• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $(\mathcal F_t)_{t\ge 0}$ be a filtration of $\mathcal A$
  • $(B_t)_{t\ge 0}$ be a real-valued Brownian motion with respect to $\mathcal F$ on $(\Omega,\mathcal A,\operatorname P)$
  • $\lambda$ be the Lebesgue integral on $[0,\infty)$

I've read the following: Let $\sigma$ and $b$ be real-valued $\mathcal F$-progressively measurable stochastic processes on $(\Omega,\mathcal A,\operatorname P)$ with $$\int_0^t\sigma_s^2+|b_s|\;{\rm d}s<\infty\tag 1\;\;\;\operatorname P\text{-almost surely}$$ for all $t\ge 0$ $\Rightarrow$ $$X_t=\int_0^t\sigma_s\;{\rm d}B_s+\int_0^tb_s\;{\rm d}s\tag 2$$ is called Itō process with diffusion coefficient $\sigma$ and drift $b$.

However, some things are weird about that definition:

  1. People state that $(1)$ and the second integral in $(2)$ are Riemann integrals. But unless the points of discontinuity of the paths of $b$ form a $\lambda$-null set, it is not Riemann integrable and hence $(1)$ is not even defined (as a Riemann integral). So, shouldn't $\int_0^t{\rm d}s$ be replaced by $\int_0^t{\rm d}\lambda(s)$? In that case $(1)$ would ensure, that almost surely the paths of $b$ are $\lambda$-integrable on $[0,t]$ for any $t\ge 0$ and thereby the second integral in $(2)$ would be almost surely well-defined (as a Lebesgue integral).
  2. But we still got a problem: Since the second integral in $(2)$ is only well-defined on $\Omega\setminus N$ for some $\operatorname P$-null set $N\subseteq\Omega$, the integral and thereby $X_t$ is undefined on $N$. So, do we in fact need to replace $X_t$ on $N$ by something well-defined?
  • $\begingroup$ "But unless $b$ has (surely) continuous paths, it is not Riemann integrable": Why do you say that? There are lots of functions that are Riemann integrable which are not continuous. For instance, every cadlag function has at most countably many discontinuities and hence is Riemann integrable. $\endgroup$ Jan 29, 2016 at 15:07
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    $\begingroup$ For your second question, sure, replace $X_t$ by $0$ on $N$, or anything else you like. Nobody cares what happens on a null set. All the theorems you want to prove say "a.s." everywhere anyway. $\endgroup$ Jan 29, 2016 at 15:08
  • $\begingroup$ @NateEldredge You're right and I've updated the question. But the point is that we still can't conclude that $t\mapsto b_t$ is Riemann integrable. So, I think we need to consider $(1)$ and the second integral in $(2)$ as a Lebesgue integral. What do you think? $\endgroup$
    – 0xbadf00d
    Jan 29, 2016 at 15:28

1 Answer 1


It should be Lebesgue integral, instead of Riemann integral. See


Second paragraph after Lemma 1, it is specifically stated: it can be shown that if V is a continuous FV process, then a predictable process {\xi} is V-integrable if and only if it is almost-surely Lebesgue-Stieltjes integrable.

In your case, $V(t) = t$, and Lebesgue-Stieltjes integrable is now simply Lebesgue integrable.


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