On a complete probability space, let $\mathcal{H}$ denote all square integrable random variables and let $\mathcal{H}_0$ denote all square integrable r.v.s with zero mean.

A stochastic process $X$ is called a second order process if $\mathbf{E}X(t)^2 < \infty$ and $\mathbf{E}X(t) = 0$, all $t$. It can be regarded as a curve in $\mathcal{H}_0$, i.e. a map $[0,T] \rightarrow \mathcal{H}_0$. Such a process is called q.m. continuous if this map is continuous, i.e. $\mathbf{E}(X(s)-X(t))^2 \rightarrow 0$ as $s \rightarrow t$. One can show that each q.m continuous process has a measurable version.

We define $\mathcal{L}(X,t)$ to be subspace of $\mathcal{H}_0$ generated by $\lbrace X(s) : s \leq t \rbrace$. For $Y \in \mathcal{H}_0$, the projection of $Y$ onto $\mathcal{L}(X,t)$ is called the best linear estimate of $Y$ given $X(s)$, $s \leq t$. One can show that, for a certain class of q.m. second order processes, one has $\mathcal{L}(X,t) = \lbrace \int_0^t f(s) \mathrm{d} X(s) : f \in L_2[0,t] \rbrace$ (this is not needed here, just for intuition).

Question. Do we have \begin{equation} \int_0^t X(s) \mathrm{d} s \in \mathcal{L}(X,t) \ \text{?} \end{equation}

Idea of proof: One can show that $X$ is q.m. continuous if and only if the covariance function $r(s,t) = \mathbf{E}X(s)X(t)$ is continuous. Using this, one can show that for every sequence $\lbrace t^n \rbrace$ of partitions of $[0,t]$, the Riemann sums \begin{equation} \xi_n = \sum_i X(t_i^n) ( t_{i+1}^n - t_i^n ) \end{equation} are converging in $\mathcal{H}_0$ to the same limit, say $\xi$. Now, the crucial question, which is even more interesting than the first one is: do we have \begin{equation} \xi = \int_0^t X(s) \mathrm{d} s \quad \text{a.s.?} \end{equation} This seems very logic, but since the right hand side is defined pathwise as a Lebesgue integral, this result is not immediate, and I find it difficult to show this formally.

It would be enough if we knew that the paths of $X$ are a.s. Riemann integrable, but this is not clear. More generally, the assertion follows from the following conjecture: if a Riemann sum converges and the corresponding Lebesgue integral exists, then the limit of the Riemann sum must be this integral. But this is also not clear to me. I also tried to use some approximation argument with continuous processes, but failed.

I found this question quite interesting, because it would be really surprising if the q.m. integral has a chance of not being equal to the pathwise Lebesgue integral, but somehow, the proof seems to be difficult. If it is true, then it is also an interesting result that Lebesgue integrals can be approximated by Riemann sums in quadratic mean.