Re: when a linear closed operator presearves measurability

From: G. A. Edgar (edgar_at_math.ohio-state.edu)
Date: 07/16/04 

Date: 16 Jul 2004 18:38:08 -0400

In article <cd92ek$a0e$1@dizzy.math.ohio-state.edu>, cervinia
<cervinia@k.ro> wrote:

> Help me, please !
>
> Problem:
>
> Let $X$ be a separable Banach space and $A:D(A)\subset X\rightarrow X$
> a closed, linear operator.
>
> Suppose that $u$ is some (Lebesgue) measurable function from, say,
> $R$ -the set of reals to $X$, such that $u(t)\in D(A)$ for almost all
> $t\in R$.
>
> Question: which are the (most general) sufficient conditions on $A$ to
> ensure the measurability of the map $t\rightarrow Au(t)$ ?
>
> Thank you very much for your help
> P.
>

We need: inverse image (under Au) of an open set U in X is
measurable in R. Well, U is open in X, so $X \times U$ is open
in $X \times X$. The graph G of A is closed in $X \times X$,
so the intersection $(X \times U) \cap G$ is a Borel set in
$X \times X$. Its projection onto the first coordinate is
therefore an analytic set in $X$. ($X$ and $X \times X$ are both
Polish spaces.) That projection is $A^{-1}(U)$.

So: what we need for $u$ is that the inverse image of an analytic
set is Lebesgue measurable. Someone who knows more descriptive set
theory will have to say if this follows from inverse image
of Borel set is Lebesgue measurable. Is the collection of Lebesgue
measurable sets closed under operation A?