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Let $B(\omega,t)$ be a Brownian motion defined on some appropriately filtered probability space $(\Omega,\mathcal{F}_{t},\mathbb{P})$, and let $f(\omega,t)$ be a stochastic process defined on $\Omega$ and adapted to $\mathcal{F}_{t}$.

The stochastic integral of $f$ with respect to $B$ is defined as (over some partition $\Pi_{n}$ of $[0,t]$ such that $\|\Pi_{n}\|\to0$ as $n\to\infty$) $$\underbrace{\int_{0}^{t}f(\omega,s)\;dB(\omega,s)}_{I(\omega,t)}:=\lim_{n\to\infty}\underbrace{\sum_{j=0}^{n-1}f(\omega,t_{j})[B(\omega,t_{j+1})-B(\omega,t_{j})]}_{I_{n}(\omega,t)},$$ where the limit is taken in $L^{2}$, $L^{1}$, probability, etc. but not pointwise (pathwise) with respect to $\omega$.

In other words, $$\|I_{n}^{t}-I^{t}\|_{L^{2}(\Omega)}\to0\;\text{as}\;n\to\infty$$ or $$\lim_{n\to\infty}\mathbb{P}\left(\omega:|I_{n}^{t}(\omega)-I^{t}(\omega)|\geq\epsilon\right)=0\;\text{for all}\;\epsilon>0.$$

Since $t$ is held fixed throughout the entire limiting process, consider just a sequence of random variables $\{X_{n}(\omega)\}$. Intuitively, the difference between convergence in probability and convergence pointwise is that while the probability of the set on which the $X_{n}$ and $X$ disagree goes to $0$ for both, convergence in probability allows this set to move around in the domain $\Omega$, whereas pointwise convergence requires the set to remain fixed for all variables in the sequence past some finite point. In other words, the total probability of all the events for which $|X_{n}(\omega)-X(\omega)|>\epsilon$ must go to $0$ for both modes of convergence, but in convergence in probability the location of the events within the sample space can change (as long as the total probability is shrinking to $0$), whereas for point wise convergence they must remain fixed past some finite point (and also shrink to $0$).

Is there a way to understand this concretely for the case of the stochastic integral? For instance, what is the nature of the "shuffling" around the sample space that causes pathwise convergence to fail, and the shrinking of probability of "bad" events to $0$ that still allows for convergence in probability? I'm willing to accept an answer for the same situation as regards $[B,B](\omega,t)=t$, which is true only when the convergence of the approximating sequence is taken in probability.

The main motivation for asking this question is to help dispel the lack of intuition behind calculations involving stochastic integrals (e.g. Ito's lemma), since you'd like to think of them as taking place pointwise, but in fact they are taking place after convergence in probability.

EDIT Added some comments to perhaps narrow the question being asked.

Kola B.
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Sargera
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    Since the question is somewhat imprecise, maybe answering this would be easier.

    Since convergence in probability implies a.s. convergence of a subsequence, if we take a rapidly decreasing sequence of partitions $\Pi_{n}$, we recover pathwise convergence. Is it possible to illustrate two explicit sequences of partitions whereby one results in convergence along both modes and the other just in probability? Furthermore, does the usual $t/n$ partition decrease rapidly enough to achieve a.s. convergence?

     – Sargera Dec 21 '14 at 02:41
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    Here's another observation that leads me to believe the usual uniform $t/n$ partition is not sufficient to force pointwise convergence. If $E_{n}$ is the "bad" set on which $X_{n}$ is not close to the limit $X$, and if it marches around the sample space, then it will obstruct pointwise convergence if $P(E_{n})\to0$ sufficiently slow such that there is enough mass to always cover a set of points of positive mass. But Borel Cantelli says that if $\sum P(E_{n})<\infty$, then $E_{n}$ loses mass sufficiently fast that it can't do this, resulting in pointwise convergence; but $\sum n^{-1}=\infty$ – Sargera Dec 21 '14 at 17:45
  • I would recommend looking at rough path theory of integration :) – Marine Galantin Aug 17 '21 at 15:35

2 Answers2

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As we know, the Riemann-Stieltjes integral $\int_0^t f(s)\,dg(s)$ converges for all continuous $f$ only if $g$ has bounded variation. But Brownian $B(\omega,s)$ has unbounded variation in $s$ for (almost) all $\omega$. So writing $\int_0^t f(s) \,dB(\omega,s)$ is (pathwise) nonsense. The amazing thing is that it converges in these weak senses.

So I think the real question is not about an obstruction to pathwise convergence. But rather the amazing, unexpected convergence in mean...

GEdgar
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    I think what's actually amazing is the fact that $\int f;dg$ exists for all continuous $f$ if and only if $g$ is of bounded variation (uniform boundedness principle). The fact that we have convergence in $d\mathbb{P}$-mean isn't really surprising because of Ito's isometry (which is the only technical part of the proof; once that's established the result is basically an application of well known $L^{p}$ approximation theorems).

    The question I'm asking is along the lines of what is the nature of the "bad" set $E_{n}$ corresponding to $\Pi_{n}$ that causes pathwise convergence to fail?

     – Sargera Dec 23 '14 at 21:29
  • @TaylorMartin Ito's isometry is surprising to me! – guest Dec 29 '14 at 08:46
  • @GEdgar Doesn't the almost sure convergence along a subsequence show exactly that $\int_{0}^{t} f(s)dB(\omega,s)$ can be given sense for almost all paths and is thu not at all nonsense? – Bananach Apr 25 '18 at 12:06
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I believe the following result answers the question raised in the comments of the original post, by the original poster, and in doing so helps "dispel the lack of intuition behind calculations involving stochastic integrals":

If the meshes are chosen regularly with meshwidth $h_n\to 0$ such that $$ \sum_{n=1}^{\infty}\mathbb{E}\left[ \int_{0}^{1}(f(\omega,x)-f(\omega,\lfloor x/h_n\rfloor h_n)^{2}\;dx\right]<\infty, $$ then the Itô sums converge almost surely to the Itô integral (under usual conditions on the integrand)

This is proven, for example, in Remark 7.7 of the textbook Brownian motion by Mörters and Peres. It is an immediate consequence of Itô's isometry.

In particular, if $f$ is $1/2+\epsilon$ Hölder continuous, then $h_n=n^{-1}$ guarantees pathwise convergence.

Bananach
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