Proof: Brownian motion has no intervals of monotonicity

Question

I quote Morters-Peres (2010). My observations/questions in $\color{red}{\text{red}}$.

Theorem Almost surely, for all $0<a<b<\infty$, Brownian motion $\left(B_t\right)_t$ is not monotone on the interval $[a,b]$.

Proof Fix a nondegenerate interval $[a,b]$. If it is an interval of monotonicity, then we pick numbers $a=a_1\le\ldots\le a_{n+1}=b$ and divide $[a,b]$ into $n$ sub-intervals $[a_i,a_{i+1}]$. Each increment $B(a_{i+1})-B(a_i)$ has to have the same sign. As the increments are independent (by definition), this has probability $2\cdot2^{-n}$, and taking $n\to\infty$ shows that probability that $[a,b]$ is an interval of monotonicity must be zero.
$\color{red}{\text{(So far so good to me and I am believing that this suffices to prove the above theorem,}}$
$\color{red}{\text{doesn't it?)}}$
Taking a countable union gives that, almost surely, there is no nondegenerate interval of monotonicity with rational endpoints, but each nondegenerate interval would have a nondegenerate rational sub-interval.
$\color{red}{\text{(I cannot really understand the immediately above statement. Is it crucial to conclude}}$
$\color{red}{\text{the proof of the above theorem? If so, why? And what does it mean?}}$
$\color{red}{\text{Why are "countable union", "rational endpoints" and "rational sub-intervals" recalled?}}$
$\color{red}{\text{Could you please help me understand this part with a detailed answer?)}}$

Answer 1

In my opinion, by taking the countable union, the author shows that the Brownian motion is not monotonic in $[a,b]$ under the assumption that $[a,b]$ can only be decomposed as a countable union of intervals with rational endpoints (rational endpoints because is a countable union). But he has to prove it for an interval without the assumption above. Technically it could exists an interval with non-rational endpoints that hasn't been considered. So by saying: "...each nondegenrate interval would have a nondegenerate rational sub-interval", he argues that that the bold statement is not possible. The reason is that $\mathbb Q$ is dense in $\mathbb R,$ so every generic interval contains an interval with rational endpoints.

Answer 2

I believe the issue here is that the theorem is claiming that, w.p. 1, the Brownian path is non-monotonic on every interval, simultaneously. What the first part shows is that, if you are given any arbitrary interval (but only one), the path is non-monotonic on it w.p. 1. Since there are uncountably many possible intervals, some additional arguments are required to show that non-monotonicity still holds for all of them, jointly, w.p. 1.

I think this has a somewhat similar flavor to the distinction between two processes being modifications of each other vs. being indistinguishable. In one case, they are equal w.p. 1 at any given time value. In the other, the joint probability of them being equal at every time value is 1.