Let us consider a diffusion process defined as $dX\_t = g(X\_t,t) \, dt + \sigma \, dW\_t$ which induces a path measure $Q$ in the time interval $[0,T]$.


Is the following expectation
$$ \left\langle \int^T\_0 \frac{f(X\_t)-g(X\_t,t)}{\sigma^2} \, dW\_t \right\rangle\_Q, $$
constant or zero?
Here $f$ is a bounded function.



Expectation of stochastic integral
As mentioned in an answer [here](https://math.stackexchange.com/questions/232932/it%C5%8D-integral-has-expectation-zero) and in the blog [here](https://almostsuremath.com/2009/12/06/martingales-and-elementary-integrals/#scn_mart_lem2),



> 
> A sufficient condition for the integral $\int\_0^t f(\omega, s)\, dB\_s$ to be a martingale on $[0,T]$ is that
> 
> 
> 1. $f(\omega,s)$ is adapted, measurable in s, *and*
> 2. $\mathbb{E}\left(\int\_0^T f^2(\omega,s)\,ds\right) < \infty$.
> In this case, indeed, $\mathsf{E} \left(\int\_0^T f(\omega,s)\, dB\_s\right)=0$.
> 
> 
> 


So if those conditions are satisfied for the given $f,g$ (eg. the integrability condition), then yes.


If we don't have the integrability, that integral will not even be well-defined/finite and so we cannot compute its expectation. Also, by taking $f=x^{p}$ for some $0<p<1/2$, we then deal with rough-integrals.


