Also of interest: We can prove (AC24) ==> (AC3) by this slight modification of the arguments presented in the book: Equip each Y-sub-lambda with the topology consisting of the cofinite sets, the empty set, and the singleton xi-sub-lambda. Thus (AC24) is still an equivalent of Choice.
It turns out that "AC25" is not an equivalent of AC at all; rather, it is an equivalent of UF. The proof is too long to include here. It's in an article that I've now submitted for publication; I'll update this page later. I'll be talking about that correction at the Joint Mathematics Meetings in January 2006.
M. Atsuji, Uniform continuity of continuous functions of metric spaces, Pacific J. Math. 8 (1958), 11-16; erratum, 941.and other papers. These papers give several equivalent characterizations of the spaces whose continuous maps are all uniformly continuous.J. Rainwater, Spaces whose finest uniformity is metric, Pacific J. Math. 9 (1959), 567-570.
the product sigma-algebra formed from the coordinate spaces' Borel sigma-algebras is a subset of the Borel sigma-algebra determined by the product topology.To see that, note that each coordinate projection map is continuous (by the definition of product topology) hence measurable, and the product sigma-algebra is the smallest sigma-algebra which makes the coordinate projections all measurable. -- -- On the other hand,
for two topological spaces that have countable bases, the Borel sigma-algebra determined by the product topology is a subset of the product sigma-algebra formed from the coordinate spaces' Borel sigma-algebras.To see that, consider any open set in the product topology; it can be written as a union of countably many open rectangles. Each open rectangle belongs to the product sigma-algebra; hence so does the countable union. -- -- By the way, it should be noted that "smallest" in this context (referring to topologies or sigma-algebras) does not merely mean "having the fewest elements;" it means "contained in any other."
As it is stated in the book, 21.28 definitely is wrong; here is a counterexample. Let Omega be the real line. Let the algebra A be the set of finite unions of intervals of the form [a,b); that is, closed on the left and open on the right. The sigma-algebra generated by such sets is the Borel sigma-algebra. Let mu be a discrete measure, defined as in the first paragraph of 21.11, with p(m/n) = 1/n whenever n is a positive integer, m is an integer, and m and n have no common factors greater than 1; let p(t)=0 for all irrational numbers t. Then mu is sigma-finite, since the rationals are countable. However, mu([a,b)) is infinity whenever a and b are real numbers with a less than b, since mu([a,b)) includes a copy of the tail of the harmonic series. Let nu = 2 mu.
Let (X,d) be a metric space, and let B be its collection of Borel subsets. Let mu be a positive, finite measure on B. Let (X,S,nu) be the completion of (X,B,mu). Then for every set M in S, we haveThe proof is not very long or difficult; one proof can be found in Billingsley, Convergence of Probability Measures. In fact, this is theorem 1 in chapter 1 of that book.Furthermore: Let T be a subset of X. Then T is a member of S if and only if there exist sets A and B such that A is an F-sigma set, B is a G-delta set, mu(A\B)=0, and A subset of T subset of B.
- nu(M) = sup{ mu(F) : F closed, F a subset of M}, and
- nu(M) = inf{ mu(G) : G open, G a superset of M}.