Does the Itô integral rely on the martingale structure of the driving Brownian motion?

Question

The proofs I have seen for the construction of the Itô integral does not use the martingale property of the driving Brownian motion to show that the integral is well-defined. But since we can extend the Itô integral to semimartingales, and this is strict, intuition says that the (semi)martingale structure is necessary. Am I overlooking something?

Answer 1

First, to be clear the Itô integral was specifically designed by Itô so that it preserves the martingale property: the process

$$S_{t}:=\int^{t} f(s)dX_{s}\approx \sum f(s_{i})(X_{s_{i+1}}-X_{s_{i}})$$

is still a martingale if $X_{t}$ is.

So all the constructions of Itô integral for Brownian motion involve the martingale property in some way. For example, if one tries to use the central-limit-theorem approach, then one uses the independent-increments property of Brownian motion which in turn implies the martingale property.

Of course, one can try different choices such as the Stratonovich integration, which lacks the martingale property but gains the chain rule. (For more general constructions see Rough paths such as the textbook "a course in Rough paths" by Friz and Hairer.)

As shown in Revuz-Yor, one can build a stochastic calculus for semimartingales $X=M+A$. However, again the focus goes to building integration for martingales because for the finite variation part $A$, we can just use the Riemann-Stieltjes integral.

So in either case the martingale-structure is essential.