Measurability of the supremum of a Brownian motion

Question

After reading some text books about Brownian Motion i often encountered the following object $$ \sup_{t \in [0, T]} B_t, $$ where $(B_t)_{t \geq 0}$ is a Brownian Motion.
But how do i see that this object is measurable? The problem is that the supremum is taken over an uncountable set. Do i miss something trivial or why is no text book mentioning why this object is well-defined as a random variable?

Answer 1

Typically a Brownian motion is defined to have continuous sample paths. If you take some countable, dense subset $S\subset [0,T]$, you then have

$$\sup_{t\in S} B_t = \sup_{t\in [0,T]} B_t$$

by continuity of $B_t$.

Answer 2

You need to use the fact that $B_t$ depends continuously on $t$. That shows that the sup is the same as the sup restricted to rational $t$.