Joint measurability of a Brownian motion
Solution 1:
Approximate $B(t,\omega)$ by the sequence of simple processes $\{B^{(n)}(t,\omega)=B(2^{-n}\lfloor 2^{n}t\rfloor,\omega)\}$. It is easy to see that $(t,\omega)\mapsto B^{(n)}(t,\omega)$ is $\mathcal{B}_{[0,\infty)}\otimes\mathcal{A}$-measurable and $B^{(n)}\to B$ pointwise as $n\to\infty$ ($\because$ $B$ is continuous). Thus, $B$ is also measurable as the limit of measurable functions.