Stochastic integral of continuous integrand with bounded variation

Question

Let me consider the $0=t_{0}<t_{1}<\ldots<t_{n}=T$ partition of the time interval $\left[0,T\right]$. I know that under very general assumptions, if $X$ is a continuous deterministic function, then

$$\sum_{i}X\left(s_{i}\right)\left[Y\left(t_{i}\right)-Y\left(t_{i-1}\right)\right]\overset{p}{\rightarrow}\int_{0}^{T}X\left(t\right)dY\left(t\right),$$

where $s_{i}\in\left[t_{i-1},t_{i}\right]$. So it is not necessary to choose $s_{i}=t_{i-1}$ (the beginning of the time interval $\left[t_{i-1},t_{i}\right]$) in the integral approximating sum, as the Itô integral construction would suggest. (See my previous question: Convergence of approximating integral sum in case of stochastic integrals).

I know that the convergence above is guaranteed if $s_{i}=t_{i-1}$ (, even in the case when $X$ is continuous, but NOT necessary has bounded variation).

Can the convergence above somehow be expanded to continuous integrand with bounded variation? Other words, does it matter where to choose $s_{i}$ in the integral approximating sum in order to reach any kind of convergence when the $X$ integrand is continuous (stochastic) process with bounded variation?

I think choosing $s_i$ does matter (so we can't choose it arbitrarily), but do we know any (counter) example?

Answer 1

By partial summation/integration it is possible to have the $s_i \in (t_i, t_i+1)$ for all $i$.

Indeed we have $$ \sum_{i=0}^{n-1} X_{s_i}(Y_{t_i} - Y_{t_{i-1}}) = \sum_{i=0}^{n-1} X_{s_i} Y_{t_{i+1}} - \sum_{i=0}^{n-1} X_{s_i} Y_{t_i} = X_{s_{n-1}} Y_{t_n} - X_{s_0} Y_{t_0} - \sum_{i=1}^{n-1} Y_{t_i} (X_{s_i} - X_{s_{i-1}}) \rightarrow X_T Y_T - X_0 Y_0 - \int_0^T Y_r d X_r = \int_0^T X_s \, d Y_s$$.

Here we used that the integral with respect to $X_r$ is simply a Stieltjes integral and there is no quadratic covariation between $X$ and $Y$ since $X$ is of bounded variation.