Proof check: Is this a valid step using limits of a sequence random variables? Let $B(t)$ be a Brownian motion on $[0,T]$. Let $V(B)$ be the total variation of this Brownian motion:
$$V(B) = \lim_{\Delta \to 0} \sum_{i=1}^{n} |B(i\Delta) - B((i-1)\Delta)|$$
where $n = T/\Delta$.
I want to prove the following statement:
With probability $1$, 
$$\lim_{\Delta \to 0} \Delta^{\frac{1}{2}} \sum_{i=1}^{n} |B(i\Delta) - B((i-1)\Delta)| = c$$
where $c$ is some positive finite constant.

My attempt:
Let 
$$V_n (B) = \sum_{i=1}^{n} |B(i\Delta) - B((i-1)\Delta)|$$
Now, since the increments $B(i\Delta) - B((i-1)\Delta)$ are I.I.D normal, it follows that
$$E[V_n(B)] = \sqrt n c_1$$ and $$Var(V_n(B)) = c_2$$
which means
$$E[\Delta^{\frac{1}{2}}V_n(B)] = c_3$$ and $$Var(\Delta^{\frac{1}{2}}V_n(B)) = \Delta c_4$$
for some positive constants $c_1,c_2,c_3,c_4$. Hence it follows (I think?) that for any $\epsilon > 0$, we have
$$P\left(\bigg|\lim_{\Delta \to 0} \Delta^{\frac{1}{2}} \sum_{i=1}^{n} |B(i\Delta) - B((i-1)\Delta)| - c\bigg| > \epsilon \right)    (1)$$
$$= \lim_{\Delta \to 0} P\left(\bigg|\Delta^{\frac{1}{2}} \sum_{i=1}^{n} |B(i\Delta) - B((i-1)\Delta)| - c\bigg| > \epsilon \right)    (2)$$
$$= \lim_{\Delta \to 0} P\left(\bigg|\Delta^{\frac{1}{2}} V_n(B) - c\bigg| > \epsilon \right) (3)$$
$$\leq \lim_{\Delta \to 0} \frac{\Delta c_4}{\epsilon^2}    (4)$$
$$= 0    (5)$$
where we use Chebyshev's inequality, proving the result

The part I'm unsure of is going from (1) to (2). Is it valid to interchange limits and probabilities like this?

Edit: So I showed it converges in probability. Now, I'm not sure if it converges almost surely. I tried applying Borel-Cantelli theorem with no luck. Any tips?

Edit2: Having thought about this more, I'm now of the opinion that it does not converge almost surely, but only in probability. My thinking being that to converge almost surely, it must be different than $c$ only a finite number of times, but every time we increase $n$ we split $B(i\Delta) - B((i-1)\Delta)$ into two new normal random variables, scaled up. This seems to contradict that it is different than $c$ a finite number of times.
 A: Well, I suspect that there is a sequence of $\Delta_k$s that actually allow this to converge almost surely. That is because if $X_n$ converges to $X$ in probability, then there is a subsequence that converges almost surely. Without knowing how you are going to zero with the deltas, I cannot say much. (I suspect since you defined it as $n = T/\Delta$, you are letting $n$ go to infinity in the "usual" sense, and thus concluding delta is going to 0, in which case, I would assume that this is in probability.)
Also, it can only be different from the limit finitely many times in almost sure convergence is actually wrong. The deterministic sequence $X_n = 1/n$ converges almost surely to 0 (for every event, it converges). What has to happen is that for any epsilon, it can only be away from that epsilon ball around the limit finitely many times (which follows from Borel-Cantelli). 
A: For 1) implies 2) you can use Fatou's Lemma: $\{(limsup |f_n|)>\epsilon\}$ is a subset of  $liminf \{|f_n|>\epsilon\}$ and Fatou's Lemma followed by your calculation  now shows $P\{limsup |f_n|>\epsilon\}=0$.
