Let
- $U$ and $H$ be separable Hilbert spaces
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $(\mathcal F_t)_{t\ge 0}$ be a filtration of $\mathcal A$
- $\mathfrak L:=\mathfrak L(U,H)$ be the space of linear operators from $U$ to $H$
- $\lambda$ be the Lebesgue measure on $[0,\infty)$ and $\operatorname P\otimes\lambda$ denote the product measure.
Let $\mathcal E$ be the set of $\mathfrak L$-valued $\mathcal F$-adapted stochastic processes $(H_t)_{t\ge 0}$ on $(\Omega,\mathcal A,\operatorname P)$ with $$H_t=\sum_{i=1}^n\xi_{i-1}1_{(t_{i-1},t_i]}(t)\;\;\;\text{for all }t\ge 0$$ for some $\mathfrak L$-valued random variables $\xi_0,\ldots,\xi_{n-1}$ with finite images and $0=t_0<\cdots<t_n$.
How can we prove that $\mathcal E\subseteq\mathcal L^2:=\mathcal L^2(\operatorname P\otimes\lambda;\mathfrak L)$?
Assuming that $U=H=\mathbb R$, it's easy to show that $H$ is productmeasurable such that we can apply Fubini's theorem to obtain $$\left\|H\right\|_{\mathcal L^2}^2=\int\int|H|^2\;{\rm d}\lambda{\rm d}\operatorname P=\operatorname E\left[\int_0^\infty |H_t|^2{\rm d}t\right]$$ (where I should note that the Riemann integral on the right-hand side is well-defined, cause $H$ is left-continuous and the right-sided limits exist). Then, it's easy to show that $$\operatorname E\left[\int_0^\infty |H_t|^2{\rm d}t\right]=\sum_{i=1}^n\operatorname E\left[\left|\xi_{i-1}\right|^2\right](t_i-t_{t-i})$$ and the claim follows.
Now, I'm curious whether something could get wrong in the more general setting described here? Can we reduce the desired statement to the case $U=H=\mathbb R$?