Brownian motion has unbounded variation. I tried to solve this problem in the following way.
Suppose $\{B_t | t\in [0,1]\}$ is our Brownian motion. Define,
$$f_n(w) = \sum_{k=1}^{2^n} \bigg|B_{\frac{k}{2^n}}(w)-B_{\frac{k-1}{2^n}}(w)\bigg|.$$
Firstly, I showed that $f_n \leq f_{n+1}$ using triangle inequality.
We have to show that $f_n \xrightarrow{a.s} \infty$.
For that I want to prove first the following,
$\mathbb{P}(f_n\geq\alpha)\rightarrow 1\ \forall \alpha > 0....(*)$.
Then from the definition of almost sure convergence we can conclude our desired result.
But I couldn't able to show the $(*)$. How to show that?
 A: Let $Z$ be a standard normal, so that $E|Z|=\sqrt{2/\pi}$. $\;$Then  $$E(f_n)= 2^{n/2}\cdot E|Z| =    2^{n/2} \cdot \sqrt{2/\pi} $$  and
$$\text{Var}(f_n) \le 1 \,,$$
so by Chebyshev's inequality,
$$P[f_n \le E(f_n)/2] \le \frac{4}{2^{n} \cdot  2/\pi}\,.$$
A: Brownnian motion is continuous and has positive finite quadratic variation, in fact
$$\lim_n\sum|B_{t^n_{j-1}}-B_{t^n_j}|^2=T$$
where the supremum is taken partitions $P_n\subset P_{n-1}$ of $[0,T]$ with $\operatorname{mesh}(P_n)\xrightarrow{n\rightarrow\infty}0$. This implies that  Brownian motion has infinite variation over $[0,T]$. This follows from a slight adaptation to the following Lemma

Lemma: Suppose $f\not\equiv0$ is  continuous on $[a,b]$ and of finite variation, that is
$V_f[a,b]=\sup_{P}\sum^{n_p}_{k=1}|f(x_k)-f(x_{k-1})|<\infty$,
where supremum is taken over all partitions $P$ of $[a,b]$. Then
$$V^2_f[a,b]:=\lim_{\|P\|\rightarrow0}\sum^{n_P}_{k=1}|f(x_k)-f(x_{k-1})|^2=0$$
where as before, the supremum is taken over partitions $P$ of $[a,b]$ such that $\|P\|\rightarrow0$.


A detailed proof of the facts mentioned in my posting can be found in Protter, P. E., Stochastic Integration and Differential Equations, 2nd edition, Springer-Verlag Berlin Heidelberg 2004, pp. 18-19.
A: Hints: Let $(X_i)$ be i.i.d. standard normal variates. Then $f_n$ has the same distribution as $2^{-n/2} \sum\limits_{k=1}^{2^{n}} |X_i|$. By CLT we have $f_n-2^{n/2}E|X_1| \to \sqrt {Var (|X_1|)}X_1$ in distribution. Thus, $f_n \to \infty$ in probability. Since $(f_n)$ is increasing it follows that $f_n \to \infty$ almost surely.
