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This is a bit vague but something I have been wondering about.

Consider a Lebesgue integral $\int f \;\text{d}\mu$ - then we know that we can change $f$ on a $\mu$-null set without changing the integral.

Now consider a stochastic integral with respect to, say, a Brownian motion $W$ (or something else of unbounded variation). Then we also know that $W$ does not induce a measure. What is in this case the stochastic counterpart to the above described Lebesgue case? Is it just with respect to the probability measure? - why?

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A natural notion of "sameness" for two continuous-time processes $X$ and $Y$ is that $P(\forall t\ge0: X_t = Y_t)=1$. If this is the case, $X$ and $Y$ are said to be indistinguishable. It is not clear that the set $(\forall t\ge0: X_t=Y_t)$ is measurable, however, if $X$ and $Y$ both are cadlag, it is measurable, and for two cadlag processes it holds that $X$ and $Y$ are indistinguishable if and only if it holds for each $t\ge0$ that $P(X_t = Y_t)=1$.

A first thing to note is that the stochastic integral is only defined up to equivalence classes of indistinguishable processes. Thus, in contrast to the Lebesgue integral, the stochastic integral $H\cdot W$ is only almost surely unique. The question you ask can therefore naturally be posed as: For two progessively measurable and locally bounded proceses $H$ and $K$, when is $H\cdot W$ and $K\cdot W$ indistinguishable? Here, $H\cdot W$ denotes the stochastic integral process, and the requirements on $H$ and $K$ are there to ensure that the integral is well-defined.

The answer is as follows: It holds that $H\cdot W$ and $K\cdot W$ are indistinguishable precisely when $H$ and $K$ are indistinguishable.

To see this, it suffices by linearity to show that $H\cdot W$ is indistinguishable from zero precisely when $H$ is indistinguishable from zero. To see this, first assume that $H\cdot W$ is indistinguishable from zero. Then the zero process satisfies the requirements to be the stochastic integral of $H$ with respect to $W$ (the integral is uniquely characterized by its quadratic covariance with other local martingales, and this can be used to show that the integral is zero. More conventionally, approximation by simple processes can also be used). Conversely, assume that $H\cdot W$ is indistinguishable from zero. Then the quadratic variation $[H\cdot W]$ is indistinguishable from zero as well. Note that

$$[H\cdot W]_t = \int_0^t H_s^2 ds$$

for all $t\ge0$ almost surely. Therefore, by ordinary results from Lebesgue integration, $H$ is indistinguishable from zero.

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This is an absolutely awesome answer! It is pretty obvious, but I did not expect it to be so. Thank you very much! – Henrik Jul 16 '13 at 12:25

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