In the definition of a Stopping Time, how important are the conditions on the Filtration being complete and right-continuous? I'm reading a book and, in its section on the definition of a stopping time(continuous), the author declares at the start that for the whole section every filtration will be complete and right-continuous.
So, in the definition of a Stopping Time, how important are these conditions? Why would they matter?
 A: If the filtration is complete and right-continuous, the condition "$\{\tau \le t \} \in \mathcal F_t$ for all $t$" is equivalent to the condition "$\{\tau < t \} \in \mathcal F_t$ for all $t$."  However  I think the main importance is that if $(\tau_n)$ is a sequence of stopping times and $\mathcal F$ is right continuous and complete, then we can also say that $\sup_n (\tau_n)$, $\inf_n(\tau_n)$, $\limsup_n(\tau_n)$, and $\liminf_n(\tau_n)$ are stopping times.  If $\mathcal F$ does not satisfy the usual conditions, we can only say that $\sup_n (\tau_n)$ is a stopping time.  I would imagine that the book you're studying from included the proofs of those statements, but let me know if you would like me to add them (and point out where right-continuity of the filtration is used).
Proof of claims:
First, the condition "$\{\tau \le t \} \in \mathcal F_t$ for all $t$" always implies $\{\tau < t \} \in \mathcal F_t$ for all $t$ regardless of whether the filtration is right continuous because $$\{\tau < t \} = \bigcup_{n=1}^\infty \left\{\tau \le t - \frac 1n \right\}$$ and $\{\tau \le t - \frac 1n\} \in \mathcal F_{t-\frac 1n} \subset \mathcal F_t$.  If the filtration satisfies the usual conditions and $\{\tau < t \} \in \mathcal F_t$ for all $t$ then we have $$\{\tau \le t \} = \bigcap_{n=1}^\infty \left\{\tau < t + \frac 1n \right\}$$ and since this is the intersection of a decreasing sequence of sets in $\mathcal F_{t+\frac 1n}$ we have that $\{\tau \le t \} \in \mathcal F_{t^+} = \mathcal F_t$
To show $\inf_n(\tau_n)$ is a stopping time when $\mathcal F$ satisfies the usual conditions, we have $$\{\inf_n(\tau_n) \le t \} = \bigcap_{m=1}^\infty \bigcup_{n=1}^\infty \left\{\tau_n \le t + \frac 1m \right\} $$ (i.e. $\inf_n \tau_n \le t$ iff "for all $m \in \mathbb N$ there exists $n \in \mathbb N$ such that $\tau_n \le t + \frac 1m$").  The set $\bigcup_{n=1}^\infty \left\{\tau_n \le t + \frac 1m \right\}$ is in $\mathcal F_{t + \frac 1m}$, and so by the same argument as above we have $\{\inf_n \tau_n \le t \} \in \mathcal F_{t^+} = \mathcal F_t$.  The $\liminf$ and $\limsup$ cases are similar.  The reason $\sup$ is different is because $$\{\sup_n \tau_n \le t \} = \bigcap_{n = 1}^\infty \{\tau_n \le t\} \in \mathcal F_t$$ so we don't need to worry about right-continuity.
A: user6247850 explains nicely the role of right continuity of a filtraton. In that discussion, "completeness" of the filtration plays no role. The main role of completeness is in showing that certain hitting times are in fact stopping times. For example, if the real-valued process $(X_t)_{t\ge 0}$ is adapted to $(\mathcal F_t)$ and has right-continuous paths, and if $A$ is a Borel subset of the real line, then
$$
D_A:=\inf\{t\ge 0: X_t\in A\}
$$
can be shown to satisfy
$$
\{D_A<t\}\in\mathcal F_t,\qquad\forall t>0,
$$
provided $(\Omega,\mathcal F,P)$ is complete and each $\mathcal F_t$ contains all of the elements of $\mathcal F$ of measure $0$. If also $(\mathcal F_t)$ is right continuous, then (referring back to the answer of user6247850) we know that
$$
\{D_A\le t\}\in \mathcal F_t,\qquad\forall t\ge 0.
$$
