Divergence of Riemann sums in the Itô integral Some motivation:
Let $W$ be a standard Brownian motion, and $X$ an integrable process with respect to $W$, i.e. progressively measurable with respect to the natural filtration of $W$ and square integrable on compacts almost surely.
It is known that if $X$ is continuous, then for any sequence of partitions $\mathcal P_n$ of $[0, T]$ with mesh going to $0$, the Riemann sums
$$\sum_{i = 0}^{K_n - 1} X_{t^n_i} (W_{t^n_{i+1}} - W_{t^n_i})$$
converge in probability to the Itô integral
$$\int_0^T X_t \, dW_t.$$
Here we have written $\mathcal P_n := \{t^n_0, \dots, t^n_{K_n}\}$.
Now it follows that if we take a suitable subsequence of partitions, the Riemann sums converge almost surely to the Itô integral. But I am wondering how almost sure convergence fails to begin with! It is unclear to me how a certain choice of partitions could lead to the limit fluctuating instead of existing.
Questions:

*

*Can anyone provide an explicit example of an almost surely continuous process $X$, integrable with respect to $W$, and a sequence of partitions $\mathcal P_n$ with mesh going to $0$ such that

$$\lim_{n \to \infty} \sum_{i = 0}^{K_n - 1} X_{t^n_i} (W_{t^n_{i+1}} - W_{t^n_i})$$
fails to exist with some positive probability, or even full probability?


*For a given $X$, are there some growth conditions on $\mathcal P_n$ that ensure the integral would converges a.s.?

 A: Here is an example of an integrand $X$ and a deterministic sequence of partitions such that almost surely, the limit fails to exist.
Indeed, consider $X = W$. The associated Riemann sums are
$$\lim_{n \to \infty} \sum_{[u, v] \in \mathcal P_n} W_u (W_{v} - W_{u}).$$
We may rewrite this as
$$\frac{1}{2} W^2_T - \frac{1}{2} \sum_{[u, v] \in \mathcal P_n} (W_u - W_v)^2.$$
Now, Exercise 1.15 in Peres and Mortens' Brownian Motion yields a sequence of deterministic partitions $\mathcal P_n$ such that $\limsup_{n \to \infty} \frac{1}{2} \sum_{[u, v] \in \mathcal P_n} (W_u - W_v)^2 = \infty$, almost surely, whence the result follows.
A: Here is a rough idea. Let
$$X_{t}^{N}=\sum_{\lvert n\rvert\leq N} e_{n}\xi_{n}\text{ and } Y_{t}^{N}=\sum_{\lvert n\rvert\leq N}\operatorname{sgn}(n) e_{n}\xi_{n},$$
for independent $\xi_{n}$ Gaussians and eigenbasis $e_n$, both of which converge to Brownian motion. Then
\begin{equation*}
\int_{0}^{2\pi }Y^{N}_{t}d X^{N}_{t}=\frac{\xi_{0}^{2}}{2}+\sum_{n=1}^{N}\frac{\xi_{n}^{2}+\xi_{-n}^{2}}{n}\to \infty   
\end{equation*}
diverges even though $X^N,Y^N \rightrightarrows \text{Brownian motion}$.
But then due to the uniform convergence to Brownian motion, we should be able to compare the above diverging integral to the sum from the main question.
In terms of the mesh, Bananach gave an answer to What is the explicit obstruction to almost sure convergence in 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.
If $f$ is Lipschitz, this basically means that any $h_n=n^{-1-\epsilon}$ guarantees pathwise convergence.

At Understand better stochastic integral through a.s. convergence, TheBridge answered:

So first why is stochastic integration impossible (*) ?
The proof of this lies on the Banach–Steinhaus Theorem which allows us to prove that:

Th. 56 of Protter's book :
If the sum $S_n=\sum_{i=1}^{n} f(t^{(n)}_i).(B_{t^{(n)}_{i+1}} - B_{t^n_i})$
converges to a limit for every continuous function $f$ then $B$ is of finite variations.

(*) You can take a look at the argument in Philip E. Protter's book "Stochastic Integration and Differential Equations" at Chapter 1 Section 8.

