Let $B=(B_t)_{t\in[0,\infty)}$ a Brownian motion (BM) and $(\Omega,\mathcal{F},P)$ be the probability on which $B$ is defined. Some define BM as a.s. continuous, e.g., Brownian motion is almost surely continuous or Continuous a.s. process, that is, for any $\omega\in\Omega_{0}\subset\Omega$ with $P(\Omega_0)=1$, we have $B_t(\omega)$ is a continuous function. Others (e.g., Billingsley's Probability and Measure) define as "$B_t(\omega)$ is continuous for each $\omega$".
Question 1 is, is this my understanding correct?: ``Brownian motion needs to be defined as 'continuous for every $\omega$' if we want the $B:[0,\infty)\times\Omega\to\mathbb{R}$ to be $\mathcal{B}([0,\infty))\otimes\mathcal{F}$-measurable.''
I think this is correct because of Example 1.11 in Medvegyev, Stochastic Integration theory p. 8, "An almost surely continuous process is not necessarily product measurable."
In Billingsley's Probability and Measure, 3rd ed, Theorem 37.2 it is proved that BM is $\mathcal{B}([0,\infty))\otimes\mathcal{F}$-measurable, and indeed his definition is consistent with Medvegyev's example (for every $\omega$ the path is continuous).
Question 2 If my understanding ``...'' above is correct why do people define BM as a.s. continuous? Why is it ok to have potentially non jointly measurable functions? Is it because people put filtration generated by BM anyway and the original $\mathcal{F}$ does not matter eventually?