Ito Math ... an appendix to Part II
First we have to consider the ordinary, garden-variety Riemann Sum ...
>Huh?
Patience. A Riemann Sum goes like so:
  1. We start with a function f(t) where t (which we can think of as time)
    lies in the interval [0,T]
  2. We subdivide this interval by equally spaced points t1, t2, t3 ... tn
    where Δt = tk+1 - tk   (k = 1, 2, ... n-1)
  3. We construct the Riemann Sum:
          f(t1)Δt+ f(t2)Δt + f(t3)Δt + ... + f(tn)Δt = Σ f(tk)Δt
Figure 1 shows this graphically, where the interval is [0,2] and Δt = 0.25.
Each term, like f(tk)Δt, represents a rectangular area (as shown in dark green, in Figure 1).

If we make the subintervals shorter and shorter, meaning that Δt is made smaller and smaller, then (as seen in Figure 2), the sum of all the rectangular areas (that's our Riemann Sum) approaches ...
>The limiting value of this Riemann Sum is the actual area.
Yes, as Δt approaches 0. This limit is a so-called Riemann Integral:
      Σ f(tk)Δt f(t) dt   as Δt 0


Figure 1


Figure 2

We need one other thing:

[f(tk+1) - f(tk)] / Δt is the slope of the chord, from the point where t = tk
to the point where t = tk+1 ... as seen in Figure 3.

However, if f(t) is a smooth (i.e. differentiable) function, then it's also the slope of the tangent at some point between tk and tk+1. Such a point is labelled sk in Figure 3. That means ...
>When we see [f(tk+1) - f(tk)] / Δt we can replace it with f '(sk).
Exactly! Or, equivalently, whenever we see [f(tk+1) - f(tk)] we can replace it by f '(sk)Δt.

     
      Figure 3
Okay. Let's forge ahead.
We'll consider the following sum, called the Quadratic Variation of the function f on the interval [0,T]:

      [f(t2) - f(t1)]2 + [f(t3) - f(t2)]2 + [f(t4) - f(t3)]2 + ... + [f(tn) - f(tn-1)]2 = Σ [f(tk+1) - f(tk)]2

Write
      [f(tk+1) - f(tk)]2 = [f '(sk)Δt] 2
and we get:
      Σ [f(tk+1) - f(tk)]2 = Σ[f '(sk)Δt] 2 = Δt Σ f '(sk)2Δt

As Δt 0 we get:   Σ f '(sk)2Δt f '(t)2 dt   and (of course!)   Δt 0
hence you can guess what their product approaches ...
If f(t) is a smooth (differentiable) function on [0,T], then Σ [f(tk+1) - f(tk)]2 0   as Δt 0

>So? Is that good?
It's to be expected, since that little guy, Δt , appears squared. One of them is incorporated into the Integral and the other ...
>Makes the result zero?
Yes. For smooth functions, the Quadratic Variation is zero.
>Is that good?
Well, the remarkable thing is this:

    If f(t) is stochastic function, its Quadratic Variation is NOT zero !

>Wow! Uh ... is that good?