# Continuous linear operator and measurability

Consider

• a probability space $$(\Omega, \mathcal{F}, \mathbb{P})$$ and a filtration $$\mathbb{F}=(\mathcal{F}_t)_{t \geq 0}$$ on $$(\Omega, \mathcal{F})$$
• separable Hilbert spaces $$H$$ and $$V$$
• an $$H$$-valued stochastic process $$(X_t)_{t \geq 0}$$ with continuous sample paths adapted to $$\mathbb{F}$$
• A random variable $$L : \Omega \rightarrow \mathcal{L}(H,V)$$, where $$\mathcal{L}(H,V)$$ is the space of linear continuous (bounded) operators from $$H$$ to $$V$$ with the operator norm.

Show that the process $$(Y_t)_{t \geq 0}$$, defined as $$Y_t (\omega) :=L(\omega)(X_t(\omega))$$, is again adapted to $$\mathbb{F}$$. If necessary one can assume that $$L$$ takes finitely many values in $$\mathcal{L}(H,V)$$.

Since for every $$\omega \in \Omega$$ the operator $$L(\omega)$$ is continuous, and so are the sample paths of $$(X_t)_{t \geq 0}$$, it is easy to see that $$(L(X_t))_{t \geq 0}$$ also has continuous paths. But why is it also measurable, and, moreover, adapted?

One can further introduce $$\hat{L} : \Omega \times H \rightarrow V \quad (\omega, h ) \mapsto \hat{L}(\omega, h) := L(\omega)h.$$ and rewrite $$L(\omega)(X_t(\omega)) = \hat{L} (\omega, X_t(\omega)) \quad \forall \omega \in \Omega, \ t \geq 0.$$ Can this be helpful for the task?

This doesn't appear to be true. Let $$H=V=\mathbb{R}$$, $$\mathbb{F}$$ by the filtration generated by a Brownian motion $$W$$, and $$X_t(\omega) = 1$$ for all $$t,\omega$$. Define $$L(\omega)(h) := hW_1(\omega)$$. Then $$L(\omega)$$ is clearly a continuous linear operator from $$\mathbb{R}$$ to $$\mathbb{R}$$, $$X$$ is clearly continuous and adapted, but $$Y_t(\omega) = X_t(\omega)W_1(\omega) = W_1(\omega)$$ is not adapted to $$\mathbb{F}$$.
If one desires that $$L$$ takes finitely many values in $$\mathcal L(H,V)$$, one could instead define $$L(\omega)(h) = h1_{W_1(\omega) > 0}$$.