I am trying to solve the following exercise:
Let $\mathcal F = \{ F(t) \}_{t\in [0,T]}$ be a complete and right-continuous filtration, $\tau$ a stopping time for $\mathcal F$ and $\sigma$ a random variable with $\tau = \sigma$ almost sure. Show that $\sigma$ is a stopping time.
Now according to my script, a filtration is complete if for every $t \in [0,T]$ $$ N \in \mathcal F(t) \mbox{ whenever } N \in \mathcal A \mbox{ with } P(N) = 0 $$ where $(\Omega, \mathcal A, P)$ is the surrounding probability space.
But this tells nothing about the completeness of $(\Omega, \mathcal A, P)$, so we cannot conclude that the probability spaces $(\Omega, \mathcal F(t), P)$ for $t \in [0,T]$ are complete. If they would be, then simply by $$ \{ \sigma \le t \} = ( \{ \tau \le t \} \cap \{ \tau = \sigma \} ) \cup ( \{ \sigma \le t \} \cap \{\tau \ne \sigma\}) $$ with $P(\tau \ne \sigma) = 0$ this would give $\{\sigma\le t\} \in \mathcal F(t)$. But in this reasoning the right-continuity does not enters, which wonders me, so either
i) the definition of completeness is incomplete, i.e. they forgot to mention that $(\Omega, \mathcal A, P)$ is complete or that the space $(\Omega, \mathcal F(t), P)$ should be complete (this is quite close to that definition from wikipedia), but then right-continuity does not matters here as I see it, or
ii) the definitions are right, but then I do not see how right-continuity of the filtration implies that the particular subsets $\{\sigma \le t\}\cap \{\sigma\ne \tau\}$ of the nullset $\{\sigma\ne\tau\}$ are themselve in $\mathcal F(t)$ for each $t$?
So, what do you think, is i) or ii) right; and then how to apply right-continuity?
In the common use of the term, a filtration $\{\mathcal F(t)\}_{t\ge 0}$ is "complete" provided (i) the probability space $(\Omega,\mathcal A,P)$ is complete (in the usual sense), and (ii) $\mathcal F(0)\supset\{N\in\mathcal A: P(N)=0\}$.
As you surmise, right continuity is not needed for the assertion about $\sigma$ to be true.