Proof that augmented filtration is right continuous Consider a filtered probability space $(\Omega, \mathcal{F}, \mathbb{F}, P)$, where $\mathbb{F}= \left\{\mathcal{F}_t:t\ge 0\right\}$ is a filtration on $(\Omega,\mathcal{F})$. Let $$\mathcal{N} = \left\{A\subseteq \Omega: A\subseteq B\text{ for some }B\in\mathcal{F}\text{ with }P(B) = 0\right\}$$ be the collection of $P$-null sets.
Often times when working with stochastic processes, one likes to consider a complete probability space with a right-continuous filtration. Therefore, they replace $\mathcal{F}$ by $\mathcal{F}^* = \sigma(\mathcal{F}\cup\mathcal{N})$ and replace $\mathcal{F}_t$, $t>0$, by $$\mathcal{F}_t^* = \bigcap_{s>t}\sigma(\mathcal{F}_s\cup\mathcal{N}).$$
The definition of a right-continuous filtration $\mathbb{G}= \left\{\mathcal{G}_t:t\ge 0\right\}$ is that for each $t\ge 0$,
$$\mathcal{G}_t = \bigcap_{s>t} \mathcal{G}_s.$$
In our case, I would like to show that for each $t\ge 0$,
$$\mathcal{F}_t^* = \bigcap_{s>t}\mathcal{F}_s^*,$$
so that I know this procedure actually produces a right-continuous filtration.
Writing things out a little bit more, we want to show that for each $t\ge 0$,
$$\bigcap_{s>t}\mathcal{F}_s^* \equiv \bigcap_{s>t}\bigcap_{r>s}\sigma(\mathcal{F}_r\cup\mathcal{N})\stackrel{\text{want}}{=}\bigcap_{s>t}\sigma(\mathcal{F}_s\cup\mathcal{N}) \equiv \mathcal{F}_t^*.$$
We see that the above entails showing that
$$\bigcap_{r>s}\sigma(\mathcal{F}_r\cup\mathcal{N}) = \sigma(\mathcal{F}_s\cup\mathcal{N}).$$
The trivial direction is showing that
$$\sigma(\mathcal{F}_s\cup\mathcal{N})\subseteq \bigcap_{r>s}\sigma(\mathcal{F}_r\cup\mathcal{N}).$$
(This is true since for all $r>s$, $\mathcal{F}_s\subseteq\mathcal{F}_r$, which implies that $\mathcal{F}_s\cup\mathcal{N} \subseteq\mathcal{F}_r\cup \mathcal{N}$ for all $r>s$.)
On the other hand, how does one show that
$$\bigcap_{r>s}\sigma(\mathcal{F}_r\cup\mathcal{N})\subseteq \sigma(\mathcal{F}_s\cup\mathcal{N}).$$
Obviously, for any $A\in\bigcap_{r>s}\sigma(\mathcal{F}_r\cup\mathcal{N})$, we have that $A\in\sigma(\mathcal{F}_{s+\epsilon}\cup\mathcal{N})$ for any $\epsilon>0$. But how does one take the final step to actually conclude that $A\in\sigma(\mathcal{F}_s\cup\mathcal{N})$? Help would be appreciated.
As a reference, I am taking this from a claim made in J. Michael Harrison's book Brownian Models of Performance and Control on page 172 (Appendix A, Section A.1).
I recognize that there have been other questions very similar to this one, but I could not find the exact question I had in terms of the way this new filtration was generated (i.e., some other posts did things in a different order, etc.).
 A: You're trying to prove the wrong thing. It's not true in general that $\sigma(\mathcal{F}_s \cap \mathcal{N}) = \bigcap_{r>s}\sigma(\mathcal{F}_r\cup\mathcal{N})$. 
If this were true, there'd be no point introducing $\mathcal{F}_t^*$ at all since then we would have $$\sigma(\mathcal{F}_s \cap \mathcal{N}) = \bigcap_{r>s}\sigma(\mathcal{F}_r\cup\mathcal{N})$$ which is the same as saying $\mathcal{F}_s^* = \sigma(\mathcal{F}_s \cup \mathcal{N})$. 
Instead the reason that the statement is true is that the indices of the double intersection of $\bigcap_{s>t}\bigcap_{r>s}\sigma(\mathcal{F}_r\cup\mathcal{N}){=}\bigcap_{s>t}\sigma(\mathcal{F}_s\cup\mathcal{N})$ run over the same set as in the single intersection. That is, we have
\begin{align*}
A \in \bigcap_{s>t}\bigcap_{r>s}\sigma(\mathcal{F}_r\cup\mathcal{N}) &\iff A \in \sigma(\mathcal{F}_r \cup \mathcal{N}) \text{ for every } r > s > t \\
& \iff A \in \sigma(\mathcal{F}_r \cup \mathcal{N}) \text{ for every } r > t \\
& \iff A \in \bigcap_{s>t}\sigma(\mathcal{F}_s\cup\mathcal{N})
\end{align*}
since the $s$ appearing in the first line doesn't actually do anything. 
