Suppose I have a function $f:X\times H_1\to H_2$, where $X$ is a measurable space with some associated $\sigma$-algebra, $\mathcal{X}$, and $H_1$ and $H_2$ are both separable Hilbert spaces (not necessarily the same). Assume I know that there is a constant $C$ such that $f$ is Lipschitz in the $h$ argument, $$ |f(x,u) - f(x,v)|_{H_2}\leq C |u-v|_{H_1} $$ and that for all $u$ and $h$, the real valued mapping $x\mapsto (f(x,u),h)_{H_2}$ is measurable.

I would like to conclude something about the joint measurability of $f$, in particular that $f$ is $(X\times H_1, \mathcal{X}\times \mathcal{B}(H_1))\to (H_2, \mathcal{B}(H_2))$, but don't quite see how to do it.

I think this can reduced to the problem to proving measurability of $$ (x,u)\mapsto(f(x,u),h_n)_{H_2} $$ where $\{h_n\}$ form an orthonormal basis of $H_2$.


1 Answer 1


I realize now this is really just an instance of a Caratheodory functions. Once we define $$ f_n(x,u) = (f(x,u),h_n)_{H_2} $$ we have a function which is continuous in one argument and measurable in the other on the separable and measurable spaces, so it is jointly measurable. Then using separability again, $$ |f(x,u)-y|_{H_2} < r\Leftrightarrow \sup_n |(f(x,u)-y,h_n)_{H_2}|<r $$ and the result follows by the countability of the $h_n$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.