Family of partitions, s.t. the quadratic variation of a BM diverges a.s. This question is about a specific step in the solution of exercise 1.13 a) of the book "Brownian Motion" by Peres and Mörters (https://www.stat.berkeley.edu/~peres/bmbook.pdf). The exercise is on page 40 and its solution on page 315. $B$ stands for a Brownian Motion.
We need to show that, almost surely, there exists a family
$0=t_0^{n} \le t_1^{n} \le ... \le t_{p(n)}^n=t$ of (random) partitions, such that $\lim\limits_{n \to \infty} \sum\limits_{j=1}^{p(n)} \left(B\left(t_j^{(n)} \right) - B\left(t_{j-1}^{(n)} \right) \right)^2 = \infty$.
The solution gives the arguments, that for given $M>0$ large, for any fixed $s\in [0,1]$, there exists $n \in \mathbb{N}$, such that the dyadic interval $I(n,s):=[k2^{-n},(k+1)2^{-n}]$ containing $s$ satisfies $\left |B\left((k+1)2^{-n}\right)-B\left(k2^{-n}\right)\right| \geq M2^{-\frac{n}{2}}$ (call it inequality 1) and call $N(s)$ the smallest integer $n$ for which this inequality holds. We think that $k$ depends on $s$, so we write $k(s)$.
The solution tells us to apply Fubini's theorem to see that $N(s) < \infty$ almost surely. My question is now, whether we applied Fubini correctly.
We thought that (call it remark 1)
$N(s) < \infty$ almost surely $\thinspace$ $\Leftrightarrow \thinspace \int_{0}^{1} \mathbb{P} \left( \underset{n \ge 0}\bigcup \left\{ \left| B \left( \frac{k(s)+1}{2^n}\right)-B \left( \frac{k(s)}{2^n}\right)\right| \geq M2^{-\frac{n}{2}}\right\}\right) \mathrm{d}s < \infty$.
Then to apply Fubini, we write
\begin{align*}&\int_{0}^{1} \mathbb{P} \left( \underset{n \ge 0}\bigcup \left\{ \left| B \left( \frac{k(s)+1}{2^n}\right)-B \left( \frac{k(s)}{2^n}\right)\right| \geq M2^{-\frac{n}{2}}\right\}\right) \mathrm{d}s \\&= 
\int_{0}^{1} \int_{\Omega} \Bbb{1}_{\left\{ \underset{n \ge 0}\bigcup \left\{ \left| B \left( \frac{k(s)+1}{2^n}\right)-B \left( \frac{k(s)}{2^n}\right)\right| \geq M2^{-\frac{n}{2}}\right\} \right\}} \mathrm{d}\mathbb{P}(\omega) \mathrm{d}s \\
&= \int_{\Omega} \int_{0}^{1}  \Bbb{1}_{\left\{ \underset{n \ge 0}\bigcup \left\{ \left| B \left( \frac{k(s)+1}{2^n}\right)-B \left( \frac{k(s)}{2^n}\right)\right| \geq M2^{-\frac{n}{2}}\right\} \right\}} \mathrm{d}s \mathrm{d}\mathbb{P}(\omega)\end{align*}
Now, since we know that there exists $n$, such that inequality 1 is fulfilled, the indicator function is equal to 1, hence the last expression is equal to $\int_{\Omega} \int_{0}^{1}  \mathrm{d}s \mathrm{d}\mathbb{P}(\omega)=1<\infty$ which implies due to remark 1 that $N(s)< \infty$.
Can anybody say something especially about the correctness of remark 1 and whether you think the author meant that by using Fubini's theorem? Thanks for any comments.
 A: 
$N(s) < \infty$ almost surely $\thinspace$ $\Leftrightarrow \thinspace \int_{0}^{1} \mathbb{P} \left( \underset{n \ge 0}\bigcup \left\{ \left| B \left( \frac{k(s)+1}{2^n}\right)-B \left( \frac{k(s)}{2^n}\right)\right| \geq M2^{-\frac{n}{2}}\right\}\right) \mathrm{d}s < \infty$

Note that the condition on the right-hand side is trivially satisfied since any bounded (measurable) function is integrable with respect to a finite measure. So, using $0 \leq \mathbb{P}(A) \leq 1$ for any measurable set $A$, it is obvious that the integral on the right-hand side is finite. However, I don't see why this is useful to prove $N(s)<\infty$.

Now, since we know that there exists $n$, such that inequality 1 is fulfilled

If you already know that there exists such $n$, then there is nothing to prove since $N(s) \leq n$ implies $N(s)<\infty$.

One possibility to prove $N(s)<\infty$ almost surely is the following: We have
$$N(s,\omega) = \infty \iff \forall n \in \mathbb{N}: |B((k+1) 2^{-n},\omega)-B(k2^{-n},\omega)| < M 2^{-n/2}.$$
Using the independence and stationarity of the increments, we find
$$\begin{align*} \mathbb{P}(N(s)=\infty) &= \mathbb{P} \left( \bigcap_{n \geq 1} \{|B((k+1) 2^{-n})-B(k2^{-n})| < M 2^{-n/2}\} \right) \\ &= \lim_{k \to \infty} \prod_{n =1}^k \mathbb{P}(|B((k+1) 2^{-n})-B(k2^{-n})| < M 2^{-n/2}) \\ &= \lim_{k \to \infty} \prod_{n =1}^k \mathbb{P}(|B(2^{-n})|<M2^{-n/2}). \end{align*}$$
Since $B_t \sim \sqrt{t} B_1$ for all $t>0$, we get
$$\mathbb{P}(N(s)=\infty) = \lim_{k \to \infty} \prod_{n =1}^k \mathbb{P}(|B_1|<M) = \lim_{k \to \infty} \mathbb{P}(|B_1|<M)^k = 0$$
as $\mathbb{P}(|B_1|<M) \in (0,1)$ for any $M>0$. This proves $N(s)<\infty$ almost surely.
