Is the supremum of an almost surely continuous stochastic process measurable? Let's take a stochastic process $(X_t)_{0\leq t \leq 1}$ and assume that the sample paths are almost surely continuous. Let us define $S \equiv \sup_{t \in [0,1]} X_t$. How can we show that $S$ is measurable?
For example, the if we take the Brownian motion $B_t$ as our stochastic process, then given the continuity of the sample paths, we can focus on the supremum over $t \in [0,1] \cap \mathbb{Q}$, which is a countable, dense subset of $[0,1]$, and we have continuity of $B_t$, therefore the supremum is measurable (see the answer here: Measurability of the supremum of a Brownian motion).
How does almost sure continuity instead of continuity change the way of proving measurability?
I would be very grateful for any hint!
 A: We have to assume that the underlying probability space is complete; otherwise the assertion might fail.
So, suppose that $(\Omega,\mathcal{A},\mathbb{P})$ is a complete probability space and $(X_t)_{t \in [0,1]}$ a process with almost surely continuous sample paths, i.e. there exists a null set $N \in \mathcal{A}$ such that $$[0,1] \ni t \mapsto X_t(\omega)$$ is continuous for all $\omega \in \tilde{\Omega} := \Omega \backslash N$. Now
$$\tilde{X}_t(\omega) := \begin{cases} X_t(\omega), & \omega \in \tilde{\Omega}, \\ 0, & \omega \in N \end{cases}$$
defines a stochastic process on $\Omega$ with continuous sample paths, and therefore
$$\sup_{t \in [0,1]} \tilde{X}_t = \sup_{t \in [0,1] \cap \mathbb{Q}} \tilde{X}_t$$
is measurable as countable supremum of measurable random variables. On the other hand, we have
$$\tilde{S}_t(\omega) = \sup_{t \in [0,1]} \tilde{X}_t(\omega) = \sup_{t \in [0,1]} X_t(\omega)= S_t(\omega) \quad \text{for all $\omega \in \tilde{\Omega} = \Omega \backslash N$}$$
and so
$$\{S_t \in B\} = \left( \{\tilde{S}_t \in B \} \cap N^c \right) \cup \bigg( \{S_t \in B \} \cap N \bigg)$$
for any Borel set $B$. Since $N \in \mathcal{A}$ and $\tilde{S}_t$ is measurable, we know that
$$\left( \{\tilde{S}_t \in B \} \cap N^c \right) \in \mathcal{A}.$$
Moreover,
$$\left\{ S_t \in B \right\} \cap N \subseteq N$$
and since the probability space is complete, this implies
$$\left\{ S_t \in B \right\} \cap N \in \mathcal{A}.$$
Combining both considerations proves $\{S_t \in B\} \in \mathcal{A}$, and this proves the measurability of $S_t$.
Remark More generally, the following statement holds true in complete probability spaces: 

Let $(\Omega,\mathcal{A},\mathbb{P})$ and $(E,\mathcal{B},\mathbb{Q})$ be two measure spaces and assume that $(\Omega,\mathcal{A},\mathbb{P})$ is complete. Let $X, Y: \Omega \to E$ be two mappings. If $X$ is measurable and $X=Y$ almost surely, then $Y$ is measurable.

