Referring to https://en.wikipedia.org/wiki/Brownian_motion, it says that for a Brownian motion $\{W_t\}_{t \geq 0}$, we have that $W_t$ is almost surely continuous. What exactly does this mean?
Does this mean that
$$\mathbb{P}(\omega\in \Omega: t \mapsto W_t(\omega) \mathrm{\ is \ continuous})= 1$$? Or is there another interpretation?
I think what they are saying is there is a set $E$ of probability $0$ such that $\omega \notin E$ implies $t \to W_t(\omega)$ is continuous. Measurability of $\{\omega: t \to W_t(\omega) \text{ is continuous} \}$ is not assumed. What you have written is correct if the probability space is assumed to be complete.