Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space. Let $H=(H\_t, t\ge 0)$ be a stochastic process with continuous trajectories. Fix $T>0$. For $n \ge 1$, we define
$$
H\_{s,n} := \sum\_{i=1}^{2^n} H\left(\frac{(i-1) T}{2^n}\right) 1\_{ \left (\frac{(i-1) T}{2^n}, \frac{i T}{2^n} \right]}(s) \quad \forall s \in [0, T].
$$


Then for $(\omega, n) \in \Omega \times \mathbb N$, the map $s \mapsto H\_{s, n} (\omega)$ is a step function. Because $H$ has continuous trajectories, we have
$$
H\_{s, n} (\omega) \underset{n \rightarrow \infty}{\longrightarrow} H\_{s} (\omega) \quad \forall (\omega, s) \in \Omega \times [0, T].
$$




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It is mentioned at page 35 of this [note](https://ipgold.epfl.ch/%7Eleveque/Lecture_Notes/sc1_final.pdf) that



> 
> **Preliminary fact** If $\mathbb E[ \int\_0^T H\_s^2 \mathrm d s] < \infty$, then
> $$
> \mathbb{E}\bigg[\int\_0^T (H\_{s,n}-H\_s)^2 \mathrm d s\bigg] \underset{n \rightarrow \infty}{\longrightarrow} 0.
> $$
> 
> 
> 


I have tried to verify this statement but got stuck. Could you elaborate on how to prove it?




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**My attempt**


* Fix $\omega \in \Omega$. We define $f\_n:[0, T] \to \mathbb R$ by $f\_n (s) := H\_{s,n} (\omega)$. We define $f:[0, T] \to \mathbb R$ by $f (s) := H\_{s} (\omega)$. Then $\sup\_n \|f\_n\|\_\infty \le \|f\|\_\infty < \infty$. Also, $f\_n-f$ converges to $0$ pointwise. By dominated convergence theorem, we have
$$
\int\_0^T (f\_n-f)^2 \mathrm d s \underset{n \rightarrow \infty}{\longrightarrow} 0.
$$
* We define $Y\_n:\Omega \to \mathbb R$ by
$$
Y\_n (\omega) := \int\_0^T (H\_{s,n} (\omega)-H\_s(\omega))^2 \mathrm d s.
$$
As proved above, $Y\_n \underset{n \rightarrow \infty}{\longrightarrow} 0$ almost surely.




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I posted [this](https://math.stackexchange.com/questions/4628234/how-is-mathbb-e-int-0t-h-s2-mathrm-d-s-infty-important-for-this-clai) question on MSE but have not received any answer so far.



How is $\mathbb E[ \int_0^T H_s^2 \mathrm d s] < \infty$ important for this claim?
This statement is false in general.


E.g., let $T=1$ and $p\_k:=2^{-k}$ for integers $k=1,2,\dots$, so that $\sum\_{k=1}^\infty p\_k=1$. Let $A\_1,A\_2,\dots$ be pairwise disjoint events with respective probabilities $p\_1,p\_2,\dots$.
Let
\begin{equation}
 H\_t:=\sum\_{k=1}^\infty 1\_{A\_k}\,2^k\,\sum\_{j=0}^{2^k}\Big(1-8^k\Big|t-\frac j{2^k}\Big|\Big)\_+,
\end{equation}
where $u\_+:=\max(0,u)$.


Then
\begin{equation}
 E\int\_0^1 H\_t^2\,dt=\sum\_{k=1}^\infty p\_k\; (2^k)^2\,2^k \int\_0^1 \Big(1-8^k\Big|t-\frac 1{2^k}\Big|\Big)\_+^2\,dt
 =\sum\_{k=1}^\infty p\_k\; (2^k)^2\,2^k \frac23\,\frac1{8^k}=\frac23<\infty. 
\end{equation}
On the other hand, for all $t\in(0,1]$ we have $H\_{t,n}\ge2^n$ on the event $A\_n$, which has probability $p\_n=\frac1{2^n}$. So,
\begin{equation}
 E\int\_0^1 H\_{t,n}^2\,dt\ge(2^n)^2\frac1{2^n}\to\infty
\end{equation}
and hence
\begin{equation}
 E\int\_0^1 (H\_{t,n}-H\_t)^2\,dt\to\infty\ne0
\end{equation}
as $n\to\infty$. $\quad\Box$




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The problem here is, of course, that the process $(H\_t)$ is not bounded (by a nonrandom constant). A standard construction of the stochastic integral -- see e.g. Proposition 2.6 in Ch. 3 of Karatzas, Ioannis; Shreve, Steven (1991), Brownian Motion and Stochastic Calculus, 2nd ed. -- is done a bit differently: first, the integrand process is truncated to obtain a bounded process, and then the truncated, bounded process is approximated by simple, piecewise-constant processes, and thus the original integrand process is approximated by simple ones (in $L^2(\Omega\times[0,T])$).


