- In 15.22, the third sentence of the proof mentions
"liminf x_\alpha" where it should say
"liminf f(x_\alpha)". Also, the last sentence of
the proof says that a certain closed set
"contains all the x_\alpha's after some \alpha_0";
instead it should say that that closed set
"contains a frequent subnet of the x_\alpha's,
and any limit of that subnet must lie
in that closed set."
- In 15.26.a, the second pair of parentheses is larger
than the first by accident, not intentionally. [JT]
- In 16.6, cl(x) should be cl({x}). (Of course, if one were
working in this topic extensively, one might find it convenient
to establish a convention of using x to also denote {x}, but
that abuse of notation is a little dangerous if one also gets
into set theory. In set theory the distinction between S
and {S} is very important.) [JT]
- In 16.16, (pi/2)arctan(f) should be (2/pi)arctan(f). [JT]
*Correction/clarification for 16.24:*The term "subordinated" is defined a little differently in different books. My book agrees with Kelley's book --- i.e., each member of the partition of unity vanishes outside some member of the open cover. But some books -- e.g., those of Dugundji, Munkres, and Bourbaki -- require that the*support*of the function is contained in some member of the open cover. This is slightly different; the support of the function is the*closure*of the set where the function doesn't vanish. However, this change in the definition doesn't matter greatly: In most applications of interest, the space being investigated is already known to be normal, and so we can apply the Shrinking Lemma (see HAF 16.26.B). In particular, we know that pseudometric spaces are normal, by HAF 16.28.a.- In 16.28.a, (j=1,2) should be (j=0,1). [JT]
*Additional material for 16.31.*A different proof of Stone's Theorem is indicated by an article by Albert Fathi in the October 1997 issue of*American Mathematical Monthly*. I'm not sure if it's shorter, but it seems to be a bit more insightful. Here it is, in three pages: 1 - 2 - 3.**17.20.**The proof of "AC25" ==> (AC3) is erroneous. In fact, that proof is essentially the same as the proof in Kelley's paper, but Kelley's proof was erroneous too. The proof asserts that the set S-sub-lambda is closed, but that's simply not true under the assumptions being used (i.e., using the cofinite topologies). Found by [MG]. Kelley's proof may be erroneous, but the main point of his paper is not; my alternate proof via (AC21) ==> (AC22) ==> (AC23) ==> (AC3) apparently is unharmed.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.

*Add this remark to 17.33:*Another characterization of compactness will be given in 19.18.*Stylistic remarks for Chapters 18 and 19:*Many of the results in these chapters are stated mainly for nets in uniform spaces. Instructors teaching from this text may find it helpful to state those results first for sequences in pseudometric spaces, and then say "We shall now generalize." In fact, I might make that alteration myself in a second edition, if there is one.*Add remark for 18.21.*The cited paper by Arala-Chaves gives an elementary and self-contained introduction to the subject. More comprehensive studies are given byM. Atsuji, Uniform continuity of continuous functions of metric spaces,

and other papers. These papers give several equivalent characterizations of the spaces whose continuous maps are all uniformly continuous.*Pacific J. Math.***8**(1958), 11-16; erratum, 941.J. Rainwater, Spaces whose finest uniformity is metric,

*Pacific J. Math.***9**(1959), 567-570.- 19.11.c should be attributed to Cantor, not Cauchy.
- In 19.21.g, "function from phi into R" should be
"function from [a,b] into R."
- In 19.33.c, the upper-case letters M and N should be
the same. (Change whichever one you prefer.) [JT]
*Add remark after 20.30:*The statement BP in section 14.74 is important for 14.76-14.77 and related material throughout the book. BP says that every subset of R has the Baire property. It follows from 20.30 that BP is equivalent to this statement: every subset of 2^N has the Baire property. (Here 2^N is the set of all sequences of 0's and 1's, with the product topology.) That reformulation is more convenient for some purposes -- e.g., for 29.38.*Informal and imprecise remarks for 21.3 and 21.4:*The calculations in these sections should show that any set which is "constructed via any reasonable algorithm" from measurable sets is measurable. That should make it plausible that (with conventional sigma-algebras) we can only form unmeasurable sets by using nonconstructive principles such as the Axiom of Choice. -- See additional hints for proof of 21.4**21.4 and 22.30.b.**The theorem and proof are correct but the proof is missing too many steps; a big hint is really needed for (A) implies (C).*Additional hints for 21.7.a.*-- -- In one direction, for*any*topologies (regardless of separable metric spaces etc.),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."*Additional result for 21.7:*Suppose (X,S) and (Y,T) are measurable spaces, and f,g:X --> Y are two measurable mappings. Define a map (f,g) : X --> YxY by (f,g)(x)=(f(x),g(x)). Then (f,g) is also measurable, where YxY has the product sigma-algebra.*Addition for 21.17:*The exercise in the last paragraph can be improved slightly. Drop the assumption of separability; then we can still prove this result: If the measure space is complete, and two functions are equal almost everywhere, and one of the functions is Borel measurable, then the other function is also Borel measurable.*Addition for 21.18:*Here is a corollary which will be helpful later: Suppose f_n --> f almost everywhere. Then we can alter the functions f_n and f, all on one measurable set that has measure 0, to obtain functions g_n --> g everywhere (not just almost everywhere). (Proof: Choose a measurable null set N that contains all the bad points. Fix some point z in X; define g_n and g to be equal to z on N.) If the f_n's are Borel measurable, it follows that the g_n's and g are Borel measurable --- but f might not be Borel measurable.*Additional details for 21.22.*The proof of 21.22 is a bit abstract, and may be difficult for some beginners. Here is a longer and more concrete-seeming presentation of the same proof.**21.24.**The proof of 21.24.d actually uses 21.25.d (but there is no circularity in the reasoning).*Additional remark for 21.25.d.*If we exclude infinity as a value, but allow measures to take values in a Banach space, then a limit of measures is a measure. That's the Vitali-Hahn-Saks-Nikodym Convergence Theorem, which is given in 29.8.**21.27, 21.28, and 22.30.c.**To correct these, assume mu is finite, or more generally (and more complicatedly) assume mu is sigma-finite on the algebra of sets in the sense of 21.26. (For 21.27 and 22.30.c, the algebra of sets is the collection of finite unions of intervals.)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.

*Additional remarks for 21.28 and 22.29.*It is useful to mention that Lebesgue's convergence theorems are valid only for sequences of functions, not for nets. For instance, for a directed set use the finite subsets of [0,1], partially ordered by inclusion. For each of those sets, take its characteristic function. This net of functions converges monotonely and dominatedly to the constant function 1, but the integrals are all 0, which does not converge to 1. -- The Monotone Convergence Theorem is true for a net of continuous functions if the limit is also continuous and the underlying space is compact and has finite measure; that follows from Dini's Theorem 17.7.j.*Additional after 21.28.*The following theorem would make a nice addition after 21.25 or after 21.28. It would also be useful in shortening the proof of 24.35 very slightly.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 have

The proof is not very long or difficult; one proof can be found in Billingsley,- 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}.

*Convergence of Probability Measures*. In fact, this is theorem 1 in chapter 1 of that book.**21.34.d.**The assertion about equivalence classes is correct without additional hypotheses. The assertion about individual functions is FALSE --- it requires the additional hypotheses that the measure is complete and the pseudometric space X is separable. (Indeed, suppose that the measure is not complete; let N be a non-measurable set that is contained in a set of measure 0. Then a sequence of zero functions converges in pseudometric to the characteristic function of N even though 0 is strongly measurable and the characteristic function of N is not. At least for some measure spaces, the assumption of separability is unavoidable too. For instance, if the measure space is [0,1] with Lebesgue measure, let N be an uncountable set with measure 0, and let f map the points of N to a non-separable range; then f is not strongly measurable but f is equivalent to strongly measurable functions.)*Additional remarks for 21.36.*Perhaps the definition of the integral given in 21.36 is a bit unmotivated. I am considering a few other definitions. One which deserves note is the one used by Kelley and Srinivasan's book,*Measure and Integral*(Springer-Verlag, 1988, Graduate Texts in Mathematics 116). That book observes that any measurable function is "sigma-simple" -- i.e., it can be represented (not uniquely) as the sum of countably many constant multiples of characteristic functions of (not necessarily disjoint) measurable sets. A function is integrable if and only if the series representation is in a certain sense absolutely convergent. Then the integral can be defined immediately: the integral of the sum is the sum of the integrals. This definition is intuitively appealing, though there are some nontrivial steps in working out the theory -- you have to prove that the resulting integral doesn't depend on the particular "sigma-simple" representation of the function, and you have to prove that the integrable functions (i.e., those with absolutely convergent series) are the same as the usual L1 functions.*Additional remark for 21.39.c.*We proved Fatou's Lemma using the Monotone Convergence Theorem, but it is also easy to go the other way.