Why is the set $\{\omega\in \Omega: \exists t \geq 0: -\infty < D^-(t,\omega) \leq D^+(t,\omega) < + \infty\}$ measurable?

Question

I'm reading a proof in Billingsley that the continuous paths of the Brownian motion $\{W_t: t \geq 0\}$ on the probability space $(\Omega, \mathcal{F}, \mathbb{P})$ are nowhere differentiable, with probability $1$. The author does this by showing that if

$$E= \{\omega\in \Omega: \exists t \geq 0: -\infty < D^-(t,\omega) \leq D^+(t,\omega) < + \infty\}$$

then $\mathbb{P}(E) = 0$.

Here $$D^- (t,\omega) = \liminf_{h \searrow 0} \frac{W(t+h,\omega)-W(t,\omega)}{h}, D^+ (t,\omega) = \limsup_{h \searrow 0} \frac{W(t+h,\omega)-W(t,\omega)}{h}$$

However, why are we sure that $E$ is a measurable set? I.e. why is $E \in \mathcal{F}$?

The following facts are definitely relevant:

(1) The maps $t \mapsto D^-(t, \omega), t \mapsto D^+(t, \omega)$ are measurable for all $\omega \in \Omega$

(2) The map $(t, \omega) \mapsto W(t,\omega)$ is jointly measurable.

We can write

$$E=\bigcup_{t \geq 0} \{\omega \in \Omega: -\infty < D^{-}(t, \omega) \leq D^+(t,\omega) < + \infty\}$$

but this does not quite help.

Any help will be appreciated!

Answer 1

Billingsley doesn't assert that $E$ is measurable. He does show that $E$ has outer measure $0$, and so will be measurable provided the measure space $(\Omega,\mathcal F,P)$ is complete. A discussion of the advantages of completion follows in the text.