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Lebesgue Measurability and Large Cardinals 1 Introduction

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Lebesgue Measurability and Large Cardinals 1 Introduction Lebesgue measurability and large cardinals 1 Introduction Lebesgue measure is motivated by the following idea. It is clear that by the \length" of an interval (a; b) or [a; b] we mean b − a. Can we extend this notion to many other subsets of R in a natural way? In other words, can we find some µ : S ! [0; 1] for some \large" subset S ⊆ P (R) with the following properties? 1. if a; b 2 R with a < b, then µ ((a; b)) = µ ([a; b]) = b − a 2. (monotonicity) if A; B 2 S with A ⊆ B, then µ (A) ≤ µ (B) 3. (countable additivity) if (Ai) is a countable collection of pairwise disjoint members S P i2I of S, then µ i2I Ai = i2I µ (Ai) 4. (translation invariance) if x 2 R and A 2 S, then µ (fx + a : a 2 Ag) = µ (A) Definition. The outer Lebesgue measure of a set A ⊆ R is ( 1 1 ) ∗ X [ µ (A) := inf (bi − ai): ai; bi 2 R and ai < bi 8i 2 !; and A ⊆ (ai; bi) : i=0 i=0 ∗ ∗ ∗ A is (Lebesgue) measurable iff µ (A) = µ (A \ B) + µ (A \ (R n B)) for every B ⊆ R, in which case the Lebesgue measure of A is µ (A) := µ∗ (A). It can be shown that Lebesgue measure satisfies properties 1{4, and that many naturally-occurring sets are Lebesgue measurable, such as Borel sets (see section 2.3). For the proof we refer the reader to Carath´eodory's extension theorem. However, the following construction shows that we can never achieve S = P (R). In particular, not every set of reals is Lebesgue measurable. Example (Vitali). Define the equivalence relation ∼ on R by x ∼ y iff y − x 2 Q. By the axiom of choice, let A ⊆ [0; 1] contain exactly one member from each equivalence class, and suppose for contradiction A 2 S. Write Aq = fa + q : q 2 Qg for each q 2 Q, where Q = Q \ [−1; 1]. Then by translation invariance, µ (Aq) = µ (A) for each q 2 Q. Now (Aq)q2Q is a collection of pairwise disjoint S P P sets, so by countable additivity µ q2Q Aq = q2Q µ (Aq) = q2Q µ (A) = 0 or 1. S S But [0; 1] ⊆ q2Q Aq ⊆ [−1; 2], so by monotonicity 1 ≤ µ q2Q Aq ≤ 3. Contradic- tion. This construction uses the axiom of choice (AC). Thus we can still hope that, in some sense, every explicitly describable set of reals is measurable. It may be natural to ask: is it consistent with ZF that every set of reals is Lebesgue measurable? But this question is not really appropriate, since AC is used in the proof that Lebesgue measure is countably additive. Indeed, if ZF is consistent then it is consistent with ZF that R is a countable union of countable sets [10, p. 142], in which case if S = P (R) then countable additivity implies µ (R) = 0! However, only a weak form of AC is required to prove the basic properties of Lebesgue measure. The following statement follows from AC via Zorn's lemma, and implies the axiom of countable choice. 1 Definition. The axiom of dependent choice (DC) is the statement that if X is a non-empty set and R is a binary relation on X such that for all x 2 X there exists y 2 X with xRy, then we can choose a sequence (xn)n2! from X with x0Rx1Rx2R::: . This axiom is sufficient to develop the theory of Lebesgue measure and for most of real analysis. By contrast, it is not enough for Vitali's example: since each equivalence class @0 was countable, we made 2 choices, each between @0 options. The purpose of this essay, then, is to address the following. Question. Is it consistent with ZF + DC that every set of reals is Lebesgue measurable? 2 Solovay's model Surprisingly, this question is connected to the existence of a strongly inaccessible cardinal. Notation. Denote by LM the statement that every set of reals is Lebesgue measurable. Denote by I the statement that there is a strongly inaccessible cardinal. This chapter is devoted to proving the following positive answer to the above question. Theorem 1. If ZFC + I is consistent, then ZF + DC + LM is consistent. To prove this, we will use forcing to prove the following in ZFC. Theorem 2 (Solovay [4]). Suppose there is a countable transitive 2-model of ZFC + I. Then there is a countable transitive 2-model of ZF + DC + LM. To prove Theorem 1 from this, we follow the approach of Kunen [1, p. 245]. This requires a modified version of Theorem 2, which will easily be seen to follow from the same proof. The following is in fact a theorem scheme rather than a theorem of ZFC (like the reflection principle). Theorem 20. For each finite Ω ⊆ ZF, there is a finite Λ ⊆ ZFC such that the following is a theorem of ZFC. Suppose there is a countable transitive 2-model of Λ + I. Then there is a countable transitive 2-model of Ω + DC + LM. Proof of Theorem 1 from Theorem 20. Suppose ZF + DC + LM is inconsistent. Then Ω + DC + LM is inconsistent for some finite Ω ⊆ ZF, and moreover this fact is provable in ZFC since it is witnessed by a finite proof of ?. Choose a corresponding finite Λ ⊆ ZFC as in Theorem 20. In this paragraph we work in ZFC + I. By the reflection principle, Λ + I has a 2-model, which by the downward L¨owenheim{Skolem theorem has a countable elementary submodel. This is a countable 2-model of Λ+I, but it need not be transitive. However, since it satisfies the axiom of extensionality and 2 is well-founded, we may apply the Mostowski collapse lemma to obtain an isomorphic transitive 2-model. This is a countable transitive 2-model of Λ + I, so by Theorem 20 there is a countable transitive 2-model of Ω + DC + LM. So by the soundness theorem, Ω + DC + LM is consistent. Thus it is possible in ZFC + I to prove both the consistency and the inconsistency of the theory Ω + DC + LM. So ZFC + I is inconsistent. 2 Outline of proof of Solovay's theorem. Working in ZFC, we will begin with a countable transitive 2-model M of ZFC + I in which (κ is a strongly inaccessible cardinal)M . We will then use a forcing poset known as the \L´evycollapse" to pass to a forcing extension M [G] in which all ordinals smaller than κ are \collapsed" to !. ∗ Finally, we will consider the submodel HODM[G] given by those sets that are \heredi- tarily ordinal-sequence-definable in M [G]", and show that this is a transitive 2-model of ZF + DC + LM. For the rest of this chapter, then, we work in ZFC unless stated otherwise. 2.1 Ordinal definability Although in the proof of Solovay's theorem we will need the notion of definability by an infinite sequence of ordinals, it will be instructive to first consider the notion of definability by finitely many ordinals. We use ideas from Kunen [1, pp. 145{147]. In this section, assume that M is a transitive 2-model of ZF. By this we mean in particular that M is a set. Definition. We say a 2 M is ordinal definable in M, and write a 2 ODM , iff there exists a [1] M formula φ (x1; x2; : : : ; xn; y) in the language of set theory and α1; α2; : : : ; αn 2 On such M that φ (α1; α2; : : : ; αn; a) and a is unique in M with this property. We say a 2 M is hereditarily ordinal definable in M, and write a 2 HODM , iff TC (fag) ⊆ ODM . Proposition 1. HODM is a transitive 2-model of ZF. The proof of Proposition 1 will be very similar to that of the corresponding result for forcing extensions, in the following sense. Since HODM is transitive, in several cases, the fact that an axiom holds in HODM will follow from the fact that it holds in M, together with a simple formula provided by the axiom itself to demonstrate ordinal definability in M. The difficulty will arise when we reach the axiom scheme of separation, and concerns the definability of ODM in M. To illustrate the difficulty, let b 2 HODM and φ (x) be a formula in the lan- n HODM o guage of set theory, and let c = d 2 b : φ (d) . Why should c 2 ODM ? We would like to define c from b by the formula 8x x 2 c , (x) ^ φ (x)HODM , where (8x ( (x) , x 2 b))M . But suppose φ (x) has the form 8y' (x; y). Then φ (x)HODM has the HODM HODM form (8y 2 HODM ) ' (x; y) , i.e. 8y TC (fyg) ⊆ ODM ) ' (x; y) . How M do we know that there is a formula χ (x) such that for every a 2 M, a 2 ODM iff χ (a) ? We certainly can't express the above definition of ODM in this fashion directly without first constructing a formula ξ (x) such that, for every formula ζ in the language of set theory, M (ξ (pζq) , ζ) , which would contradict Tarski's undefinability theorem. We nonetheless have the following result, which is analogous to the definability lemma for forcing exten- sions. [1] It is implicit that the free variables of a formula φ (x1; x2; : : : ; xn) are exactly x1, x2,..., xn.
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