How is $\mathbb E[ \int_0^T H_s^2 \mathrm d s] < \infty$ important for this claim?

Question

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 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 question on MSE but have not received any answer so far.

Answer 1

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 $L^2(\Omega\times[0,T])$).