Points of bounded variation of Wiener process May be my luck of knowledge made me ask this question. And I'm sorry for if a question is inappropriate.
I rigorously understand that Wiener process is not one of bounded variation. That is for some $\omega$-s the path of the process is not of bounded variation. But I would like to know how large is the size of the set of that $\omega$-s. More precisely $P(\omega | W_{\omega}(t) \ is \ not \ a \  function \ of \ bounded \ variation)=?$. 
I can't guess whether this property holds almost everywhere.
Question has arised in finance. More exactly if the property holds almost everywhere, wouldn't that mean that making right trades (or holding right positions), we can at finite time make infinite profit. (As it seem to me, the total variation represents the maximum profit we could gain).
 A: Suppose that $W$ is of bounded variation on $[0,T]$ for $\omega\in A$, where $P(A)>0$, and call it $V$. Consider partitions $\Pi_n=\{0=t_0^n<t_1^n<\ldots<t_{r_{n}}^n=T\}$ such that their mesh tends to $0$ when $n$ grows. We have $$\lim_n \sum_{k=1}^{r_{n}}|W(t_k^n)-W(t_{k-1}^n)|=V$$ on $A$. Now,$$\sum_{k=1}^{r_{n}}(W(t_k^n)-W(t_{k-1}^n))^2\leq   \left(\sup_{1\leq k\leq r_{n}}|W(t_k^n)-W(t_{k-1}^n)|\right)\sum_{k=1}^{r_{n}}|W(t_k^n)-W(t_{k-1}^n)|.$$ As $W$ is almost sure continuous and of quadratic variation on $[0,T]$, and $P(A)>0$, there must exist $\omega\in A$ such that $W(\omega)$ is continuous and of quadratic variation on $[0,T]$ (which is known to be $T$). Then, taking limits, $T\leq 0\cdot V(\omega)=0$, which is a contradiction. Thus, as you thought, $$P(\{\omega:\,W(\omega)\text{ is not of bounded variation on }[0,T]\})=1.$$
Concerning the finance part:
Usually, in a discrete time setting $\{0=t_0<t_1<\ldots<t_n=T\}$, $$\text{money}=\sum_{j=1}^n H_j\cdot(S(t_j)-S(t_{j-1})),$$
where $\{H_j\}_{j=1}^n$ is a predictable process (mathematically, this means that $H_j$ is $\mathcal{F}_{j-1}$-measurable, where $\mathcal{F}_j=\sigma(S(t_0),\ldots,S(t_j))$, but do not worry if you do not understand this). In a real setting, $H_j$ represents your bet for time $j$, and it has to be decided at time $j-1$ based on how the price of stocks evolved until time $j-1$ (this is the idea of predictable process).
You are describing the following situation: $H_j=\text{sign}(S(t_j)-S(t_{j-1}))$. Here, $H_j$ depends on $S(t_j)$, so you have not decided your bet at time $j-1$ (your process $\{H_j\}_{j=1}^n$ is not predictable). Thus, in your situation, you decide your bet at time $j$ being based on $S(t_j)$, so obviously you will always win. 
Example: Imagine a game, in which you start with $0$ money, and:


*

*choose heads or tails,

*toss a coin: if the result corresponds to what you chose, you win $1$ unit of money, otherwise you lose $1$ unit of money.

*come back to 1.
Then your "stocks" are: $S_0=0$, $S_j=S_{j-1}+\xi_j$, where $\xi_j$ is $1$ if the $j$-th coin is heads or $-1$ if it is tails. Your bet for the $j$-th coin (the bet is decided at time $j-1$) is $H_j$, which is $1$ if you think that the $j$-th coin will be heads and $-1$ if you think it will be tails. Your final money will be $\sum_{j=1}^n H_j\xi_j$.
