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. In connection structed without the Axiom of Choice but using the with this result, Waclaw Sierpifiski (in 1936) posed the induction allowed by DC (this is the case in all con- following question "'Does there exist any maximal in- structions in classical analysis) then all sets we are in- variant measure?" (see [Sz]). Let us notice that by terested in are Lebesgue measurable. Thus, in view of property (b) any such measure should extend Le- the above arguments and the naturalness of Le- besgue measure. besgue's construction, the Lebesgue measure is the unique reasonable candidate to be a canonical in- variant measure. How far can we extend Lebesgue measure and Another idea to improve Lebesgue measure was to what properties can such an extension pre- extend it to a universal measure it: !~(Rn) ~ [0,oo] by serve? weakening some part of properties (a)-(d), thus avoiding Vitali's argument. The most popular ap- proach is to drop isometric invariantness and to con- This question was examined by several mathemati- sider as measures the functions it: ~ff~~ [0,oo] that sat- cians under different assumptions. The answer was isfy conditions (a), (b), and (c). This definition, of mea- always the same: there is no maximal invariant mea- sure is common today mostly because we can easily sure. The firstresult was noticed by Andrzej Hulanicki generalize it from R" to an arbitrary set. To answer the in 1962 (see [Hu]) under the additional set-theoretical question of when there exists a universal measure ex- assumption that the continuum 2 ~ is not RVM (this tending Lebesgue measure, let us consider the fol- assumption will be discussed later in this article).This lowing sentence: "There exists a measure it: !~(R) result was also obtain by S. S. Pkhakadze (see [Pk]) [0,oo] satisfying (a)-(c) such that m({x}) = 0 for all using similar methods. In 1977 A. B. Harazi~vili (from x E R." If this sentence holds, we say that 2'* is RVM, Georgia, USSR) got the same answer in the one-di- i.e., on the continuum there is a real-valued measure. mensional case without any set-theoretical assump- It is well known that if 2 `0 is RVM, then there exists a (see [HarD. Finally, in 1982 Krzysztof Ciesielski tion universal extension of Lebesgue measure (the con- and Andrzej Pelc generalized Harazi~vili's result to all verse implication is obvious). We shall s[
56 THE MATHEMATICAL INTELLIGENCER VOL. 11, NO. 2, 1989 binary representation of a. It is not difficult to see that disjoint balls B1 and B 2 each having the same volume this )~ really extends Lebesgue measure on [0,1]. 1I There is even more: we can do this by cutting the The systematic investigation of the statement "20) is ball B into only five pieces. This result is so paradox- RVM'" was started in 1930 by Stanislaw Ulam, when ical that our discomfort with it can probably be com- he proved that it implies the existence of a weakly in- pared only with the Pythagorean "logical scandal" accessible cardinal (i.e., axiom WIC), so it cannot be connected with the discovery of incommensurable line proved under the usual axioms of set theory (see [U1] segments. This is also one of the strongest arguments or Ue], Thin. 66, p. 297). Although this result was fun- against the use of the Axiom of Choice (see [Mo], p. damental for one of the most interesting parts of 188). modern set theory, namely the theory of large car- But what about our measures on R3? It is easy to see dinals, for our discussion it is only a disadvantage. It for every universal finitely additive isometrically in- means that in order to have a universal countably ad- variant measure ~ that if X C R" is congruent to Y C ditive measure extending Lebesgue measure we not R", then ~(X) = )L(Y). In particular ~(B) = la(Bt U B2). only lose isometric invariantness, we must also as- But for 3-dimensional Lebesgue measure m(B) = 1 # 2 sume a very strong additional axiom that is usually not = re(B1 U B2), so ~ doesn't extend Lebesgue mea- accepted by mathematicians. sure m. Connected with this paradox is a nice open Finitely Additive Extensions of problem, the Tarski Circle-Squaring Problem from Lebesgue Measure 1925, about which Paul ErdOs wrote (see [Ma], p. 39): It is a very beautiful problem and rather well known. If it Let us consider another approach to the universal ex- were my problem I would offer $1000 for it--a very, very tension of Lebesgue measure by asking: "Is it really nice question, possibly very difficult. necessary to assume that measure is countably addi- Tarski noticed that although in 3-dimensional Eu- tive? Isn't it enough to deal with finitely additive mea- clidean space all bounded sets with nonempty interior sures?" More precisely, let us consider the properties: are congruent, this is not the case on the plane where (a') Sp is an algebra, i.e., S~ is dosed under unions (if sets with different Lebesgue measure are not con- A,B E ~ then A U B E S~) and under complementation gruent. But what about subsets of the plane with the (A E S~ implies R"\A E S~); same measure? In particular he formulated his Circle- (c') m is finitely additive, i.e., for every disjoint pair Squaring Problem: "Is a square with unit measure A and B from S~, congruent to a circle with the same measure?" This m(A U B) = m(A) + m(B). problem seems to be so difficult that Stan Wagon wrote about it (see [Wa], p. 101): "The situation seems A function m: ~/~ --) [0,o0] satisfying the properties not so different from that of the Greek geometers who (a'), (b'), (c') and (d) will be called a finitely additive considered the classical straight-edge-and-compass isometrically invariant measure. Does there exist a form of the circle-squaring problem." universal finitely additive isometrically invariant mea- sure extending Lebesgue measure? Conclusion Stefan Banach (see [Ba] or [Wa]) proved in 1923 that such a measure exists on the plane and on the line So, what is the answer to the question "How good is (i.e., for n = 2 and n = 1). This beautiful result seems Lebesgue measure?" In the class of invariant mea- to be the only reasonable improvement of Lebesgue sures, Lebesgue measure seems to be the best candi- measure. But what is going on with n-dimensional Eu- date to be a canonical measure. In the class of count- clidean spaces for n _-__3? The answer, due to Stefan ably additive not necessarily invariant measures, to Banach and Alfred Tarski (1924; see [BT] or [Wa]), is find a universal measure we have to use a strong addi- surprising: there is no universal finitely additive iso- tional set-theoretical assumption and this seems to be metrically invariant extension of Lebesgue measure for too high a price. Thus the best improvement of Le- n I 3. But undoubtedly more surprising is the result besgue measure seems to be the Banach construction that leads to this conclusion: the Banach-Tarski Par- of a finitely additive isometrically invariant extension adox. of Lebesgue measure on the plane and line. However, To state this paradox let us introduce the following such a measure does not exist on R" for n 1 3, and to terminology. We say that a set A c Rn is congruent to keep the theory of measures uniform for all dimen- B C R" if we can cut A into finitely many pieces and sions we cannot accept the Banach measure on the rearrange them (i.e., transform each of these pieces plane as the best solution to the measure problem. using some isometry of R n) to form the set B. The From this discussion it seems clear that there is no Banach-Tarski Paradox is the theorem that a ball B reason to depose Lebesgue measure from the place it with volume I is congruent to the union of two similar has in modern mathematics. Lebesgue measure also
THE MATHEMATICALINTELLIGENCER VOL. 11, NO. 2, 1989 ~7 has a nice topological property called regularity: for every E ~ ~e and every e > 0, there exists an open set V D E and closed set F C E such that m(V\F) < ~. It is not difficult to prove that Lebesgue measure is the richest countably additive measure having this prop- erty (see [Ru], Thm. 2.20, p. 50). B Kpyry ~Ipysefi 3a npaaJInnqnbIM CTOJIOM MbI BbIXOJI KHHr~I namefi aaMeqadivi. References Jlepxca B pyKax xpycT~mnfi SOBbtfi TOM, [Ba] S. Banach, Sur le probl6me de la mesure, Fund. Bce OT ;IytuH Hac C H3efi no3JIpaBa~an. Math. 4 (1923), 7-33. ~a, nacTynna Ham JIoaroxoannbIfi ,~ac, [BT] S. Banach and A. Tarski, Sur la d6composition des H nax:e CKeIITHKMpaqHbIfi ynbt6nyac~. ensembles de points en parties respectivement con- OH noJIoIIIeJI, qTOSbI no3JIpaBnTb Mac, gruents, Fund. Math. 6 (1924), 244-277. A s OT pa~OCTH.., npocuyaca. [Ci] K. Ciesielski, Algebraically invariant extensions of tr-finite measures on Euclidean spaces, to appear. [CP] K. Ciesielski and A. Pelc, Extensions of invariant Ha cna M. F. Kpefina B HoaoroamoIo HOqb 1963 r. measures on Euclidean spaces, Fund. Math. 125 Ho CBItJIeTeJIbCTByI/I. C. HOXBIa~oaa. (1985), 1-10. [Ev] H. Eves, An Introduction to the History of Mathematics, Saunders College Publishing, 1983. Around the festive table all our friends [Har] A.B. Harazi~vili, On Sierpifiski's problem con- cerning strict extendability of an invariant measure, Have come to mark our new book's publication. Soviet. Math. Dokl. 18 no. 1 (1977), 71-74. The new and shiny volume in their hands, [Haw] T. Hawkins, Lebesgue Theory of Integration, The Univ. They offer Is and me congratulation. of Wisconsin Press, 1970. Our long-awaited hour has come at last. [Hu] A. Hulanicki, Invariant extensions of the Lebesgue The sourest skeptic sees he was mistaken measure, Fund. Math. 51 (1962), 111-115. Ue] T. Jech, Set Theory, Academic Press, 1978. And smiling comes to cheer us like the rest; [Le] H. Lebesgue, Contribution a l'6tude des correspon- And I am so delighted... I awaken. dances de M. Zermelo, Bull. Soc. Math. de France 35 (1907), 202-221. From M. G. Krein's dream, New Year's 1963, as [Ma] R.D. Mauldin, The Scottish Book, Birkh/iuser, reported by I. S. Iokhvidov. Boston, 1981. [Mo] G. Moore, Zermelo's Axiom of Choice, Springer- Veflag, New York, 1982. English version by Chandler Davis. (The book [Pk] S.S. Pkhakadze, K teorii lebegovskoi mery, Trudy dreamt of, Introduction to the theory of linear non- Tbilisskogo Matematiceskogo Instituta 25, Tbilisi selfadjoint operators by I. C. Gohberg and M. G. 1958 (in Russian). Krein, was published only in 1965.) [Ra] J. Raisonnier, A mathematical proof of S. Shelah's theorem on the measure problem and related re- sults, Isr. J. Math. 48 (1984), 48-56. [Ru] W. Rudin, Real and Complex Analysis, McGraw-Hill Book Company, 1987. [So] R. Solovay, A model of set theory in which every set of reals is Lebesgue measurable, Ann. of Math. 92 (1970), 1-56. [Sz] E. Szpilrajn (alias E. Marczewski), Sur l'extension de la mesure lebesguienne, Fund. Math. 25 (1935), 551-558. [Ul] S. Ulam, Zur Mass-theorie in der allgemeinen Men- genlehre, Fund. Math 16 (1930), 140-150. [Vi] G. Vitali, Sul problema della mesure dei gruppi di punti di una retta, Bologna, 1905. [Wa] S. Wagon, The Banach-Tarski Paradox, Cambridge Univ. Press, 1985. Dept. of Mathematics, Mechanics and Computer Science Warsaw University Warsaw, Poland and
Dept. of Mathematics The University of Louisville Louisville, KY 40292 USA
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