Let $\{X, \mathscr{F}_t^X; 0\leq t<\infty\}$ be a process with initial distribution $\mu$ on the space $(\Omega, \mathscr{F}_{\infty}^X, P^\mu)$, where $P^\mu [X_0\in\Gamma] = \mu (\Gamma)$, $\Gamma\in\mathscr{B} (\mathbb{R}^d)$. Here $\mathscr{F}_t^X=\sigma(X_s; 0\leq s\leq t)$. For $0\leq t<\infty$, $$ \mathscr{N}_t^\mu = \{ F\subseteq\Omega; \; \exists G\in\mathscr{F}_{t}^X \; with\; F\subseteq G, P^\mu (G) = 0\}. $$ Let ${\mathscr{F}}_t^\mu = \sigma ( \mathscr{F}_{t}^X \cup \mathscr{N}^\mu ) $ be the augmentation of $\mathscr{F}_t^X$ under $P^\mu$. For $t=\infty$ we set $\mathscr{F}^\mu = \sigma ( \mathscr{F}_{\infty}^X \cup \mathscr{N}^\mu )$.
Useful property: If $X$ is left-continuous, then the filtration $\{\mathscr{F}_{t}^X\}$ is left-continuous.
First problem (problem 2.7.5 from Karatzas&Shreve)
Show that the $\sigma$-field $\mathscr{F}^\mu$ agrees with $$ \mathscr{F}_{\infty}^\mu = \sigma \big(\bigcup\limits_{t\geq0} \mathscr{F}_{t}^\mu \big). $$ Solution from book: $\mathscr{F}_t^\mu = \sigma ( \mathscr{F}_{t}^X \cup \mathscr{N}^\mu ) \subseteq \sigma ( \mathscr{F}_{\infty}^X \cup \mathscr{N}^\mu ) = \mathscr{F}^\mu$ holds for every $0\leq t<\infty$, so $\mathscr{F}_\infty^\mu \subseteq \mathscr{F}^\mu$. For the opposite inclusion, take any $F\in\mathscr{F}^\mu$. By another problem (pr. 2.7.3) we know the there exists an event $G\in\mathscr{F}_\infty^X$ such that $N = F\bigtriangleup G \in\mathscr{N}^\mu$. But now $\mathscr{F}_\infty^X \subseteq \mathscr{F}_\infty^\mu$ (we have $\mathscr{F}_t^X \subseteq \sigma ( \mathscr{F}_{t}^X \cup \mathscr{N}^\mu ) = \mathscr{F}_t^\mu \subseteq \sigma \big(\bigcup\limits_{t\geq0} \mathscr{F}_{t}^\mu \big) = \mathscr{F}^\mu_\infty$ for every $0\leq t<\infty$), and thus $G\in\mathscr{F}^\mu_\infty$, $N\in\mathscr{N}^\mu\subseteq\mathscr{F}^\mu_\infty$ imply $F=G\bigtriangleup N \in\mathscr{F}_\infty^\mu.$
Second problem (problem 2.7.6 from Karatzas&Shreve)
If the process $X$ has left-continuous paths, then the filtration $\{\mathscr{F}_{t}^\mu\}$ is left-continuous.
Hint: Repeat the argument employed in the previous solution, replacing $\mathscr{F}^\mu$ by $\mathscr{F}_t^\mu$ and $\mathscr{F}_\infty^\mu$ by $\mathscr{F}_{t-}^X$ and using the left-continuity of the filtration $\{\mathscr{F}_t^X\}$.
I understand the solution of the first problem, but I am not able to solve the second even with the given hint. Could somebody help? Thanks in advance for any help!
In fact, $\mathscr F^X_{t-} = \mathscr F^X_{t}$ for each $t>0$. Clearly $\mathscr F^X_{t-}\subset\mathscr F^X_{t}$ for each $t>0$. Fix $t>0$ and suppose $A\in\mathscr F^X_t$. Then there is an infinite sequence $\{t_n\}$ of distinct times contained in $[0,t]$ such that $A\in\sigma(X_{t_n}:n\in\Bbb N)$. For later reference call this $\sigma$-field $\mathscr G$. Because $X$ is left-continuous we can arrange that $t$ is not an element of $\{t_n\}$. Indeed, by said left-continuity, $X_t$ is measurable with respect to $\sigma(X_{t(1-1/k)}: k=1,2,\ldots)$, so adding the times $(1-1/k)t$, $k=1,2,\ldots$ to the list if necessary, we can remove $t$ if it's on the list. Because now each $t_n$ is strictly less than $t$ you have $\sigma(X_{t_n})\subset \mathscr F^X_{t-}$ for each $n$, and so $\mathscr G=\sigma(X_{t_n}: n=1,2,\ldots)\subset\mathscr F^X_{t-}$. Consequently, $A\in\mathscr F^X_{t-}$.