Unusual augmentation of a filtration

Question

consider a probablity space $(\Omega,\mathcal{F}, \mathcal{P})$ and a filtration $(\mathcal{F}^0_t)$. In general $(\mathcal{F}^0_t)$ doesn't satisfy the usual conditions (it is not both complete at any $t$ and right continuous). To overcome this problem one usualy set for all $t$ $$\mathcal{F}_t= \mathcal{F}^0_{t+}V \mathcal{N}$$ where $\mathcal{N}$ are the negligible sets for $\mathbb{P}$. The new filtration $(\mathcal{F}_t)$ satisfies the usual conditions and it is called the usual augmentation of $(\mathcal{F}_t^0)$. On the other hand one could also consider the filtration $(\mathcal{G}_t)$ defined by $$\mathcal{G}_t = \cap_{\epsilon>0} (\mathcal{F}_{t+\epsilon}^0 V \mathcal{N})$$ It is also right continuous and complete. Obviously for all $t$ $\mathcal{F}^0_t \subset \mathcal{F}_t \subset \mathcal{G}_t$, so $(\mathcal{G}_t)$ does not seem optimal. It has however some good properties. My question is the following : do you have an example of what we loose by using $(\mathcal{G}_t)$ instead of $(\mathcal{F}_t)$. For instance a Theorem which would fail ?

Answer 1

They're the same, $\mathcal G_t=\mathcal F_t$.

Indeed, suppose $A\in\mathcal G_t$. So in particular $A\in\bigcap_{n=1}^\infty(\mathcal F_{t+1/n}\vee\mathcal N)$.

Note that for any $\sigma$-algebra $\mathcal M$, the $\sigma$-algebra $\mathcal M\vee\mathcal N$ consists of all sets whose symmetric difference with a set in $\mathcal M$ is null, i.e., sets that are almost equal to an element of $\mathcal M$.

Thus $ A $ is almost equal to some $ B_n \in\mathcal F^0_{t+1/n} $ for each $ n $.

But then $ A $ is almost equal to $B:=\bigcup_{n=1}^\infty\bigcap_{m\ge n} B_m=\bigcup_{n=N}^\infty\bigcap_{m\ge n} B_m\in \mathcal F^0_{t+1/N}$ for each $ N $, hence this set $B\in \mathcal F^0_{t+} $. Thus, $A\in\mathcal F^0_{t+}\vee\mathcal N$.

Answer 2

They are indeed the same, i.e. $\sigma(\cap_{s>t}\mathcal{F}_s,\mathcal{N})=\cap_{s>t}\sigma(\mathcal{F}_s,\mathcal{N})$.

Let $(\Omega,\overline{\mathcal{F}},\overline{P})$ be the completion of the original measure space so $\overline{\mathcal{F}}=\sigma(\mathcal{F},\mathcal{N})$where $\mathcal{N}:=\{N\subseteq \Omega| \exists A\in \mathcal{F},s.t. N\subseteq A,P(A)=0\}$.

***First let's show $\sigma(\cap_{s>t}\mathcal{F}_s,\mathcal{N})\subseteq\cap_{s>t}\sigma(\mathcal{F}_s,\mathcal{N})$. For any $E\in\sigma(\cap_{s>t}\mathcal{F}_s,\mathcal{N})$. Notice $\sigma(\cap_{s>t}\mathcal{F}_s,\mathcal{N})$ is the completion of $\cap_{s>t}\mathcal{F}_s$ relative to $\overline{\mathcal{F}}$ so there exist $A,B\in\cap_{s>t}\mathcal{F}_s$ such that $$A\subseteq E\subseteq B \mbox{ and }\overline{P}(B\setminus A)=0 $$ Because $A,B\in\cap_{s>t}\mathcal{F}_s$ so $A,B\in\mathcal{F}_s$ for each $s>t$. Thus (1) with $A,B\in\mathcal{F}_s$ implies that $E\in\sigma(\mathcal{F}_s,\mathcal{N})$ for each $s>t$ (by the property of $\sigma(\mathcal{F}_s,\mathcal{N})$ as the relative completion of $\mathcal{F}_s$ with respect to $\overline{\mathcal{F}}$) So $E\in \cap_{s>t}\sigma(\mathcal{F}_s,\mathcal{N})$.

***On the other hand, let's show $\sigma(\cap_{s>t}\mathcal{F}_s,\mathcal{N})\supseteq\cap_{s>t}\sigma(\mathcal{F}_s,\mathcal{N})$.

For any $E\in\cap_{s>t}\sigma(\mathcal{F}_s,\mathcal{N})$, $E\in\sigma(\mathcal{F}_s,\mathcal{N})$ for each $s>t$. Let $\{s_k\}_{k=1}^\infty$ be a sequence that decreases to $t$ but larger than $t$ (i.e. $t<\dots\le s_3 \le s_2 \le s_1$ and $s_k\to t$ as $k\to \infty$) so $E\in\sigma(\mathcal{F}_{s_k},\mathcal{N})$. Thus, for each $k=1,2,\dots$, there exist $A_k,B_k\in\mathcal{F}_{s_k}$ such that $A_k\subseteq E \subseteq B_k$ and $P(B_k\setminus A_k)=0$. It is easy to see that we also have $\cap_{k\ge m}A_k \subseteq E\subseteq \cup_{k\ge m}B_k$ for $m=1,2,\dots$, and hence $\cup_{m\ge 1}\cap_{k\ge m}A_k \subseteq E\subseteq \cap_{m\ge1}\cup_{k\ge m}B_k$. Define $A=\cup_{m\ge 1}\cap_{k\ge m}A_k$ and $B=\cap_{m\ge1}\cup_{k\ge m}B_k$.

Claim 1: $A\in\cap_{s>t}\mathcal{F}_s$. To see why, notice that $\cap_{k\ge m}A_k$ is non-decreasing in $m$ so $A=\cup_{m\ge 1}\cap_{k\ge m}A_k=\cup_{m\ge 2}\cap_{k\ge m}A_k=\cup_{m\ge 3}\cap_{k\ge m}A_k=\dots$. Let $\epsilon>0$ be arbitrary. Since $s_k$ decreases to $t$, there exists $M>0$ such that $t<s_k<t+\epsilon$ for all $k\ge M$. So $A=\cup_{m\ge M}\cap_{k\ge m}A_k\in\mathcal{F}_{t+\epsilon}$ because in $\cup_{m\ge M}\cap_{k\ge m}A_k$, all $k$ are greater than or equal to $M$ and thus $A_k\in\mathcal{F}_{s_k}\subseteq\mathcal{F}_{t+\epsilon}$. Since $\epsilon>0$ is arbitrary, so $A\in \cap_{\epsilon>0}\mathcal{F}_{t+\epsilon}=\cap_{s>t}\mathcal{F}_s$.

Claim 2: $B\in\cap_{s>t}\mathcal{F}_s$. It can be proved using the fact $\cup_{k\ge m}B_k$ is non-increasing in $m$ and similar argument in the proof of claim 1.

Claim 3: $P(B\setminus A)=0$. To show this claim, notice that $$\begin{aligned}B\setminus A&=(\cap_{m\ge1}\cup_{k\ge m}B_k)\setminus (\cup_{m\ge 1}\cap_{k\ge m}A_k)=(\cap_{m\ge1}\cup_{k\ge m}B_k)\cap (\cap_{m\ge 1}\cup_{k\ge m}A_k^c)\\&=\cap_{m\ge 1}[(\cup_{k\ge m}B_k)\cap(\cup_{k\ge m}A_k^c)]=\cap_{m\ge 1}\cup_{k\ge m}(B_k\cap A_k^c)\\&=\cap_{m\ge 1}\cup_{k\ge m}(B_k\setminus A_k)\subseteq\cup_{k\ge 1}(B_k\setminus A_k).\end{aligned}$$ So $0\le P(B\setminus A)\le P (\cup_{k\ge 1}(B_k\setminus A_k))\le \sum_{k=1}^\infty P(B_k\setminus A_k)=0.$

Now using the fact that $A\subseteq E \subseteq B$ and claims 1,2,3, we conclude $E\in \sigma(\cap_{s>t}\mathcal{F}_s,\mathcal{N})$.