Let $(X,\mathcal{A})$ be a measure space and $K$ a compact metric space wiht the Borel $\sigma$-álgebra $\mathcal{B}$. Is it true that if $f:X\times K\rightarrow \mathbb{R}$ is $(\mathcal{A}\otimes\mathcal{B},\mbox{Borel of }\mathbb{R})$-measurable then $x\rightarrow f_x \in C(K)$ is $(\mathcal{A},\mbox{borel algebra generated by the uniform topology on C(K)})$-measurable?

P.S. Here $C(K)$ is the space of all continuous functions $g:K\rightarrow \mathbb{R}$ wih the uniform norm $\|g\|=\sup_{v\in K} |g(v)|$.


1 Answer 1


This is indeed true.

First observe that since $\mathcal C(K)$ is separable ($K$ is $metric$ compact), the Borel sets are the same for the norm and the $weak$ topology of $\mathcal C(K)$. This is because every norm-open set is a countable union of closed balls and closed balls are weakly closed. So it is enough to show that the map $x\mapsto f_x$ is ``weakly measurable" from $(X,\mathcal A)$ into $\mathcal C(K)$, i.e. that $x\mapsto L(f_x)$ for every continuous linear functional $L$ on $\mathcal C(K)$.

By the Riesz representation theorem, this amounts to showing that for any complex Borel measure $\mu$ on $\mathcal C(K)$, the map $x\mapsto \int_K f_x d\mu=\int_K f(x,t) d\mu (t)$ is measurable from $(X,\mathcal A)$ into $\mathbb C$. But this is well known (see the "product spaces" part in any book on integration theory). In the present situation, one can prove it as follows. There is a sequence of finitely supported measures $(\mu_n)$ such that $\mu_n\xrightarrow{w^*} \mu$, i.e.$\int_K gd\mu_n\to \int_K g d\mu$ for every $g\in\mathcal C(K)$. So it is enough to show that the map $x\mapsto \int f(x,t)d\mu (t)$ is measurable when $\mu$ is finitely supported. This reduces in fact to the case of a point mass $\mu=\delta_a$, but then it is obvious since $\int f(x,t)d\delta_a(t)=f(x,a)$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .