I am confused about how the following three facts are not in contradiction.
The question was essentially already raised here, but while I understand the proof in the link, I do not understand how the below is not a contradiction.
Let $B = (B_t)_{t \in [0,1]}$ be a standard Brownian motion. Then
- the quadratic variation of $B$ along the sequence $(\Pi_n)_{n \in \mathbb{N}} := \left\{ \frac{k}{n}: 0 \leq k \leq n\right\}_{n \in \mathbb{N}}$ of finite partitions of $[0,1]$ converges to $1$ in probability, i.e.
$$ V(\Pi_n) := \sum_{k = 0}^n \vert B(t_{i+1}) - B(t_{i}) \vert^2 \rightarrow 1, ~~~ \text{in probability} $$
- the $2$-variation of $B$ goes to $+ \infty$ almost surely, i.e.
$$ \Vert B_{\omega}(\cdot) \Vert_{2-var}^2 := \sup_{P ~~ \text{fin. partition}} \sum_{t_i \in P} \vert B(t_i) - B(t_{i+1}) \vert^2 = + \infty, ~~~ \mathbb{P}-a.e. \omega \in \Omega $$
(here the $\sup$ goes over all finite partitions of $[0,1]$).
- for a continuous function $f: [0,1] \rightarrow \mathbb{R}$ the $2$-variation is equal to the limit of the quadratic variation along any sequence of finite partitions whose mesh size goes to $0$, i.e. for any sequence of partitions $\{P_n\}_{n \in \mathbb{N}}$ with
$$ \max_{t_i \in P_n} \vert t_{i+1} - t_i \vert \rightarrow 0, ~~~ n \rightarrow \infty $$
we have
$$ \sup_{P ~~ \text{fin. partition}} \sum_{t_i \in P} \vert f(t_{i+1}) - f(t_{i}) \vert^2 = \lim_{n \rightarrow \infty} \sum_{t_i \in P_n} \vert f(t_{i+1}) - f(t_{i}) \vert^2 $$
Thus there should be a contradiction since 1. implies that there exists a subsequence $(P_{n_k})_{k \in \mathbb{N}}$ s.t.
$$ \max_{t_i \in P_{n_k}} \vert t_{i+1} - t_i \vert \rightarrow 0, ~~~ k \rightarrow \infty $$
and $V(P_{n_k}) \rightarrow 1$ almost surely. But then, since Brownian motion has sample paths almost surely, 3) implies that the limit of $V(P_{n_k})$ coincides with the quadratic variation i.e. with the $\sup$, which is infinite almost surely.
