From: "G. A. Edgar" Subject: Re: measure theory problem Date: Wed, 07 Jun 2000 09:05:40 -0400 Newsgroups: sci.math.research Summary: [missing] No. With their product topologies your spaces are Polish, and your sigma-algebras are their Borel sets. However, the projection pi(X) need not be Borel (it is "analytic" or "Suslin"), and pi^{-1}(pi(X)) therefore need not be Borel. [reference... Donald L. Cohn, Measure Theory, Chapter 8.] In article , Marielle Stoelinga wrote: > Can someone help me with the following problem on measure theory? > Given two finite alphabets A and B, I consider the following > sigma-field (Omega, F) > > Omega is the set of infinite, alternating sequences a1 b1 a2 b2 ... > where ai is from A and bi from B > > F is the smallest sigma field that contains the set Cw, for any > finite, alternating sequence w = a1 b1 a2 b2 ... an bn > where again, ai is from A and bi from B. > Cw is the cone above w, defined by > > Cw = {v in Omega | w is a prefix of v} > > > Furthermore, I have a projection function pi which projects a finite or > infinite alternating sequence on B: > > pi(a1 b1 a2 b2 a3 b3 ...) = b1 b2 b3 > > > My question is: Given a measurable set X in F, is the set > > pi^-1(pi(X)) = > {w in Omega | there exists a v in X such that pi(w) = pi(v)} > > also measurable? -- Gerald A. Edgar edgar@math.ohio-state.edu