A probability space $(\Omega, \mathcal{A}, \mathbb{P})$ is called complete iff every subset $\tilde{N}$ of a set $N \in \mathcal{A}$ of measure $0$ is measurable, i.e. if $\mathbb{P}(N) = 0$ and $\tilde{N} \subseteq N$ implies that $\tilde{N}\in \mathcal{A}$. In other words, for $(\Omega, \mathcal{A}, \mathbb{P})$ to be complete, the inclusion $$ \mathcal{N}:= \{ \tilde{N} \subseteq \Omega \mid \exists N \in \mathcal{A}: \mathbb{P}(N) = 0 \ \land \ \tilde{N} \subseteq N \} \subseteq \mathcal{A}$$ has to hold.
(A) On Wikipedia, a filtration $(\mathcal{A}_t)_{t \in I}$ on $(\Omega, \mathcal{A}, \mathbb{P})$ is defined to be complete iff, for all $t \in I$, it holds that $\mathcal{N} \subseteq \mathcal{A}_t$.
(B) Furthermore, it is claimed that definition (A) is equivalent to the statement that $(\Omega, \mathcal{A}_t, \mathbb{P})$ is complete for all $t \in I$.
I don't think that this equivalence holds and this is a mistake. It is easy to show that (A) implies (B), but I don't see how the converse can be true. Do you agree, or am I missing something?
Kind regards, Joker
The claim enunciated in (B) is incorrect, as you have noted. Example: Suppose that a standard 1-dimensional Brownian motion $(B_t)_{t\in[0,\infty)}$ is defined on a probability space $(\Omega,\mathcal F,\Bbb P)$. For each $t\ge 0$ let $\mathcal A_t$ be the $\Bbb P$-completion of $\sigma\{B_s:0\le s\le t\}$, and let $\mathcal A$ be the $\Bbb P$-completion of $\sigma\{B_s:s\ge 0\}$. The event $C:=\{\omega\in\Omega: \lim_{r\to\infty,r\in\Bbb Q}{B_r(\omega)\over r}$ fails to exist$\}$ is an element of $\mathcal A$ with $\Bbb P(C)=0$; but $C\notin\mathcal A_t$, for each $t>0$. In this case, (B) holds but (A) does not.