I'm reading Le Gall's book Brownian Motion, Martingales, and Stochastic Calculus and can't understand the remark after Theorem 6.15 on page 164.
Let $(\Omega, \mathscr{F}, (\mathscr{F}_t)_{t\in[0,\infty]}, P)$ be a filtered probability space. Assume that the filtration $(\mathscr{F}_t)_{t\in[0,\infty]}$ is right-continuous. Let $\tilde{\mathscr{F}_\infty}=\mathscr{F}_\infty$ and $$\tilde{\mathscr{F}_t}=\sigma\left(\mathscr{F}_t\bigcup\sigma(\mathscr{N})\right), \ \ \forall t\geq0,$$ where $\mathscr N$ denotes the class of all $\mathscr F_\infty$-measurable sets that have zero probability. Then the filtration $(\tilde{\mathscr{F}_t})_{t\in[0,\infty]}$ is right-continuous.
To show the right-continuity of $(\tilde{\mathscr{F}_t})$, we need to show that for every $t\geq0$,
$$\bigcap_{s>t}\sigma\left(\mathscr{F}_s\bigcup\sigma(\mathscr{N})\right)=\sigma\left(\mathscr{F}_t\bigcup\sigma(\mathscr{N})\right).$$
It is clear that the RHS is included in the LHS. For the converse inclusion, let $A\in\bigcap_{s>t}\sigma\left(\mathscr{F}_s\bigcup\sigma(\mathscr{N})\right)$ so $A\in\sigma\left(\mathscr{F}_s\bigcup\sigma(\mathscr{N})\right)$ for all $s>t$. How should I continue from here?
Any help would be appreciated.
For the other inclusion let $A\in \bigcap_{n\ge 1}\sigma(\mathcal{F}_{t+1/n},\mathcal{N})$. Then for each $n\ge 1$ there exists $B_n\in\mathcal{F}_{t+1/n}$ s.t. $A\triangle B_n\in\mathcal{N}$. Let $B\equiv\limsup_{n\to\infty}B_n\in\mathcal{F}_t$. It suffices to show that $A\triangle B\in \mathcal{N}$. First, $$ B\setminus A\subseteq \bigcup_{n\ge 1}B_n\setminus A \in\mathcal{N}. $$ Similarly, $$ A\setminus B\subseteq \bigcup_{n\ge 1}A\setminus B_n \in\mathcal{N}. $$ Therefore, $A\in \sigma(\mathcal{F}_t,\mathcal{N})$.
Here we used the fact that for a $\sigma$-algebra $\mathcal{G}\subseteq \mathcal{F}_{\infty}$, $$ \sigma(\mathcal{G},\mathcal{N})=\{F\in \mathcal F_\infty:\exists G\in\mathcal{G} \text{ s.t. } F\triangle G\in\mathcal{N}\}. $$