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How Good Is Lebesgue Measure? Krzysztof Ciesielski*

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How Good Is Lebesgue Measure? Krzysztof Ciesielski* The problem of determining the distance between two For a given subset of R", what do we mean by the points, the area of a region, and the volume of a solid statement that some number is its area (for n = 2), are some of the oldest and most important problems in volume (for n = 3) or, more generally, its n-dimen- mathematics. Their roots are in the ancient world. For sional measure? Such a number must describe the size example, the formula for the area of a circle was al- of the set. So the function that associates with some ready known to the Babylonians of 2000 to 1600 B.C., subsets of R" their measure must have some "good" although they used as "rr either 3 or 31/K The Egyptians properties. How can we construct such a function and of 1650 B.C. used "rr = (4/3)4 = 3.1604 .... The first what reasonable properties should it have? crisis in mathematics arose from a measurement This question, in connection with the theory of inte- problem. The discovery of incommensurable magni- gration, has been considered since the beginning of tudes by the Pythagoreans (before 340 B.C.) created the nineteenth century. It was examined by such well- such a great "logical scandal" that efforts were made known mathematicians as Augustin Cauchy, Lejeune for a while to keep the matter secret (see [Ev], pages Dirichlet, Bernhard Riemann, Camille Jordan, Emile 25, 31, 85, 56). Borel, Henri Lebesgue, and Giuseppe Vitali (see [Haw]). Lebesgue's solution of this problem, dating from the turn of the century, is now considered to be the best answer to the question, which is not com- pletely settled even today. Lebesgue Measure In his solution Lebesgue constructed a family Y of subsets of Euclidean space R" (where n = 1, 2 .... ) and a function m: ~ --* [0,oo] that satisfies the following properties: (a) ~ is a (r-algebra, i.e., ~ is closed under countable unions (if A k E ~ for every natural number k then U {Ak: k E N} E ~) and under complementation (A E ~ implies R"\A E ~e); (b) [0,1]" ~ 5~ and m([0,1]") = 1; * I wish to thank my colleagues at Bowling Green State University for their hospitality in 1986/87, while writing this article. I am most grateful to V. Frederick Rickey for his kind help. THE MATHEMATICALINTELLIGENCER VOL. tl, NO. 2 91989 Springer-VerlagNew York (c) m is countably additive, i.e., for every family which contradicts property (b) of Lebesgue measure. {Ak: k E IV} of pairwise disjoint sets from S~, Thus V is not in ~. (For the general case, modify the above proof using V x [0,1]"-1.) Notice that if ~: ~ --~ [0,~] satisfies conditions m A k = m(Ak); (a)-(d) (we will call such a function an invariant mea- sure) then the above proof shows that V ~ ~. Thus, V is not measureable for any invariant measure. In (d) m is isometrically invariant, i.e., for every iso- particular, there is no invariant measure ~: ~ --* [0,oo] metry i of R" which is universal, i.e., for which ~ equals ~(R"). (1) A E ~f if and only if i(A) E ~; Another remark we wish to make is that in Vitali's (2) re(A) = m(i(A)) for every A in 5~. proof we used the Axiom of Choice (hereafter abbre- Recall that a function i: R" --~ R" is an isometry pro- viated AC). At the beginning of the twentieth century vided it preserves distance, in other words, i is com- the Axiom of Choice was not commonly accepted (see posed of translations and rotations. [Mo]) and Lebesgue had his reservations about Vitali's The properties (a)-(d) seem to capture perfectly our construction (see [Le] or [Haw], p. 123). Today we ac- intuitive notion of area and volume. Lebesgue mea- cept the Axiom of Choice, so we cannot support sure is not sensitive to moving sets around without Lebesgue's complaints. But was Lebesgue completely distorting them. To find the measure of a set it suffices wrong? to split it up into a finite or countable family of disjoint If we do not accept the Axiom of Choice then Vitali's sets and then to add up their measures (provided, of proof will not work. In 1964 Robert Solovay (see [So] course, that each piece is measurable). However, we or [Wa], Ch. 13) showed something much stronger: should ask at least one more question: "With how we cannot prove that 5~ # ~(R") without AC. More many sets does Lebesgue measure deal?" or "'How big precisely, he proved that there exists a model ("a is the family ~?" In particular, we should decide mathematical world") in which the usual Zermelo- whether m is universal, i.e., whether ~ is equal to the Frankel set theory (ZF) is true and where all subsets of family ~(R") of all subsets of R". R" are Lebesgue measurable, i.e., ~f = ~(R"). More- Unfortunately the answer to this question is nega- over, although the full power of the Axiom of Choice tive. In 1905 Vitali constructed a subset V of R" such must fail in this "world," the so-called Axiom of De- that V is not in S~ (see [Vi], [Haw], p. 123, or [Ru], pendent Choice (DC) is true in the model. In fact, DC Thm. 2.22, p. 53). His proof is easy, so we shall repeat tells that we can use inductive definitions and there- it here for n = 1. fore all classical theorems of analysis remain true in For x in [0,1] let E x = {y E [0,1]: y - x E Q} and put this "world." % = {E=: x E [0,1]}. Clearly % is a nonempty family of Solovay's theorem has only one disadvantage. Be- non-empty pairwise disjoint sets. Therefore, by the sides the usual axioms of set theory (ZF), Solovay's Axiom of Choice there exists a set V C [0,1] such that proof uses an additional axiom: there exists a weakly V n E has exactly one element for every E E %. We inaccessible cardinal (in abbreviation WIC, see Lie], p. prove that V is not in ~e. 28). The theory ZF + WIC is essentially stronger than First notice that if V + q = {x + q: x E V} for q E Q, ZF. For over 20 years mathematicians wondered then V + q and V + p are disjoint for distinct p, q E whether it is possible to eliminate the hypothesis re- Q. Otherwisex + q = y + pforsomex, yEVsox garding WIC from Solovay's theorem, but in 1980 and y both belong to the same E E %. Then V n E Saharon Shelah showed that it cannot be done, contains more than one element, contradicting the proving that the consistency of the theory ZF + DC + definition of V. "~e = ~(R") '" implies the consistency of the theory ZF Now if V is in ~, then so is V + q for every q E Q, + WIC (see IRa] or [Wa], p. 209). and re(V) = m(V + q). But U {v + q: q E Q N [0,1]} C [0,2] and so Extensions of Lebesgue Measure X m(V + q) = m(U{v + q: q EQ n [0,1]}) q ~Qn[0,1] Let us go back to doing mathematics with the Axiom m([0,2]) = 2. of Choice. We know that Lebesgue measure is not uni- versal, i.e., Y # ~(R"). So let us examine the problem This implies re(V) = m(V + q) = 0 for every q E Q. of how we can improve Lebesgue measure. The prop- On the other hand, [0;1] C R = U {V + q:q E Q} erties (a)-(d) of Lebesgue measure seem to be which implies most desirable. Therefore, we will try to improve m: ~ ---* [0,~] by examining its extensions, the func- m([0,1]) ~ m(U{v + q: q EQ}) =~P m(V + q) = 0, tions p,: ~ ~ [0,oo] such that ~ C ~J~ C ~(R n) and q~Q ~(X) = m(X) for all X E ~. THE MATHEMATICAL INTELLIGENCER VOL. 11, NO. 2, 1989 55 By Vitali's theorem, if it is an invariant measure then 2J~ # ~(R'). In this context the following question it([O,1] n) ~ it(Rn) = it ~ it(Nj) = O. naturally appears: "How far can we extend Lebesgue j=l measure and what properties can such an extension Therefore the invariant measure v as in (ii) is a proper preserve?" This question has been investigated care- extension of it and so Wis not maximal. fully by members of the Polish Mathematical School At this stage of our discussion we are able to give a and all but one of the results presented here were partial answer to the question of our title. If we restrict proved by this group. our search to invariant measures, then Lebesgue mea- Let us first concentrate on the extensions that are sure is not the richest. However, any other invariant invariant measures. The first result in this direction is measure has the same defect. This means that if we due to Edward Szpilrajn (who later changed his name would like to compare invariant measures only by the to Edward Marczewski). In 1936 he proved that Le- size of their domain then the best solution to the mea- besgue measure is not a maximal invariant measure, sure problem simply does not exist. On the other i.e., that there exists an invariant measure that is a hand, if we deal only with the subsets of R" con- proper extension of Lebesgue measure.
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