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Pointwise compact and stable sets

arXiv:math/9209218v1 [math.LO] 15 Sep 1992amsppt.styFSp90a.texPointwise compact and stable setsof measurable functionsS.Shelah & D.H.FremlinHebrew University, JerusalemUniversity of Essex, Colchester, England[University of Essex Mathematics Department Research Report 91-3]Introduction In a series of papers culminating in [Ta84], M.Talagrand, the second author and othersinvestigated at length the properties and structure of pointwise compact sets of measurable functions. Anumber of problems, interesting in themselves and important for the theory of Pettis integration, were solvedsubject to various special axioms.

It was left unclear just how far the special axioms were necessary. Inparticular, several results depended on the fact that it is consistent to suppose that every countable relativelypointwise compact set of Lebesgue measurable functions is ‘stable’ in Talagrand’s sense; the point being thatstable sets are known to have a variety of properties not shared by all pointwise compact sets.

In the presentpaper we present a model of set theory in which there is a countable relatively pointwise compact set ofLebesgue measurable functions which is not stable, and discuss the significance of this model in relation tothe original questions. A feature of our model which may be of independent interest is the following: in it,there is a closed negligible set Q ⊆[0, 1]2 such that whenever D ⊆[0, 1] has outer measure 1 thenQ−1[D] = {x : ∃y ∈D, (x, y) ∈Q}has inner measure 1 (see 2G below).1.

The model We embark immediately on the central ideas of this paper, setting out a construction of apartially ordered set which forces a fairly technical proposition in measure theory (1S below); the relevanceof this proposition to pointwise compact sets will be discussed in §2. The construction is complex, andrather than give it in a single stretch we develop it cumulatively in 1E, 1I, 1Q below; it is to be understoodthat each notation introduced in these paragraphs, as well as those in the definitions 1A, 1K, 1L, is to standfor the remainder of the section.

After each part of the construction we give lemmas which can be dealtwith in terms of the construction so far, even if their motivation is unlikely to be immediately clear.When we come to results involving Forcing, we will try to follow the methods of [Ku80]; in particular, ina p.o.set, ‘p ≤q’ will always mean that p is a stronger condition than q.1A Definition If A is any family of sets not containing ∅, setdp(A) = min{#(I) : I ∩A ̸= ∅∀A ∈A}.Observe that dp(A) = 0 iffA = ∅and that dp(A ∪B) is at most the cardinal sum of dp(A) and dp(B). (Ofcourse much more can be said.

)1B Lemma Suppose that n, l, k ∈N, with n, l not less than 2, and that ǫ is such that 0 < ǫ ≤12 andlǫk ≥(k + 2) ln n. Then there is a set W ⊆n × n (we identify n with the set of its predecessors) such that#(W) ≤ǫn2 and whenever I ∈[n]l and J0, . .

. , Jl−1 ∈[n]≤k are disjoint, there are i ∈I, j < l such that{i} × Jj ⊆W.proof If k = 0 this is trivial; suppose that k > 0.

Set Ω= P(n × n). Give Ωa probability for which theevents (i, j) ∈W, as (i, j) runs over n × n, are independent with probability ǫ.

If W ∈Ωis a random set,thenPr(#(W) ≤ǫn2) > 14because ǫ ≤12 and #(W) has the binomial distribution B(n2, ǫ). On the other hand, if J ∈[n]≤k and i < n,Pr({i} × J ⊆W) ≥ǫk.

So if I ∈[n]l and J0, . .

. , Jl−1 are disjoint members of [n]≤k,1

2Pr({i} × Jj ̸⊆W ∀i ∈I, j < l) ≤(1 −ǫk)l2≤exp(−l2ǫk).AccordinglyPr(∃I ∈[n]l, disjoint J0, . .

. , Jl−1 ∈[n]≤k such that {i} × Jj ̸⊆W ∀i ∈I, j < l)≤#([n]l)#([n]≤k)l exp(−l2ǫk)≤nlnkl exp(−l2ǫk)= exp((k + 1)l ln n −l2ǫk) ≤14becausel2ǫk −(k + 1)l ln n ≥l ln n ≥2 ln 2.There must therefore be some W ∈Ωof the type required.Remark Compare the discussion of cliques in random graphs in [Sp87], pp.

18-20.1C Lemma Let m and l be strictly positive integers and A a non-empty family of non-empty sets. LetT be the family of non-empty sets T ⊆Am.

For T ∈T write T ∗= {t↾j : t ∈T , j ≤m} ⊆Sj≤m Aj. ForT , T0 ∈T say that T ≼T0 if T ⊆T0 anddp({u : t⌢u ∈T ∗}) ≥dp({u : t⌢u ∈T ∗0 })/2lfor every t ∈T ∗\ T .

Fix T0 ∈T and a cover ⟨Si⟩i<2l of T0. Then there is a T ≼T0 such that T ⊆Si forsome i < 2l.

[Notation: In this context we use ordinary italics, ‘u’, for members of A, and bold letters, ‘t’, for finitesequences of members of A. ]proof For t ∈T ∗0 \ T0 setαt = dp({u : t⌢u ∈T ∗0 })/2l > 0.For i < 2l define ⟨S(j)i⟩j≤m by setting S(m)i= Si ∩T0,S(j)i= {t : t ∈Aj ∩T ∗0 , dp({u : t⌢u ∈S(j+1)i}) ≥αt}for j < m. An easy downwards induction (using the fact that dp is subadditive) shows that T ∗0 ∩Aj =Si<2l S(j)ifor every j ≤m.

In particular, there is some i < 2l such that ∅∈S(0)i. Now define T byT = {t : t ∈Am, t↾j ∈S(j)i∀j ≤m} ⊆T0 ∩Si,and see that T ≼T0, as required.1D Corollary Let n, l, k and W be as in Lemma 1B.

Take r ≤k, let Z be the cartesian product nr andset˜W = {(i, z) : i < n, z ∈Z, (i, z(j)) ∈W ∀j < r}.Let m, A, T and ≼be as in Lemma 1C, and take T0 ∈T, H : T0 →n any function. Theneither there are i < n, T ≼T0 such that H(t) = i for every t ∈Tor there is a J ∈[n]≤rl such that for every z ∈(n \ J)r there is a T ≼T0 such that (H(t), z) ∈˜W forevery t ∈T .proof SetA = {z : z ∈Z, ∃T ≼T0 such that (H(t), z) ∈˜W ∀t ∈T }.If A ⊇(n \ J)r for some J ∈[n]≤rl, we have the second alternative; suppose otherwise.

Then we can findz0, . .

. , zl−1 ∈Z \ A such that the sets Jj = {zj(i) : i < r} are all disjoint.

Each Jj belongs to [n]≤k, so bythe choice of W,I = {i : {i} × Jj ̸⊆W ∀j < l}has cardinal less than l. Now observe that if t ∈T0 then either H(t) ∈I or (H(t), zj) ∈˜W for some j < l.So we have a cover of T0 by the setsSi = {t : H(t) = i} for i ∈I,S′j = {t : (H(t), zj) ∈˜W} for j < l.

3By Lemma 1C, there is a T ≼T0 such that either T ⊆Si for some i ∈I or T ⊆S′j for some j < l. But wecannot have T ⊆S′j, because zj /∈A; so T ⊆Si for some i, and we have the first alternative.Remark 1C-1D are of course elementary, but their significance is bound to be obscure; they will be usedin 1R below.An essential feature of 1C is the fact that the denominator ‘2l’ in the definition of ≼isindependent of the size of A.1E Construction: part 1 (a) Take a sequence ⟨nk⟩k∈N of integers increasing so fast that(i) n0 ≥4;(ii) nk > 2k+1;(iii) writing ˜cl = Qi

Then T is afinitely-branching tree of height ω in which the Tk are the levels and ‘rank’ is the rank function. For t ∈Twrite T (t) for the subtree {t′ : t′ ≤t or t ≤t′}, suc(t) for {t′ : t ≤t′, rank(t′) = rank(t) + 1}.

(c) For k ∈N, setγk = (k + 1)/ ln(⌈2−k−1nk⌉);2−k−1nk > 1 by (a)(ii) above. For t ∈T define dt : PT →R ∪{−∞} by writingdt(S) = γrank(t) ln(dp({C : t × C ∈S}))for every S ⊆T , allowing dt(S) = −∞if S ∩suc(t) = ∅.

Observe that dt(T ) ≥k + 1 whenever rank(t) = k(becausedp({C : C ⊆nk, #(C) ≥(1 −2−k−1)nk}) ≥⌈2−k−1nk⌉. )(d) Let Q be the set of subtrees q ⊆T such thatq ̸= ∅;if t ≤t′ ∈q then t ∈q;if t ∈q then q ∩suc(t) ̸= ∅;writing δk(q) = min{dt(q) : t ∈q ∩Tk} for k ∈N, limk→∞δk(q) = ∞.Observe that δk(T ) ≥k + 1 for each k ∈N, so that T ∈Q and Q ̸= ∅.

(e) For q, q′ ∈Q say that q ≤q′ if q ⊆q′. Then (Q, ≤, T ) is a p.o.set (that is, a pre-ordered set with atop element, as in [Ku80]).

Observe that if t ∈q ∈Q then q ∩T (t) ∈Q and q ∩T (t) ≤q. (f) For q, q′ ∈Q and k ∈N say that q ≤k q′ if q ≤q′ and q ∩Tk = q′ ∩Tk anddt(q) ≥min(k, dt(q′)) −2−kfor every t ∈q.

Note that ≤k is not transitive unless k = 0.Remarks Of course the point of the sequence ⟨nk⟩k∈N on which the rest of this construction will dependis that it increases ‘as fast as we need it to’. The exact list given in (a) above is of no significance and willbe used only as a list of clues to the (elementary) arguments below which depend on the rapidly-increasingnature of the sequence.

This is why we have made no attempt to make the list as elegant or as short aspossible.Three elements may be distinguished within the construction of Q.First, it is a p.o.set of rapidlybranching trees; that is, if t ∈q ∈Q, q ∩suc(t) is large compared with Trank(t), except for t of small rank.This is the basis of most of the (laborious but routine) work down to 1P below. Second, there is a naturalQ-name for a subset of X = Qk∈N nk of large measure; a generic filter in Q leads to a branch of T and

4hence to the Ψ of 1Q(d). Third, the use of dp in the definition of ‘rapidly branching’ ((c)-(d) above) is whatmakes possible the side-step in the last part of the proof of 1R.1F Lemma Q is proper.proof This is a special case of Proposition 1.18 in [Sh326].

(In fact, the arguments of 1G-1H below showthat Q satisfies Axiom A, and is therefore proper; see [Ba84], 2.4. )1G Lemma Let k ∈N and let ζ be an ordinal.

Suppose that A is a set with #(A) ≤exp(2−k/γj) −1for every j ≥k, and that τ is a Q-name for a member of A. Let ∆be a Q-name for a countable subset ofζ.

Then for every q ∈Q there are a q′ ≤k q, a function H : Tk →A and a countable (ground-model) setD ⊆ζ such thatq′ ∩T (t) ⊩Q τ = H(t) ∀t ∈q′ ∩Tk,q′ ⊩Q ∆⊆D.proof (a) Set m = #(A). The point is that if j ≥k and t ∈Tj and ⟨Si⟩i≤m is a family of subsets of T ,thendt([i≤mSi) = γj ln(dp([i≤m{C : t × C ∈Si}))≤γj ln(Xi≤mdp({C : t × C ∈Si}))≤γj ln((m + 1) maxi≤m dp({C : t × C ∈Si}))= γj ln(m + 1) + maxi≤m γj ln(dp({C : t × C ∈Si}))≤2−k + maxi≤m dt(Si).

(b) For each a ∈A, let Sa be the set{t : t ∈q, rank(t) ≥k, ∃p ∈Q, D ∈[ζ]≤ω,p ≤k q ∩T (t), p ⊩Q τ = a & ∆⊆D}.If t ∈q \ Sa and rank(t) ≥k, then dt(Sa) < min(k, dt(q))−2−k. For if Sa ∩suc(t) = ∅, dt(Sa) = −∞.

Whileif Sa ∩suc(t) ̸= ∅, then for each s ∈suc(t) ∩Sa we can find ps ∈Q and Ds ∈[ζ]≤ω such that ps ≤k q ∩T (s),ps ⊩Q τ = a and ps ⊩Q ∆⊆Ds. If we now setp = Ss∈suc(t)∩Sa ps, D = Ss∈suc(t)∩Sa Ds,then p ⊆q ∩T (t) and p ⊩Q τ = a and p ⊩Q ∆⊆D.

Because t /∈Sa, p ̸≤k q ∩T (t) and there must be ans ∈p such that ds(p) < min(k, ds(q ∩T (t))) −2−k; evidently s = t anddt(Sa) < min(k, dt(q)) −2−k,as claimed. (c) Suppose, if possible, that there is a t0 ∈q ∩Tk \ Sa∈A Sa.

Setp = {t : t ∈q ∩T (t0), t′ /∈Sa∈A Sa ∀t′ ≤t}.Then p is a subtree of T . For every t ∈p with t ≥t0,dt(q) ≤max({dt(p)} ∪{dt(Sa) : a ∈A}) + 2−kbecause #(A) = m. But dt(Sa) < dt(q) −2−k for every a ∈A, by (b) above, so dt(p) ≥dt(q) −2−k (andp ∩suc(t) ̸= ∅).

This shows both that p has no maximal elements and that δi(p) ≥δi(q) −2−k for everyi ≥k, so that p ∈Q. Because Q is proper, we can find a p′ ≤p and a countable D ⊆ζ such that p′ ⊩∆⊆D([Sh82], p. 81, III.1.16).

Next, there are p′′ ≤p′, a ∈A such that p′′ ⊩Q τ = a. Let j ∈N be such thatδi(p′′) ≥k whenever i ≥j, and take t ∈p′′ such that rank(t) ≥max(k, j).

Then p′′ ∩T (t) witnesses thatt ∈Sa; which is impossible. (d) Accordingly we have for every t ∈q ∩Tk an H(t) ∈A, a countable set Dt and a pt ∈Q such thatpt ⊩Q τ = H(t) & ∆⊆Dt,pt ≤k q ∩T (t).

5Set q′ = St∈q∩Tk pt, D = St∈q∩Tk Dt; then q′ ≤k q, D is countable, q′ ⊩Q ∆⊆D and q′ ∩T (t) ⊩Q τ = H(t)for every t ∈q ∩Tk.1H Lemma Let ⟨qk⟩k∈N be a sequence in Q such that q2k+2 ≤k+1 q2k+1 ≤k q2k for every k ∈N. Thenˆq = Tk∈N qk belongs to Q and is accordingly a lower bound for {qk : k ∈N} in Q; also ˆq∩Tk+1 = q2k+1∩Tk+1for each k ∈N.proof Because each qk is a finitely-branching subtree of T with no maximal elements, so is ˆq, and dt(ˆq) =limk→∞dt(qk) for every t ∈ˆq.

Moreover, if t ∈ˆq and k ≤l ∈N,dt(q2l) ≥min(k, dt(q2k)) −3.2−k + 3.2−l,(induce on l, using the definition of ≤l), so we haveδi(ˆq) = liml→∞δi(q2l) ≥min(k, δi(q2k)) −3.2−kfor every i, k ∈N; consequently limi→∞δi(ˆq) = ∞and ˆq ∈Q. Now if k ∈N and 2k + 1 ≤l, ql+1 ∩Tk+1 =ql ∩Tk+1, so ˆq ∩Tk+1 = q2k+1 ∩Tk+1.amsppt.styFSp90b.texVersion of 16.9.921I Construction: part 2 Let κ be the cardinal c+ (evaluated in the ground model).

(a) Let (⟨Pξ⟩ξ≤κ, ⟨Qξ⟩ξ<κ) be a countable-support iteration of p.o.sets, as in [Ku80], chap. 8, such thateach Qξ is a Pξ-name for a p.o.set with the same definition, interpreted in V Pξ, as the p.o.set Q of 1E.

(Notethat T is absolute, and so, in effect, is ⟨dt⟩t∈T , because each dt is determined by its values on the finite setP(suc(t)); so that the difference between Q and Qξ subsists in the power of Pξ to add new subsets of T .Also each Qξ is ‘full’ in Kunen’s sense.) Write P = Pκ.

(b) If ζ ≤κ, K ∈[ζ]<ω, k ∈N and p, p′ belong to Pζ, say that p ≤K,k p′ if p ≤p′ andp↾ξ ⊩Pξ p(ξ) ≤k p′(ξ) ∀ξ ∈K,taking ≤k here to be a Pξ-name for the relation on Qξ corresponding to the relation ≤k on Q as defined in1E(f). Of course ≤K,k is not transitive unless K = ∅or k = 0.1J Lemma (a) Pζ is proper for every ζ ≤κ.

(b) If ξ < κ, ζ ≤κ then Pξ+ζ may be identified with a dense subset of the iteration Pξ ∗P′ζ, where P′ζ isa Pξ-name with the same definition, interpreted in V Pξ, as the definition of Pζ in V . (c) For every ζ < κ,11Pζ ⊩Pζ 2ω < κ.

(d) If ζ ≤κ has uncountable cofinality, A is a (ground-model) set, ˙f is a Pζ-name for a sequence in Aand p ∈Pζ, then we can find ξ < ζ, p′ ≤p and a Pξ-name ˙g such thatp′ ⊩Pζ ˙f = ˙g. (e) If A is a (ground-model) set and ˙f is a P-name for a sequence in A, then we can find a ξ < κ and aPξ-name ˙g such that11P ⊩P ˙f = ˙g.proof (a) This is just because Q is proper, as noted in 1F; see [Sh82], p. 90, Theorem III.3.2.

(b) This now follows by induction on ζ. The inductive step to a successor ordinal is trivial, because ifwe can think of Pξ+ζ as dense in Pξ ∗P′ζ then we can identify Qξ+ζ with Q′ζ.

As for the inductive step tolimit ζ, any member of Pξ+ζ can be regarded as (p, p′) where p ∈Pξ and p′ is a Pξ-name for a member ofP′ζ. On the other hand, given (p, p′) ∈Pξ ∗P′ζ, we have a Pξ-name ˙J for the support of p′ which in V Pξ is acountable subset of ζ.

But because Pξ is proper there are a p1 ≤p and a countable ground-model set I ⊆ζsuch that p1 ⊩Pξ˙J ⊆I ([Sh82], p. 81, III.1.16). Now (p1, p′) can be re-interpreted as a member of Pξ+ζstronger than (p, p′).

Thus Pξ+ζ is dense in Pξ ∗P′ζ, as claimed. (c) [Sh82], p. 96, III.4.1.

(d) [Sh82], p. 171, V.4.4. (e) By [Sh82], p. 96, III.4.1, P satisfies the κ-c.c.

; because κ is regular, (d) gices the result.1K Definition Let ζ ≤κ, p ∈Pζ.

6(a) Define U(p), ⟨p(u)⟩u∈U(p) as follows. A finite function u ⊆ζ × T belongs to U(p) ifeither u = ∅, in which case p(u) = p,or u = v ∪{(ξ, t)} where v ∈U(p), dom(v) ⊆ξ < ζ, andp(v)↾ξ ⊩Pξ t ∈p(v)(ξ),in which case p(u) is defined by writingp(u)(η) = p(v)(η) ∀η ∈ζ \ {ξ},p(u)(ξ) = p(v)(ξ) ∩T (t).

(b) Observe that if u ∈U(p) then p(u)(ξ) = p(ξ) for ξ ∈ζ\dom(u), p(u)(ξ) = p(ξ)∩T (u(ξ)) if ξ ∈dom(u);U(p) is just the set of finite functions u for which these formulae define such a p(u) ∈Pζ. Of course p(u) ≤pfor every u ∈U(p).

(c) Note that if ξ ≤ζ, p ∈Pζ, u ∈U(p) then u↾ξ ∈U(p↾ξ) and (p↾ξ)(u↾ξ) = p(u)↾ξ. (d) If p ∈Pζ, u ∈U(p) and v ⊆ζ ×T is a finite function such that dom(u) ⊆dom(v) and u(ξ) ≤v(ξ) inT for every ξ ∈dom(u), then v ∈U(p) iffv ∈U(p(u)), and in this case p(v) = (p(u))(v) (induce on #(v)).

(e) We shall mostly be using not the whole of U(p) but the sets U(p; K, k) = U(p) ∩T Kkfor K ∈[ζ]<ω,k ∈N, writing T Kkfor the set of functions from K to Tk.1L Definition For ζ ≤κ, K ∈[ζ]<ω, k ∈N and p ∈Pζ, say that p is (K, k)-fixed if for every η ∈K,u ∈U(p; K ∩η, k) there is a (ground-model) set A ⊆Tk such thatp(u)↾η ⊩Pη p(η) ∩Tk = A.Equivalently, p is (K, k)-fixed if U(p; K, k) ⊇U(p1; K, k) for every p1 ≤p.1M Lemma Suppose ζ ≤κ, K ∈[ζ]<ω, k ≥1 and that A is a finite set with 2cmacm−1 ≤exp(2−k/γi)for every i ≥k, where c = #(Tk), m = #(K) and a = #(A). Let τ be a Pζ-name for a member of A, and∆a Pζ-name for a countable subset of κ.

Then for any p ∈Pζ there are p1 ≤K,k p, a function H : T Kk →Aand a countable (ground-model) set D ⊆κ such thatp1 is (K, k)-fixed,p(u)1⊩Pζ τ = H(u) ∀u ∈U(p1; K, k),p1 ⊩Pζ ∆⊆D.proof Induce on m = #(K). If m = 0 we may take any a ∈A, p′1 ≤p such that p′1 ⊩τ = a, and (againusing [Sh82], III.1.16, this time based on 1Ja) a countable D and a p1 ≤p′1 such that p1 ⊩Pζ ∆⊆D; nowset H(∅) = a.For the inductive step to #(K) = m ≥1, let ξ be max K. As explained in 1Jb, Pζ may be regarded as adense subset of Pξ+1 ∗P′; arguing momentarily in V Pξ+1 we can find a Pξ+1-name ˆr0 for a member of P′, aPξ+1-name τ ′ for a member of A and a Pξ+1-name ∆′ for a countable set such that(p↾ξ + 1, ˆr0) ≤p in Pξ+1 ∗P′,(p↾ξ + 1, ˆr0) ⊩Pξ+1∗P′ τ ′ = τ,(p↾ξ + 1, ˆr0) ⊩Pξ+1∗P′ ∆⊆∆′.Now let ∆′0 be a Pξ+1-name for a countable subset of ζ \ (ξ + 1) such that11Pξ+1 ⊩Pξ+1 supp(ˆr0) = ∆′0.Because #(A) = a < 2cmacm−1 ≤exp(2−k/γi) for every i ≥k, we can use Lemma 1G in V Pξ to find ˜H, ˜q,˜∆such that˜H is a Pξ-name for a function from Tk to A,˜∆is a Pξ-name for a countable subset of κ,˜q ∈Qξ,p↾ξ ⊩Pξ ˜q ≤k p(ξ),p↾ξ ⊩Pξ˜q ∩T (t) ⊩Qξ τ ′ = ˜H(t) ∀t ∈˜q ∩Tk,p↾ξ ⊩Pξ˜q ⊩Qξ ∆′ ∪∆′0 ⊆˜∆.Now consider the pair ( ˜H, ˜q ∩Tk).

This can be regarded as a Pξ-name for a member of A1 = ATk × PTk,and a1 = #(A1) = 2cac, so2cm−1acm−21= 22cm−1acm−1 ≤2cmacm−1 ≤exp(2−k/γj) ∀i ≥k.

7The inductive hypothesis therefore tells us that there are ˆp1 ≤K∩ξ,k p↾ξ, H∗: T K∩ξk→ATk, F ∗: T K∩ξk→PTk and a countable D ⊆κ such thatˆp1 is (K ∩ξ, k)-fixed,ˆp(u)1⊩Pξ ˜H = H∗(u) & ˜q ∩Tk = F ∗(u)for every u ∈U(ˆp1; K ∩ξ, k), andˆp1 ⊩Pξ ˜∆⊆D.At this point we observe thatˆp1 ⊩Pξ˜q ⊩Qξ supp(ˆr0) ⊆D.Now the only difference between Pξ+1 ∗P′ and Pζ is that for members of the former their supports haveto be regarded as Pξ+1-names for countable subsets of ζ, and these are not always reducible to countableground-model sets. But in the present case this difficulty does not arise and we have a p1 ∈Pζ defined bysaying that p1↾ξ = ˆp1, p1↾ξ ⊩Pξ p1(ξ) = ˜q, and p1↾η ⊩Pη p1(η) = ˆr0(η) for ξ < η < ζ; then supp(p1) ⊆supp(ˆp1) ∪{ξ} ∪(D ∩ζ) is countable.Consequently p1 ∈Pζ is well-defined and now, setting H(u⌢t) = H∗(u)(t) for u ∈T K∩ξk, t ∈Tk,p1 ≤K,k p,p1 ⊩P ∆⊆D,U(p1; K, k) = {u⌢t : u ∈U(ˆp1; K ∩ξ, k), t ∈F ∗(u)},p(v)1⊩Pζ τ = H(v) ∀v ∈U(p1; K, k)and finallyp(u)1 ↾ξ ⊩Pξ p1(ξ) ∩Tk = F ∗(u) ∀u ∈U(p1; K ∩ξ, k),so that p1 is (K, k)-fixed, and the induction proceeds.1N Lemma Suppose ζ ≤κ, ⟨Kk⟩k∈N is an increasing sequence of finite subsets of ζ, ⟨pk⟩k∈N is a sequencein Pζ; suppose thatp2k+2 ≤Kk,k+1 p2k+1 ≤Kk,k p2kfor every k ∈N and that Sk∈N supp(pk) ⊆Sk∈N Kk.

Then there is a ˆp ∈Pζ such that ˆp ≤pk for everyk ∈N, supp(ˆp) ⊆Sk∈N Kk andˆp↾ξ ⊩Pξ ˆp(ξ) ∩Tk = p2k+1 ∩Tk ∀k ∈N, ξ ∈Kk,ˆp↾ξ ⊩Pξ ˆp(ξ) ∩Tk+1 = p2k+2 ∩Tk+1 ∀k ∈N, ξ ∈Kk+1,so thatU(ˆp; Kk, k) ⊇U(p2k+1; Kk, k) and U(ˆp; Kk, k + 1) ⊇U(ˆpk+2; Kk, k + 1)for every k ∈N.proof For each ξ < ζ choose ˆp(ξ) such that11Pξ ⊩Pξ ˆp(ξ) = Tk∈N pk(ξ).An easy induction on ξ shows that ˆp↾ξ ∈Pξ for every ξ ≤ζ; for if ξ ∈ζ \ Sk∈N Kk then11Pξ ⊩Pξ ˆp(ξ) = T = 11Qξ,while if k ∈N and ξ ∈Kk thenˆp↾ξ ⊩Pξ p2l+2(ξ) ≤l+1 p2l+1(ξ) ≤l p2l(ξ) ∀l ≥k,so that by Lemma 1H,ˆp↾ξ ⊩Pξ ˆp(ξ) ∈Qξ & ˆp(ξ) ∩Tk+1 = p2k+1 ∩Tk+1 = p2k+2 ∩Tk+1& ˆp(ξ) ∩Tk = p2k+1 ∩Tk.It follows at once that U(ˆp; Kk, k) ⊇U(p2k+1; Kk, k), U(ˆp; Kk, k +1) ⊇U(p2k+2; Kk, k +1) for every k ∈N.1O Lemma Suppose that 0 < ζ ≤κ, σ is a Pζ-name for a member of Qk∈N nk, and p ∈Pζ. Then wecan find a ˆp and sequences ⟨Kk⟩k∈N, ⟨Hk⟩k∈N such thatˆp ∈Pζ, ˆp ≤p;⟨Kk⟩k∈N is an increasing sequence of subsets of ζ, #(Kk) ≤k + 1 for every k, K0 = {0};supp(ˆp) ⊆Sk∈N Kk;ˆp is (Kk, k)-fixed and (Kk, k + 1)-fixed for every k;Hk is a function from T Kkk+1 to nk for every k;ˆp(u) ⊩Pζ σ(k) = Hk(u) whenever k ∈N and u ∈U(ˆp; Kk, k + 1).proof Using Lemma 1M, we can find sequences ⟨pk⟩k∈N, ⟨Kk⟩k∈N and ⟨Hk⟩k∈N such that

8p = p0, K0 = {0};#(Kk+1) ≤k + 2, Kk+1 ⊇Kk;p2k+1 ≤Kk,k p2k, p2k+1 is (Kk, k)-fixed;Hk : T Kkk+1 →nk is a function;p2k+2 ≤Kk,k+1 p2k+1, p2k+2 is (Kk, k + 1)-fixed,p(u)2k+2 ⊩Pζ σ(k) = Hk(u) ∀u ∈U(p2k+2; Kk, k + 1)for every k ∈N. Furthermore, we may do this in such a way that Sk∈N Kk includes Sk∈N supp(pk).

Weneed of course to know that the nk are rapidly increasing; specifically, that2ck+1k≤exp(2−k/γi) ∀i ≥k(when choosing p2k+1) and that2ck+1k+1nckk+1k≤exp(2−k−1/γi) ∀i ≥k + 1(when choosing p2k+2), where we write ck = #(Tk). But as ck ≤Qi

Similarly, ˆp is (Kk, k +1)-fixed for every k. Moreover, if u ∈U(ˆp; Kk, k +1) =U(p2k+2; Kk, k + 1) we have ˆp(u) ≤p(u)2k+2, soˆp(u) ⊩Pζ σ(k) = Hk(u)as required.1P Lemma Suppose that ζ ≤κ, p ∈Pζ, k ∈N, K ∈[ζ]<ω and V is a non-empty subset of U(p; K, k).Then we have a p1 = Wv∈V p(v) defined (up to ≤-equivalence in Pζ) by sayingif ξ ∈ζ \ K then p1(ξ) = p(ξ);if ξ ∈K then(p1↾ξ)(u) ⊩Pξ p1(ξ) = S{p(ξ) ∩T (t) : ∃v ∈V such that v↾ξ + 1 = u⌢t}for u ∈{v↾ξ : v ∈V}.Now p1 ≤p and if ξ < ζ, t ∈p1(ξ), rank(t) ≥k we shall havep1↾ξ ⊩Pξ suc(t) ∩p1(ξ) = suc(t) ∩p(ξ);so if ξ < ζ, i ≥k we havep1↾ξ ⊩Pξ δi(p1(ξ)) ≥δi(p(ξ)).If p2 ≤p1 there is some v ∈V such that p2 is compatible with p(v)1= p(v). If k ≤l ∈N, K ⊆L ∈[ζ]<ωthenU(p1; L, l) = {w : w ∈U(p; L, l), ∃v ∈V such that v(ξ) ≤w(ξ) ∀ξ ∈K},and p(w)1= p(w) for every w ∈U(p1; L, l); consequently, p1 is (L, l)-fixed if p is.proof Requires only a careful reading of the definitions.Remark Note that 1G-1P are based just on the fact that Q is a p.o.set of rapidly branching trees; the exactdefinition of ‘rapidly branching’ in 1E(c) is relevant only to some of the detailed calculations.

Similar ideasmay be found in [BJSp89]and [Sh326].1Q Construction: part 3 (a) Set X = Qk∈N nk. Then X, with its product topology, is a compactmetric space.

Let µ be the natural Radon probability on X, the product of the uniform probabilities on thefactors. (b) For each k ∈N set lk = ⌈(ln nk)2⌉.

Take W ′k ⊆nk × nk such that #(W ′k) ≤2−k−1n2k and wheneverI ∈[nk]lk and J0, . .

. , Jlk−1 are disjoint members of [nk]≤k, there are i ∈I and j < lk such that {i} × Jj ⊆W ′k.

(This is possible by Lemma 1B and 1E(a)(iv).) Set Wk = W ′k ∪{(i, i) : i < nk}.Write R for{(x, y) : x, y ∈X, (x(k), y(k)) ∈Wk ∀k ∈N, {k : x(k) = y(k)} is finite};then R is negligible for the product measure of X × X.

For r ∈N write Rr for the set{(x, ⟨yi⟩i

9L = {⟨Lk⟩k∈N : Lk ⊆nk ∀k ∈N, Qk∈N #(Lk)/nk > 0}.Again, we shall wish to distinguish between the ground-model set L and a corresponding P-name ⌜L⌝. (d) For each k ∈N let Φk be the P-name for a subset of nk defined (up to equivalence) by saying thatp ⊩P Φk = Ck(t)whenever rank(t) > k and p(0) ⊆T (t).

(Here Ck(t) is the kth factor of t, as described in 1E(b).) Let Ψk, Ψbe P-names for the subsets of ⌜X⌝given by11P ⊩P Ψk = {σ : σ ∈⌜X⌝, σ(i) ∈Φi ∀i ≥k}, Ψ = Sk∈N Ψk.Then we have11P ⊩P #(Φk) ≥(1 −2−k−1)nk ∀k ∈N,so that11P ⊩P ⌜µ⌝(Ψ) = 1.1R Main Lemma If r ∈N and D ⊆Xr is a (ground-model) set such that D ∩(Qk∈N Lk)r ̸= ∅for every(ground-model) sequence ⟨Lk⟩k∈N ∈L, then for every (ground-model) sequence ⟨Lk⟩k∈N ∈L11P ⊩P Ψ ∩Qk∈N Lk ⊆⌜Rr⌝−1[D ∩(Qk∈N Lk)r].proof (a) Let ⟨Lk⟩k∈N ∈L, let σ be a P-name such that11P ⊩P σ ∈Ψ ∩Qk∈N Lk,and let p ∈P.

Write D′ for D ∩(Qk∈N Lk)r. Let k0 ≥r, p1 ≤p be such that p1 ⊩P σ ∈Ψk0. By Lemma1O, we have a p2 ≤p1, an increasing sequence ⟨Kk⟩k∈N of finite subsets of κ, and a sequence ⟨Hk⟩k∈N offunctions such thatp2 is (Kk, k)-fixed and (Kk, k + 1)-fixed for every k ∈N,p(u)2⊩P σ(k) = Hk(u) whenever u ∈U(p2; Kk, k + 1), k ∈N;Sk∈N Kk ⊇supp(p2);0 ∈K0, #(Kk) ≤k + 1 for every k ∈N.

(b) For k ≥k0, let Zk be the cartesian product set nrk and take ˜Wk to be{(i, z) : i < nk, z ∈Zk, (i, z(j)) ∈Wk ∀j < r}.Set Ak = Pnk \ {∅} and Tk = P(AKkk ) \ {∅}; define ≼k on Tk as in Lemma 1C, taking lk and Kk (with theorder induced by that of κ) in place of l and m there.For each u ∈U(p2; Kk, k) setTu = {c : u∧c ∈U(p2; Kk, k + 1)} ∈Tk,where for u ∈T Kk , c ∈(Pnk)K we writeu∧c = ⟨u(ξ) × c(ξ)⟩ξ∈K.By Corollary 1D, we may find for each such u a wu < nk and a set Ju ⊆nk such that #(Ju) ≤rlk andeither there is a T ≼k Tu such that Hk(u∧c) = wu for every c ∈Tor for every z ∈(nk \ Ju)r there is a T ≼k Tu such that (Hk(u∧c), z) ∈˜Wk, that is,(Hk(u∧c), z(j)) ∈Wk ∀j < r,for every c ∈T .Set˜Ik = {wu : u ∈U(p2; Kk, k)},˜Jk = S{Ju : u ∈U(p2; Kk, k)},so that#(˜Ik) ≤#(U(p2; Kk, k)) ≤#(T Kkk) ≤(Qi

(d) We have

10p3 ⊩P σ(i) = Hi(v∗i ) ∀i < k1,where v∗i is that member of T Kii+1 such that v∗i (η) ≤v∗(η) for every η ∈Ki. Set L′k = {Hk(v∗k)} for k < k1,L′k = Lk \ ˜Jk for k ≥k1; then Qk∈N #(L′k)/nk > 0, because Qk∈N #(Lk)/nk > 0 and Pk∈N #( ˜Jk)/nk < ∞.So there is a ˜z ∈D ∩(Qk∈N L′k)r ⊆D′.

Writing zk = ⟨˜z(j)(k)⟩j

An easy induction on k shows thatT ′u = {c : u∧c ∈U(˜pk+1; Kk, k + 1)}whenever u ∈U(˜pk; Kk, k), k ≥k1, that ˜pk is (Kl, l)-fixed and (Kl, l + 1)-fixed whenever k1 ≤k ≤l, andthat ˜p(v)k= p(v)3whenever k1 ≤k ≤l and v ∈U(˜pk; Kl, l) ∪U(˜pk; Kl, l + 1). Also supp(˜pk) ⊆Sl∈N Kl forevery k ≥k1.

(f) It is likewise easy to see that, for k ≥k1,˜pk+1 ≤˜pk,˜pk+1↾ξ ⊩Pξ ˜pk+1(ξ) ∩Tk = ˜pk(ξ) ∩Tk ∀ξ < κ,˜pk+1↾ξ ⊩Pξ ˜pk+1(ξ) ∩suc(t) = ˜pk(ξ) ∩suc(t) ∀t ∈˜pk+1(ξ) ∩Tiunless i = k and ξ ∈Kk,˜pk+1 ⊩P σ(k) ∈˜Ik or (σ(k), zk) ∈˜Wk. (g) On the other hand, if k ≥k1 and ξ ∈Kk,˜pk+1↾ξ ⊩Pξ dp({C : t × C ∈˜pk+1(ξ)}) ≥dp({C : t × C ∈˜pk(ξ)})/2lk∀t ∈˜pk+1(ξ) ∩Tk.To see this, take any q ≤˜pk+1↾ξ and t such thatq ⊩Pξ t ∈˜pk+1(ξ) ∩Tk = ˜pk(ξ) ∩Tk.We may suppose that v0 ∈Sk is such that q ≤˜p(v0)k↾ξ = ˜p(v0)k+1↾ξ = q1.

Now ˜pk+1 is (Kk, k + 1)-fixed so theremust be a t′ ≥t such thatq1 ⊩Pξ t′ ∈Tk+1 ∩˜pk+1(ξ).There is accordingly a v1 ∈Sk such that v0↾ξ = v1↾ξ and v1(ξ) = t′. Express v1 as u∧c1 where u ∈U(˜pk; Kk, k) and c1 ∈T ′u.

Of course u(ξ) = t.Nowq1 ⊩Pξ {C : t × C ∈˜pk+1(ξ)} ⊇{c(ξ) : c ∈T ′u, c↾ξ = c1↾ξ},q1 ⊩Pξ {C : t × C ∈˜pk(ξ)} = {c(ξ) : c ∈Tu, c↾ξ = c1↾ξ}because ˜pk and ˜pk+1 are both (Kk, k + 1)-fixed, while v0↾ξ = (u↾ξ)∧(c1↾ξ). But because T ′u ≼k Tu,dp({c(ξ) : c ∈T ′u, c↾ξ = c1↾ξ}) ≥dp({c(ξ) : c ∈Tu, c↾ξ = c1↾ξ})/2lk.So we getq ≤q1 ⊩Pξ dp({C : t × C ∈˜pk+1(ξ)}) ≥dp({C : t × C ∈˜pk(ξ)})/2lk.As q and t are arbitrary, we have the result.

(h) Because γk ln(2lk) ≤2−k (by 1E(a)(v)),˜pk+1 ≤Kk,k ˜pkfor every k ≥k1. Also, supp(˜pk) ⊆Sl∈N Kl for every k ≥k1.

By Lemma 1N, there is a p4 ∈P such thatp4 ≤˜pk for every k ≥k1. Moreover, we may take it thatp4(0) = Tk≥k1 ˜pk(0)(as in Lemma 1H), so that

11δi(p4(0)) ≥δi(p3(0)) −γi ln(2li) ≥1whenever i ≥k1. Note that for k ≥k1,p4 ⊩P σ(k) ∈˜Ik or (σ(k), zk) ∈˜Wk,while for k < k1,p4 ⊩P (σ(k), zk) ∈˜Wk.

(i) Now define p5 ∈P by setting˜I′i = ˜Ii ∪{˜z(j)(i) : j < r} ∀i ∈N,p5(0) = {t : t ∈p4(0), Ci(t) ∩˜I′i = ∅whenever k1 ≤i < rank(t)},p5(ξ) = p4(ξ) if 0 < ξ < κ.Of course we must check that p5(0), as so defined, belongs to Q0 ∼= Q; but because δk(p4(0)) ≥1 for k ≥k1,we havedp({C : t × C ∈p4(0)}) ≥exp(1/γk) ≥2(Qi

(Here at last is the key step which depends on using dp in our measure of ‘rapidly branching’given in 1E(d).) Thus p5(0) ∈Q and p5 ∈P.

But alsop5 ⊩P Φk ∩˜I′k = ∅∀k ≥k1,so thatp5 ⊩P σ(k) /∈˜Ik, σ(k) ̸= ˜z(j)(k) ∀j < r, (σ(k), zk) ∈˜Wkfor k ≥k1; finallyp5 ⊩P (σ, ˜z) ∈⌜Rr⌝, σ ∈⌜Rr⌝−1[D′];as p5 ≤p and p, σ are arbitrary,11P ⊩P Ψ ∩Qk∈N Lk ⊆⌜Rr⌝−1[D′],as claimed.1S Theorem For each r ∈N,11P ⊩P if Di ⊆⌜X⌝and Di ∩Yk∈NLk ̸= ∅∀⟨Lk⟩k∈N ∈⌜L⌝, i < r,then ∀⟨Lk⟩k∈N ∈⌜L⌝∃⟨xi⟩i

Because members of L can be coded by simple sequences, we maysuppose that L is a Pα-name for some α < κ (1Je). The inductive hypothesis tells us that11P ⊩P ∀⟨Lk⟩k∈N ∈⌜L⌝∃⟨xi⟩i

. .⌝(β) to indicate that we are interpreting some formula in V Pβ.

12Now we remark that by 1Jb P can be regarded, for forcing purposes, as an iteration Pβ ∗P′, where P′ isa Pβ-name for a p.o.set with the same definition, interpreted in V Pβ, as P has in the ground model. So wemay use Lemma 1R in V Pβ to say that11Pβ ⊩Pβ11P′ ⊩P′ Ψ(β) ∩Q L ⊆⌜Rr⌝−1[∆∩(Q L)r],using the notation Ψ(β) to indicate which version of the P-name Ψ we are trying to use.

Moving to V P fora moment, we have ⌜µ⌝Ψ(β) = 1 and ⌜µ⌝(Q L) > 0, so11P ⊩P ∃l ∈N, ⌜µ⌝(Ψ(β)l∩Q L) > 0.Also, of course, every Ψ(β)l∩Q L can be regarded (in V P) as the product of a sequence belonging to ⌜L⌝.By the original hypothesis on ∆r,11P ⊩P ∆r ∩Ψ(β) ∩Q L ̸= ∅.We can therefore find a P-name σr for a member of ∆r ∩Ψ(β) ∩Q L, and now further Pβ-names σi, fori < r, such that11P ⊩P ⟨σi⟩i

We begin with some definitions and results taken from [Ta84].2A Definitions (a) Let (X, Σ, µ) be a probability space. Write L0 = L0(Σ) ⊆RX for the set of Σ-measurable real-valued functions on X.

Let Tp be the topology of pointwise convergence, the usual producttopology, on RX. Let Tm be the (non-Hausdorff, non-locally-convex) topology of convergence in measureon L0, defined by the pseudometricρ(f, g) =Rmin(|f(x) −g(x)|, 1)µ(dx)for f, g ∈L0.

(b) A set A ⊆RX is stable if whenever α < β in R, E ∈Σ and µE > 0 there are k, l ≥1 such thatµ∗k+l{(x, y) : x ∈Ek, y ∈El, ∃f ∈A, f(x(i)) ≤α & f(y(j)) ≥β ∀i < k, j < l}< (µE)k+l,writing µ∗k+l for the usual product outer measure on Xk × Xl. (See [Ta84], 9-1-1.

)2B Stable sets Suppose that (X, Σ, µ) is a probability space and that A ⊆RX is a stable set. (a) If (X, Σ, µ) is complete, then A ⊆L0(Σ).

([Ta84], §9.1. )(b) The Tp-closure of A in RX is stable.

(c) If A is bounded above and below by members of L0, its convex hull is stable ([Ta84], 11-2-1). (d) If A ⊆L0 (as in (a)), then Tm↾A, the subspace topology on A induced by Tm, is coarser than Tp↾A.

([Ta84], 9-5-2. )For more about stable sets, see [Ta84]and [Ta87].2C Pettis integration Let (X, Σ, µ) be a probability space and B a (real) Banach space.

(a) A function φ : X →B is scalarly measurable if gφ : X →R is Σ-measurable for every g ∈B∗, thecontinuous dual of B. (b) In this case, φ is Pettis integrable if there is a function θ : Σ →B such thatRE gφ dµ exists = g(θE) ∀E ∈Σ, g ∈B∗.

(c) If φ : X →B is bounded and scalarly measurable, thenA = {gφ : g ∈B∗, ∥g∥≤1} ⊆L0is Tp-compact. In this case φ is Pettis integrable ifff 7→RE f : A →R

13is Tp↾A-continuous for every E ∈Σ ([Ta84], 4-2-3). In particular (by 2B(d)) φ is Pettis integrable if A isstable.2D The rivals Write µL for Lebesgue measure on [0, 1], and ΣL for its domain.

Consider the followingtwo propositions:(*) [0, 1] is not the union of fewer than c closed negligible sets;(†) there are sequences ⟨nk⟩k∈N, ⟨Wk⟩k∈N such thatnk ≥2k, Wk ⊆nk × nk, #(Wk) ≤2−kn2k ∀k ∈N;taking X = Qk∈N nk, µ the usual Radon probability on X,R = {(x, y) : x, y ∈X, (x(k), y(k)) ∈Wk ∀k ∈N, {k : x(k) = y(k)} is finite},then whenever D ⊆X, µ∗D = 1 and r ∈N there are x0, . .

. , xr ∈D such that (xj, xi) ∈R wheneveri < j ≤r.Evidently (*) is a consequence of CH, while in the language of §1, 11P ⊩P (†), this being a slightly weakerversion of Theorem 1S.Thus both (*) and (†) are relatively consistent with ZFC.

Consequences of (*) are explored in [Ta84],where it is called Axiom L; we list a few of them in 2E below. Our purpose in this paper is to show that (†)leads to a somewhat different world.2E Theorem Assume (*).

Write L0 for L0(ΣL). (a) If A ⊆L0 is separable and compact for Tp, it is stable.

(b) If A ⊆L0 is separable and compact for Tp, its closed convex hull in R[0,1] lies within L0. (c) If A ⊆L0 is separable and compact for Tp, then Tm↾A is coarser than Tp↾A.

(d) If (Y, S, T, ν) is a separable compact Radon measure space and f : [0, 1] × Y →R is measurable inthe first variable and continuous in the second, then it is measurable for the (completed) product measureµL × ν. (e) If ⟨En⟩n∈N is a stochastically independent sequence of measurable subsets of [0, 1], with limn→∞µEn =0 but Pn∈N(µEn)k = ∞for every k ∈N, then there is an ultrafilter F on N such thatlimn→F En = {x : {n : x ∈En} ∈F}is non-measurable.proof (a) See [Ta84], 9-3-1(b).

(b) Use (a) and 2B(c). (c) Use (a) and 2B(d).

(d) Use (a) and [Ta84],10-2-1. (e) Observe that, writing χEn for the characteristic function of En, the set {χEn : n ∈N} is notstable, and use (a).2F Theorem Assume (†).

(a) There is a bounded Pettis integrable function φ : [0, 1] →ℓ∞such that {gφ : g ∈(ℓ∞)∗, ∥g∥≤1} isnot stable in L0(ΣL). (b) There is a separable convex Tp-compact subset of L0(ΣL) which is not stable.proof (We write ℓ∞for the Banach space of bounded real sequences.) Take ⟨nk⟩k∈N, ⟨Wk⟩k∈N, X, µ, Rfrom the statement of (†).

Because ([0, 1], µL) is isomorphic, as measure space, to (X, µ), we may work withX rather than with [0, 1]. Write Σ for the domain of µ, L0 = L0(Σ).

(a) For k ∈N writeIk = {I : I ⊆nk, #(I) ≤k, (i, j) /∈Wk for all distinct i, j ∈I},For k ∈N, I ⊆nk setHkI = {x : x ∈X, x(k) ∈I}.Let A be{χHkI : k ∈N, I ∈Ik},writing χH : X →{0, 1} for the characteristic function of H ⊆X; let Z be the Tp-closure of A inRX.Because A is uniformly bounded, Z is Tp-compact.For E ∈Σ define fE : Z →R by settingfE(h) =RE h(x)µ(dx) for h ∈A, fE(u) = 0 for u ∈Z \ A.Enumerate A as ⟨hm⟩m∈N, and defineφ : X →ℓ∞, θ : Σ →ℓ∞by settingφ(x)(m) = hm(x) ∀m ∈N, x ∈X,θ(E)(m) = RE hm(x)µ(dx) ∀m ∈N, E ∈Σ.We aim to show

14(i) that A is not stable;(ii) that if ν is a Radon probability on A′ = Z \ A then R u(x) ν(du) = 0 for µ-almost every x;(iii) fE : Z →R is continuous for every E ∈Σ;(iv) θ is the indefinite Pettis integral of φ, so that φ is Pettis integrable;(v) K = {gφ : g ∈C(Z)∗, ∥g∥≤1} includes A so is not stable.ad (i) Suppose that k, l ≥1. Take any m ≥l.

SetG = {y : y ∈Xl, ∃I ∈Im, y(j) ∈HmI ∀j < l}⊇{y : y ∈Xl, (y(i)(m), y(j)(m)) /∈Wm for distinct i, j < l}.Because #(Wm) ≤2−mn2m, µlG ≥(1 −2−m)l2. If y ∈G, set I = {y(j)(m) : j < l} ∈Im; thenµk{x : x ∈Xk, x(i)(m) /∈I ∀i < k} ≥(1 −n−1m l)k.So we conclude thatµk+l{(x, y) : x ∈Xk, y ∈Xl, ∃f ∈A, f(x(i)) = 0 ∀i < k, f(y(j)) = 1 ∀j < l}≥(1 −2−m)l2(1 −n−1m l)k(by Fubini’s theorem).

Because k, l and m are arbitrary, A cannot be stable.ad (ii) Because each Im is finite, any member of A′ must be of the form χE where E ⊆X andx ∈E, x′ ∈X, {k : x(k) ̸= x′(k)} is finite ⇒x′ ∈E.Note also that if x, y ∈E then (x, y) /∈R; because either x(k) = y(k) for infinitely many k, or there are k,I such that x(k) ̸= y(k), I ∈Ik and x, y both belong to HkI, in which case (x(k), y(k)) /∈Wk.Now let ν be a Radon probability on A′, and set w(x) =Ru(x) ν(du) for each x ∈X, so that w belongs tothe closed convex hull of A′ in RX. If x, x′ are two members of X differing on only finitely many coordinates,then u(x) = u(x′) for every u ∈A′; consequently w(x) = w(x′).

Also 0 ≤w(x) ≤1 for every x ∈X.Take δ > 0 and set D = {x : w(x) ≥δ}. By the zero-one law, µ∗D must be either 0 or 1.

Suppose,if possible, that µ∗D = 1. Let r ∈N be such that rδ ≥1.

By (†), there are x0, . .

. , xr ∈D such that(xj, xi) ∈R for i < j ≤r.

But in this case Pi≤r u(xi) ≤1 for every u ∈A′, while Pi≤r w(xi) ≥(r+1)δ > 1,and w cannot belong to the closed convex hull of A′.Accordingly µ∗D must be 0. As δ is arbitrary, w = 0 a.e.ad (iii) Because fE(χHkI) ≤kn−1kfor every I ∈Ik, limm→∞fE(hm) = 0 and fE is continuous.ad (iv) We need to show thatRE g(φ(x)) µ(dx) exists = g(θ(E)) ∀g ∈(ℓ∞)∗, E ∈Σ.It is enough to consider positive linear functionals g of norm 1.

For any such g we have a Radon probabilityν on Z such thatg(⟨f(hm)⟩m∈N) =RZ f(u) ν(du) for every f ∈C(Z),using the Riesz representation of positive linear functionals on C(Z). Set ǫm = ν{hm}, ǫ = 1 −Pm∈N ǫm =νA′.

Then we can find a Radon probability ν′ on A′ such thatg(⟨f(hm)⟩m∈N) = Pm∈N ǫmf(hm) + ǫRA′ f(u) ν′(du)for every f ∈C(Z). Now an easy calculation (using (ii)) shows thatg(θ(E)) = g(⟨fE(hm)⟩m∈N) = Pm∈N ǫmRE hm(x) µ(dx) =RE g(φ(x)) µ(dx)for every E ∈Σ.ad (v) If m ∈N then hm = emφ ∈K where em ∈(ℓ∞)∗is defined by setting em(z) = z(m) for everyz ∈ℓ∞.

This completes the proof. (b) The unit ball of (ℓ∞)∗is w∗-separable and its continuous image K ⊆L0 is separable; so K witnessesthe truth of (b).2G Further properties of the model Returning to 1R/1S, we see that the model of §1 has somefurther striking characteristics closely allied to, but not obviously derivable from, (†).

Consider for instance(‡) there is a closed negligible set Q ⊆[0, 1]2 such that whenever D ⊆[0, 1] and µ∗LD = 1 thenµLQ−1[D] = 1;(‡)′ there is a negligible set Q′ ⊆[0, 1]2 such that whenever C, D ⊆[0, 1] and (C × D) ∩Q′ = ∅thenone of C, D is negligible.

15Then 11P ⊩P (‡). For start by taking Q1 to be{(x, y) : x, y ∈X, (x(k), y(k)) ∈Wk ∀k ∈N},the closure of R in X × X.

Then the argument for 1S shows that11P ⊩P if D ⊆⌜X⌝and D ∩Yk∈NLk ̸= ∅∀⟨Lk⟩k∈N ∈⌜L⌝then ∃β < κ such that Ψ(β) ⊆⌜Q1⌝−1[D].Consequently11P ⊩P if D ⊆⌜X⌝and ⌜µ⌝∗D = 1 then ⌜µ⌝(⌜Q1⌝−1[D]) = 1.Accordingly we have in V P the version of (‡) in which ([0, 1], µL) is replaced by (X, µ). However there isnow a continuous inverse-measure-preserving function f : X →[0, 1], and takingQ = {(f(x), f(y)) : (x, y) ∈Q1}we obtain (‡) itself.

Evidently (‡) implies (‡)′, taking Q′ to be{(x + q, y + q′) : (x, y) ∈Q, q, q′ are rational} ∩[0, 1]2.Of course (∗) and (‡) are mutually incompatible (the argument for 2E(a) from (∗), greatly simplified,demolishes (‡) also). The weaker form (‡)′ is incompatible with CH or MA, but not with (∗), both (‡)′ and(∗) being true in Cohen’s original model of not-CH (see [Frp89]).2H Problems The remarkable results quoted in 2E depend on the identification of separable relativelypointwise compact sets with stable sets (‘Axiom F’ of [Ta84]).

In models satisfying (†), this identificationbreaks down. But our analysis does not seem to touch any of 2E(b)-(e).

We therefore spell out the obviousproblems still outstanding. Write L0 for L0(ΣL).

(a) Is it relatively consistent with ZFC to suppose that there is a separable Tp-compact set A ⊆L0 suchthat the closed convex hull of A in R[0,1] does not lie within L0? (b) Is it relatively consistent with ZFC to suppose that there is a separable Tp-compact set A ⊆L0 suchthat Tm↾A is not coarser than Tp↾A?

Does it make a difference if A is assumed to be convex? (This questionseems first to have been raised by J.Bourgain and F.Delbaen.

)(c) Is it relatively consistent with ZFC to suppose that there are a separable compact Radon measurespace (Y, S, T, ν) and a function f : [0, 1] × Y →R which is measurable in the first variable, continuous inthe second variable, but not jointly measurable for µL × ν? (d) Is it relatively consistent with ZFC to suppose that there is a stochastically independent sequence⟨En⟩n∈N in ΣL such that Pn∈N(µLEn)k = ∞for every k ∈N, but µL(limn→F En) = 0 for every non-principal ultrafilter F on N?

(This question is essentially due to W.Moran; see also [Ta84], 9-1-4 for anotherversion. )Here we note only that a positive answer to (a) would imply the same answer to (c), and that the word‘separable’ in (a)-(c) is necessary, as is shown by examples 3-2-3 and 10-1-1 in [Ta84].References[BJSp89] T.Bartoszy´nski, H.Judah & S.Shelah, ‘The Cicho´n diagram’, preprint, 1989 (MSRI 00626-90).

[Ba84] J.E.Baumgartner, ‘Applications of the proper forcing axiom’, pp. 913-959 in [KV84].

[Frp89] H.Friedman, ‘Rectangle inclusion problems’, Note of 9 October 1989. [Ku80] K.Kunen, Set Theory.

North-Holland, 1980. [KV84] K.Kunen & J.E.Vaughan (eds.

), Handbook of Set-Theoretic Topology. North-Holland, 1984.

[Sh82] S.Shelah, Proper Forcing. Springer, 1982 (Lecture Notes in Mathematics 940).

[Sh326] S.Shelah, ‘Vive la diff´erence!’, submitted for the proceedings of the set theory conference at MSRI,October 1989; notes of July 1987, preprint of October 1989; abbreviation ‘ShCBF’. [Sp87] J.Spencer, Ten Lectures on the Probabilistic Method, S.I.A.M., 1987.

[Ta84] M.Talagrand, Pettis integral and measure theory. Mem.

Amer. Math.

Soc. 307 (1984).

[Ta87] M.Talagrand, ‘The Glivenko-Cantelli problem’, Ann. of Probability 15 (1987) 837-870.Acknowledgements Part of the work of this paper was done while the authors were visiting the M.S.R.I.,Berkeley; we should like to thank the Institute for its support.

The first author was partially supported by

16the Fund for Basic Research of the Israel Academy of Sciences. The second author was partially supportedby grants GR/F/70730 and GR/F/31656 from the U.K. Science and Engineering Research Council.

We aremost grateful to M.Burke for carefully checking the manuscript.To appear in J.S.L.

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