Total variation of almost all Brownian motion paths is infinite. Some doubts along the proof I quote Schilling, Partzsch (2012).

Let $(B_t)_{t\ge0}$ be a one-dimensional Brownian motion and $(\Pi_n)_{n\ge 1}$ be any sequence of finite partitions of $[0,t]$ satisfying $\lim\limits_{n\to\infty}|\Pi_n|=0$. Define $$S_2^{\Pi}(B;t)=\sum_{t_{j-1}, t_j\in\Pi}|B(t_j)-B(t_{j-1})|^2$$ and $$\text{VAR}_p(B;t)=\sup\{S_p^{\Pi}(B;t): \Pi\text{ finite partition of }[0,t])\}$$ as the p-variation of a Brownian motion.



Statement Almost all Brownian paths are of infinite total variation. In fact we have $\text{VAR}_p(B;t)=\infty$ a.s. for all $p<2$. $\color{red}{(1.)}$





Proof Let $p=2-\delta$ for some $\delta>0$. Let $\Pi_n$ be any sequence of partitions of $[0,t]$ with $|\Pi_n|\to0$. Then
\begin{align}\sum_{t_{j-1}\text{, }t_j\in\Pi_n}\left(B(t_j)-B(t_{j-1})\right)^2&=\sum_{t_{j-1}\text{, } t_j\in\Pi_n}\left(B(t_j)-B(t_{j-1})\right)^{2-\delta}\left(B(t_j)-B(t_{j-1})\right)^{\delta}\\&\le\max_{t_{j-1},\text{ }t_j\in\Pi_n}\left|B(t_j)-B(t_{j-1}\right|)^{\delta}\sum_{t_{j-1},\text{ }t_j\in\Pi_n}\left|B(t_j)-B(t_{j-1})\right|^{2-\delta}\\&\le\max_{t_{j-1},\text{ }t_j\in\Pi_n}\left|B(t_j)-B(t_{j-1}\right|)^{\delta}\text{ VAR}_{2-\delta}(B; t)\end{align}
The left-hand side converges, at least for a subsequence, almost surely to $t$. $\color{red}{(2.)}$
On the other hand, $\lim_{\Pi_n\to0}\max_{t_{j-1}, t_j\in\Pi_n}\left|B(t_j)-B(t_{j-1})\right|^{\delta}=0$, since Brownian paths are (uniformly) continuous on $[0,t]$. $\color{red}{(3.)}$
This shows that $\text{VAR}_{2-\delta}(B;t)=\infty$ almost surely. $\color{red}{(4.)}$





Questions:
$\color{red}{(1.)}$ I know that, by definition, a function $f$ is said to be of finite total variation if $\text{ VAR}_1(f; t)< \infty$. So, why here are we trying to show that "almost all Brownian paths are of infinite total variation", by considering $\text{ VAR}_p(B; t)$ with a generic $p<2$ and not straight with $p=1$?;
$\color{red}{(2.)}$ I suspect that the property $B(t)\sim\mathcal{N}\left(0,\sqrt{t}\right)$ is somehow involved in the fact that $$\lim\limits_{|\Pi_n|\to0}\sum_{t_{j-1}\text{, }t_j\in\Pi_n}\left(B(t_j)-B(t_{j-1})\right)^2=t\text{ a.s.}\tag{1}$$but I cannot see how one can explicitly show the almost sure convergence in $(1)$ immediately above (at least for some subsequence);
$\color{red}{(3.)}$ Doesn't this contradict point $\color{red}{(2.)}$? That is, point $\color{red}{(3.)}$ seems to be saying that $\lim\limits_{|\Pi_n|\to0}\max_{t_{j-1}\text{, }t_{j}\in\Pi}\left|B(t_j)-B(t_{j-1})\right|^{\delta}=0$, while point $\color{red}{(2.)}$ states that $\lim\limits_{|\Pi_n|\to0}\sum_{t_{j-1}\text{, }t_j\in\Pi_n}\left(B(t_j)-B(t_{j-1})\right)^2=t\text{ a.s.}$;
$\color{red}{(4.)}$ Does that follow since for $\left|\Pi_n\right|\to0$, according to all proof passages, one would have
$$t\le\left(0\cdot\text{ VAR}_{2-\delta}(B;t)\right)\iff \text{ VAR}_{2-\delta}(B;t)\ge\displaystyle{\frac{t}{0}}=+\infty\iff\text{ VAR}_{2-\delta}(B;t)=+\infty\tag{2}$$? 
Finally, is the result stated in terms of "almost surely" since in general a Brownian motion is such that $t\mapsto B_t(\omega)$ is continuous for at least almost all $\omega$?
 A: *

*The Statement they want to prove is "Brownian paths are of infinite variation on $[0,t]$ almost surely". However, they say that in fact the stronger result holds. Not only for $p=1$ (that is in usual sense of definition) the total variation is infinite almost surely on that interval, but also any $p-$variance of brownian paths on $[0,t]$ is infinite almost surely (in fact, even stronger result holds, that is, brownian paths has infinite $p-$variation on ANY interval almost surely.) If you like it more, you can re-read this proof taking everytime $p=1$ and you'll prove statement "Brownian paths are of infinite variation on $[0,t]$" (but not the one about $p-$variation).


*The point is, that if you have partitions of $[0,t]$, call them $\Pi_n$ such that diameter of partitions tends to $0$ (thas is $|\Pi_n| \to 0$), then $S(\Pi_n) = \sum_{t_j \in \Pi_n} (B(t_j) - B(t_{j-1}))^2$ converges in $L_2$ to $t$ as $n \to \infty$.
Indeed $$ \mathbb E[ S(\Pi_n)] = \sum_{t_j \in \Pi_n}\mathbb E[(B(t_j) - B(t_{j-1}))^2] = \sum_{j \in \Pi_n} t_j - t_{j-1} = t$$ so that $$ \mathbb E[ (S(\Pi_n) - t)^2 ] = Var(S(\Pi_n)) = \sum_{t_j \in \Pi_n} Var( (B(t_j) - B(t_{j-1}))^2) = \sum_{t_j \in \Pi_n}(t_j-t_{j-1})^2Var( B(1)^2)$$
Where we used independence of incremets (variance of sum = sum of variances) and stationarity of increments. Now, $(t_j - t_{j-1}) \le |\Pi_n|$, and everything is positive, hence:
$$ \mathbb E[ (S(\Pi_n) - t)^2 ] \le Var(B(1)^2)|\Pi_n|\sum_{t_j \in \Pi_n} (t_j-t_{j-1}) = tVar(B(1)^2) |\Pi_n| \to 0$$
And since $S(\Pi_n) \to t$ in $L_2$, it implies $S(\Pi_n) \to t$ in probability, which then implies existence of subsequence $(n_k)$ such that $S(\Pi_{n_k}) \to t$ almost surely.


*You know (via definition) that almost all brownian paths are continuous. Continuity on compact set implies uniform continuity. Hence on $[0,t]$ almost all brownian paths are uniformly continuous, so that almost surely $\lim_n \max_{t_j \in \Pi_n}| B(t_j) - B(t_{j-1})|^\delta \to 0$ for any $\delta > 0$. (Indeed, just definition of uniform continuity).
It does not contradict point (2), cause even though $(B(t_j) - B(t_{j-1}))^2 \to 0$ almost surely as $n \to \infty$ (where $t_j \in \Pi_n$), our sum can have many and many terms in it. For example $\sum_{k=1}^n \frac{1}{n}$ converges to $1$ as $n \to \infty$ (well, it's just $1$ for any $n$), but obviously, every term goes to zero as $n \to \infty$.


*Yes, we showed (using my notation from (2)) that $$ S(\Pi_n) \le \max_{t_j}|B(t_j)-B(t_{j-1})|^\delta \cdot Var_{2-\delta}(B,t) $$
Now, using (3) we see that first term on the right goes to $0$ (almost surely, due to almost surely uniform continuity). What's more, using (2), we see that for some subsequence (and we can always take only that subsequence) $S(\Pi_{n_k}) \to t$ (again, almost surely, but now it's not the case of only almost surely continuity). Hence for $\omega \in A$, where $A$ is an intersection of sets where $B$ is uniformly continuous and  limit (in almost surely sense for subsequence $(n_k)$) is satisfied (that is $\mathbb P(A) = 1$, we must have $Var_{2-\delta}(B(\omega),t) = \infty$, cause otherwise by letting $k \to \infty$ in our inequality, we would get $$ t \le 0 \cdot Var_{2-\delta}(B(\omega),t) = 0 $$


*As for the last question, I think that some of the answers is in (4). The result is stated in "almost surely" sense because of two things. One is as you mentioned, sometimes brownian motion is just defined to have at least almost surely continuous paths (so that $ \max_{t_j}|B(t_j)-B(t_{j-1})|^\delta \to 0$ only almost surely, and not always), but even though we would define brownian motion to have always continuous paths, there would be a problem with $S(\Pi_{n_k}) \to t$ (cause we have only almost surely convergence on subsequence, and not on full sequence (at least it is not so clear that such thing would hold for whole sequence (if for example you assume $\sum_{n=1}^\infty |\Pi_n| < \infty$ , then by borel cantelli you can prove $S(\Pi_n) \to t$ almost surely).

