is important when, for example we want to work with processes that are right-continuous. We don't know from the elements of in a given set in the filtration what's "true or false" for a given process - that is more akin to the notion of the sample path $t\mapsto X_t(\omega)$ for a fixed $\omega\in\Omega$.