Re: when a linear closed operator presearves measurability

From: Thomas Mautsch (mautsch_at_math.ethz.ch)
Date: 07/19/04 

Date: Mon, 19 Jul 2004 14:44:01 +0000 (UTC)

In ne:<cd9teu$an7$1@dizzy.math.ohio-state.edu> schrieb Robert E. Beaudoin:
> G. A. Edgar wrote:
> ) In article <cd92ek$a0e$1@dizzy.math.ohio-state.edu>, cervinia wrote:
> )>
> )>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 $t\rightarrow Au(t)$ ?
> )
> ) 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?
> )
> This seems easy enough: Let M_t be Lebesgue measurable for each
> (finite sequence) t, and let M be the result of applying operation
> A to the M_t's. For each t one can choose a Borel set B_t and
> a Borel measure-zero set N_t so that B_t - N_t is a subset of M_t
> which in turn is a subset of the union of B_t and N_t. Let S be the
> result of applying operation A to the sets B_t - N_t and let N be
> the union of the N_t's. As there are only countably many t's, N has
> measure zero, and as S is analytic it is Lebesgue measurable. But
                      ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
> M is sandwiched between S and the union of S with N, so M is
> measurable. So I think the answer to your question is yes, and
> hence the answer to the OP's question is: no additional conditions
> beyond measurability of u and closedness of A are needed.

S is a subset of X, an *arbitrary* seperable Banach space.
What does "S is Lebesgue measurable" mean?