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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$1@dizzy.math.ohiostate.edu> schrieb Robert E. Beaudoin: >> >> >>>G. A. Edgar wrote: >>>) In article <cd92ek$a0e$1@dizzy.math.ohiostate.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 measurezero 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? >> > > > 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.
Sorry about the peccadillo of replying to my own post, and even sorrier about any confusion this terminology may have caused. Hope this clears things up.
Bob Beaudoin



