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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.

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1 Answer 1

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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.

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  • $\begingroup$ But why is $\omega \mapsto L(\omega)$ measurable? It seems that this question is also relevant for the previous post. $\endgroup$
    – Harry
    Dec 7, 2022 at 0:36
  • $\begingroup$ Sorry, I didn't see the assumption that $L$ was measurable. What is the $\sigma$-algebra on $\mathcal L(H,V)$ you are using? $\endgroup$ Dec 7, 2022 at 0:46

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