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Topic: when a linear closed operator presearves measurability
Replies: 5   Last Post: Jul 20, 2004 4:00 PM

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 Robert E. Beaudoin Posts: 52 Registered: 12/8/04
Re: when a linear closed operator presearves measurability
Posted: Jul 20, 2004 4:00 PM

Robert E. Beaudoin wrote:
> Thomas Mautsch wrote:
>

>>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
>>>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?
>>

>
>
> No, the sets M_t, B_t, and N_t are subsets of the reals, and
> so S is a subset of the reals. What is being proved is that
> the collection of Lebesgue measurable subsets of the reals is
> closed under operation A. From that it follows that the
> preimage of an analytic subset of X via a measurable function
> from the reals to X is a Lebesgue measurable subset of the reals.
> (The analytic subset of X can be obtained by applying operation
> A to a family of closed subsets of X, and so its preimage is
> the result of applying operation A to the preimages of the closed
> sets.)
>
> Hope that helps.
>
> Bob Beaudoin
>

It occurred to me this morning that there is an unfortunate
collision of terminology here that may be a source of confusion.
Namely, the term "operation A" in my replies (and G. A. Edgar's
question) refers not to the closed linear operator A of the
original post but to Suslin's operation A: If M_t is a family
of sets indexed by finite sequences of natural numbers then
operation A applied to this family is the union, over all
infinite sequences s of natural numbers, of the intersection
of M_<>, M_<s_0>, M_<s_0, s_1>, .... (So if each M_t is a
set of reals then so is S.) This is standard terminology in
descriptive set theory, as this operation is commonplace in that
subject. For example, every analytic set in a Polish
space is the result of applying operation A to a family of
closed sets.

even sorrier about any confusion this terminology may have
caused. Hope this clears things up.

Bob Beaudoin

Date Subject Author
7/16/04 cervinia
7/16/04 G. A. Edgar
7/16/04 Robert E. Beaudoin
7/19/04 Thomas Mautsch
7/19/04 Robert E. Beaudoin
7/20/04 Robert E. Beaudoin