# Follow-up: continuous linear operator and measurability

Consider

• a probability space $$(\Omega, \mathcal{F}, \mathbb{P})$$
• separable Hilbert spaces $$H$$ and $$V$$
• an $$H$$-valued random variable $$X$$ defined on $$(\Omega, \mathcal{F}, \mathbb{P})$$
• 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.

Is the mapping $$Y : \Omega \rightarrow V, \quad Y (\omega) := L(\omega)(X(\omega))$$ a random variable, i.e., $$\mathcal{F}-\mathcal{B}(V)$$-measurable. If necessary one can assume that $$L$$ takes finitely many values in $$\mathcal{L}(H,V)$$.

One can further introduce $$\hat{L} : \Omega \times H \rightarrow V, \quad (\omega, h ) \mapsto \hat{L}(\omega, h) := L(\omega)h.$$ and rewrite $$L(\omega)(X(\omega)) = \hat{L}(\omega, X(\omega)) \quad \forall \omega \in \Omega.$$ The mapping $$\Omega \rightarrow \Omega \times H \quad \omega \mapsto (\omega, X(\omega))$$ is $$\mathcal{F}-\mathcal{F}\otimes\mathcal{B}(H)$$-measurable, since $$\omega \mapsto \omega$$ is $$\mathcal{F}-\mathcal{F}$$-measurable and $$\omega \mapsto X(\omega)$$ is $$\mathcal{F}-\mathcal{B}(H)$$-measurable. Is it possible to show that $$\hat{L}$$ is $$\mathcal{F}\otimes\mathcal{B}(H)-\mathcal{B}(V)$$-measurable. If so, then the composition $$\omega \mapsto \hat{L}(\omega, X(\omega))$$ will have the desired measurability.

Some further thoughts:

For every fixed $$\omega \in \Omega$$ the mapping $$h \mapsto \hat{L}(\omega, h)$$ is continuous by assumption. If one can show that for every fixed $$h \in H$$ the mapping $$\omega \mapsto \hat{L}(\omega, h)$$ is $$\mathcal{F}$$-measurable, then $$\hat{L}$$ would be jointly measurable as a Carathéodory function.

I think this is also false. Let $$H=V=\mathbb{R}$$, $$Y: \Omega \rightarrow \mathbb{R}$$ be any function on $$\Omega$$ that is not $$\mathcal F$$-measurable, $$X=1$$ be a constant random variable, and $$L(\omega)(h) = hY(\omega)$$. $$L(\omega)$$ is a continuous linear operator from $$\mathbb{R}$$ to $$\mathbb{R}$$, but $$L(\omega)(X(\omega)) = Y(\omega)$$ is not $$\mathcal F$$-measurable.
• But why is $\omega \mapsto L(\omega)$ measurable? It seems that this question is also relevant for the previous post. Dec 7, 2022 at 0:36
• Sorry, I didn't see the assumption that $L$ was measurable. What is the $\sigma$-algebra on $\mathcal L(H,V)$ you are using? Dec 7, 2022 at 0:46