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We give an example of a measurable set E ⊆R such that the set

arXiv:math/9211206v1 [math.LO] 25 Nov 1992Measurable RectanglesArnold W. Miller1Department of MathematicsUniversity of WisconsinMadison, WI 53706miller@math.wisc.eduOct 92, revised Nov 92Abstract.We give an example of a measurable set E ⊆R such that the setE′ = {(x, y) : x + y ∈E} is not in the σ-algebra generated bythe rectangles with measurable sides. We also prove a strongerresult that there exists an analytic (Σ11) set E such that E′ is notin the σ-algebra generated by rectangles whose horizontal side ismeasurable and vertical side is arbitrary.

The same results aretrue when measurable is replaced with property of Baire.The σ-algebra generated a family F of subsets of a set X is he smallestfamily containing F and closed under taking complements and countableunions.In Rao [12] it is shown that assuming the continuum hypothesisevery subset of the plane R2 is in the σ-algebra generated by the abstractrectangles, i.e. sets of the form A × B where A and B are arbitrary sets ofreals.

In Kunen [5] it is shown that it is relatively consistent with ZFC thatnot every subset of the plane is in the σ-algebra generated by the abstractrectangles. He shows that this is true in the Cohen real model.

It also followsfrom a result of Rothberger [14] that if for example 2ℵ0 = ℵ2 and 2ℵ1 = ℵω2,then not not every subset of the plane is in the σ-algebra generated by theabstract rectangles. For a proof of these results see Miller [11] remark 4 and5 page 180).A set is analytic or Σ11 iffit is the projection of a Borel set.

Answering aquestion of Ulam, Mansfield [7][8] showed that not every analytic subset ofthe plane is in the σ-algebra generated by the analytic rectangles. Note thata rectangle A × B ⊆R × R is analytic iffboth A and B are analytic.1Research partially supported by NSF grant DMS-9024788.1

He did this by showing that, in fact, any universal analytic set is not inthe σ-algebra generated by the rectangles with measurable sides. This doesthe trick because analytic sets are measurable (see Kuratowski [6]).

Thistheorem was also proved by Rao [13]. Their argument shows a little more sowe give it next.

A set U ⊆R2 is a universal analytic set iffit is analytic andfor every analytic set A ⊆R there exist a real x such thatA = Ux = {y : ⟨x, y⟩∈U}.Theorem 1 (Mansfield [8] and Rao [13]) Suppose U is a universal analyticset, then U is neither in the σ-algebra generated by rectangles of the formA × B with A ⊆R arbitrary and B ⊆R measurable; nor in the σ-algebragenerated by rectangles of the form A × B with A ⊆R arbitrary and B ⊆Rhaving the property of Baire.proof:For any set U in the σ-algebra generated by rectangles of the form A×Bwith A ⊆R arbitrary and B ⊆R measurable there is a countable family{An ×Bn : n ∈ω} such that each Bn is measurable and U is in the σ-algebragenerated by {An × Bn : n ∈ω}.Let Z be a measure zero set and Cn be Borel sets such that for everyn we have Bn∆Cn ⊆Z where ∆is the symmetric difference. Since Z isa measure zero set its complement must contain a perfect set P, i.e.

a sethomeomorphic to the Cantor space 2ω. Now for any real x and any set Vin the σ-algebra generated by {An × Bn : n ∈ω} we have that Vx ∩P isBorel.

This is proved by noting that it is trivial if V = An × Bn (since wehave P ∩Bn = P ∩Cn), and it is preserved when taking complements andcountable unions. But the set P being perfect must contain a subset A whichis analytic but not Borel.

Since U is universal for some real x we have thatUx = A and Ux ∩P is not Borel. A similar proof works for the σ-algebragenerated by sets of the form A×B where B has the property of Baire, sinceif B has the property of Baire, then for some G, an open set, B∆G is meager.■In Miller [10] it is shown that it is relatively consistent with ZFC that nouniversal analytic set is in the σ-algebra generated by the abstract rectangles,answering a question raised by Mansfield.2

James Kuelbs raised the following question2: If E ⊆R is measurable,then isE′ = {(x, y) : x + y ∈E}in the σ-algebra generated by the rectangles with measurable sides?One can think of E′ as a parallelogram tipped 45 degrees, so it is clearby rotation and dilation that E′ is measurable if E is. Note also that sinceE′ is the continuous preimage of E, if E is Borel then E′ is Borel also.We give a negative answer to Kuelbs’ question by showing:Theorem 2 For any set E ⊆R we have that E is Borel iffE′ is in theσ-algebra generated by rectangles of the form A × B where A and B aremeasurable.Similarly, E is Borel iffE′ is in the σ-algebra generated byrectangles of the form A × B where A and B have the property of Baire.proof:Suppose E′ = σ⟨Xn × Yn : n ∈ω⟩, where Xn and Yn are measurablefor every n. Here σ is a recipe which describes how a particular set is builtup (using countable intersections and complementation), i.e., it is the Borelcode of E′.

Since Xn and Yn are measurable we can obtain An and Bn Borelsets and Z a Borel set of measure zero such that:Xn∆An ⊆Z and Yn∆Bn ⊆Z for each n ∈ω.Claim. u ∈E iff∃x, y /∈Z (x + y = u and ⟨x, y⟩∈σ⟨An × Bn : n ∈ω⟩).The implication ←is clear because if x, y /∈Z, then since [x ∈Xn iffx ∈An] and [y ∈Yn iffy ∈Bn] we have that (x, y) ∈σ⟨An × Bn : n ∈ω⟩iff(x, y) ∈σ⟨Xn × Yn : n ∈ω⟩and hence u ∈E.The implication →is true because of the following argument.

Supposeu ∈E is given. Choose x /∈Z ∪(u −Z).

Since these are measure zero setsthis is easy to do. But now let y = u −x, then y /∈Z since x /∈u −Z.

Sinceu ∈E it must be that ⟨x, y⟩∈E′ and so ⟨x, y⟩∈σ⟨Xn × Yn : n ∈ω⟩andthus ⟨x, y⟩∈σ⟨An × Bn : n ∈ω⟩. This proves the Claim.By the Claim, E is the projection of a Borel set and hence analytic.

Butnote that (Ec)′ = (E′)c where Ec denotes the complement of E. It follows2I want to thank Walter Rudin for telling me about this question and also for encour-aging me to write up the solution.3

that Ec is also analytic and so by the classical theorem of Souslin, E is Borel(see Kuratowski [6]).A similar proof works for the Baire property since we need only that everyset with the property of Baire differs from some Borel set by a meager set.■This answers Kuelbs’ question since if E is analytic and not Borel (or anymeasurable set which is not Borel), then E′ is not in the σ-algebra generatedby rectangles with measurable sides. The argument also shows, for example,that a set E ⊆R is analytic iffit can be obtained by applying operation Ato the σ-algebra generated by the rectangles with measurable sides.Next we show that a slightly stronger result holds for the sets of the formE′.

The following lemma is the key.Lemma 3 There exists an analytic set E ⊆R such that for any Z whichhas measure zero or is meager there exists x ∈R such that E \ (x+ Z) is notBorel.proof:Note that we may construct two such sets, one for category and one formeasure, and then putting them into disjoint intervals and taking the unionwould suffice to prove the lemma.We first give the proof for category. We may assume that the set Z is thecountable union of compact nowhere dense sets.

This is because, if Z′ is anymeager set, then there is such a Z ⊇Z′. But now if E \ (x + Z) is not Borel,then neither is E \ (x + Z′) since E \ (x + Z) = (E \ (x + Z′)) \ (x + Z).It is a classical result that the set of irrationals is homeomorphic to theBaire space, ωω, which is the space of infinite sequences of integers with theproduct topology (see Kuratowski [6]).

Let h : R \ Q →ωω be a homeo-morphism. For f, g ∈ωω define f ≤∗g ifffor all but finitely many n ∈ωwe have f(n) ≤g(n).

It is not hard to see the for any countable union ofcompact sets F ⊆ωω there exists f ∈ωω such that F ⊆{g ∈ωω : g ≤∗f}.Therefore, if G ⊆R is a Gδ set (countable intersection of open sets) whichcontains the rationals, then R\G is a σ-compact subset of R\Q and thereforethere exists f ∈ωω such that for every g ≥∗f we have h−1(g) ∈G. (This isa trick going back at least to Rothberger [14]).4

For p : ωω →2ω (the parity function) byp(g)(n)(0if g(n) is even1if g(n) is oddLet E0 ⊆2ω be an analytic set which is not Borel. DefineE = {g ∈R \ Q : p(h(g)) ∈E0}.Now suppose that Z ⊆R is a meager set which is the countable union ofcompact sets and let G = R\ Z.

Let x0 ∈Tq∈Q(q −G) be arbitrary (this setis nonempty since each q −G is comeager), and note that Q ⊆(x0 + G).Hence there exists f ∈ωω such that for every g ∈ωω with g ≥∗f wehave h−1(g) ∈x0 + G. Without loss we may assume that for every n ∈ωthat f(n) is even. LetP = {g ∈ωω : ∀n g(n) = f(n) or g(n) = f(n + 1)}.Then P is homeomorphic to 2ω and Q = h−1(P) ⊆x+G and so Q∩(x+Z) =∅.

But clearly Q∩E is homeomorphic to E0 via p◦h and therefore E\(x+Z)cannot be Borel.Next we give the proof for measure. Here we use a coding technique dueto Bartoszynski and Judah [1] who used it to show that a dominating realfollowed by a random real gives a perfect set of random reals (Theorem 2.7[1]).

We begin by giving the proof in a slightly different situation, namely 2ωinstead of the reals and where + denotes pointwise addition modulo 2 on 2ωand the usual product measure on 2ω. Afterwards we will indicate how tomodify the proof for the reals with ordinary addition and Lebesgue measure.Define I ⊆2ω to be the set of all x ∈2ω which have infinitely manyones and infinitely many zeros.

It is easy to see that I is a Gδ set. Anyx ∈I can be regarded as a sequence of blocks of consecutive ones, i.e., blocksof consecutive ones each separated by blocks of one or more zeros.

Defineq : I →ωω by q(x)(n) is the length of the nth block of consecutive ones.As above let E0 ⊆2ω be an analytic set which is not Borel and let p(f) forf ∈ωω be the parity function. LetE = {x ∈I : p(q(x)) ∈E0}.5

We claim that this works, i.e. given a measure zero set Z ⊆2ω thereexists a real x such that E \ (x + Z) is not Borel.

Let dn for n ∈ω be thedominating sequence as given in the proof of Theorem 2.7 [1]. Without losswe may assume that for every n ∈ω that d2n+2 −d2n+1 is even if n is evenand odd if n is odd.According to Bartoszynski and Judah there exists sufficiently randomreals r, r′ ∈2ω such that following holds.Let r′′ be defined by r′′(n) =r′(n) + 1 mod 2 for each n. Define P to be the set of all x ∈2ω such that forevery n we have thatx ↾[d2n, d2n+1) = r ↾[d2n, d2n+1)and eitherx ↾[d2n+1, d2n+2) = r′ ↾[d2n+1, d2n+2)orx ↾[d2n+1, d2n+2) = r′′ ↾[d2n+1, d2n+2).The main difficulty of Bartoszynski and Judah’s proof is to show that theperfect set P is disjoint from Z.

Let x1 ∈2ω be defined byx1 ↾[d2n, d2n+1) = r ↾[d2n, d2n+1)andx1 ↾[d2n+1, d2n+2) = r′ ↾[d2n+1, d2n+2).Let Q ⊆2ω be the perfect set of all x ∈2ω such that x ↾[d2n, d2n+1) isconstantly zero and x ↾[d2n+1, d2n+2) is constantly zero or constantly one. Itthen follows that P = x1 + Q and that Q is disjoint from x1 + Z.

But E ∩Qis not Borel, because d2n+2 −d2n+1 is even if n is even and odd if n is oddand so it easy to see the E0 is coded into E ∩Q.Now we indicate how to modify the above proof so as to work for the realswith ordinary addition and Lebesgue measure. First we modify it to work forthe unit interval [0, 1] with ordinary addition modulo one.

Let s : 2ω →[0, 1]be the map defined bys(x) = Σ∞n=0x(n)2n+1.This map is continuous, measure preserving, and one-to-one except on count-ably many points where it is two-to-one. On the points x where it is two-to-one let us agree that s−1(x) denotes the preimage of x which is eventually6

zero. The main difficulty is that addition mod 1 in [0, 1] is quite differentthan point-wise addition modulo 2 in 2ω.

Define for x, y ∈2ω the operationx ⊕y to be s−1(s(x) + s(y)) where s(x) + s(y) is the ordinary addition in[0, 1] modulo 1. The operation ⊕just corresponds to a kind of pointwise ad-dition with carry.

Instead of r′′ being the complement of r′ as in the proof ofBartoszynski and Judah we will take a sparser translate. Let Q ⊆2ω be theset of x ∈2ω such that x(m) = 1 only if for some n we have m = d2n+2 −1,i.e., the last element of the interval [d2n+1, d2n+2).Note that the set of all r ∈2ω such that for all but finitely many n thereexists i ∈[dn, dn+1) such that r(i) = 0 has measure one.

Hence, by changingour dn if necessary we may assume that our random real r has the propertythat for every n there is an i ∈[d2n+1, d2n+2) such that r(i) = 0. This meansthat when we calculate r ⊕x for any x ∈Q the carry digit on each interval[d2n+1, d2n+2) does not propagate out of that interval.

Now by the argumentof Bartoszynski and Judah there exists r ∈2ω sufficiently random so thatr ⊕Q is disjoint from s−1(Z).We also use a different coding scheme. We may assume that for everyn that d2n+2 −1 is even for even n and odd for odd n. Let J be the setof all x ∈2ω such that there are infinitely many n with x(n) = 1.

Defineq : 2ω →2ω be defined by q(x) = y where {in : n ∈ω} lists in order all isuch that x(i) = 1 and y(n) = 1 iffin is even. Now letE = {x ∈[0, 1] : q(s−1(x)) ∈E0}.Let x = s(r) and note that (x+ E) \ Z is not Borel, since x+ s(Q) is disjointfrom Z.

Also note that we can assume r(0) = 0 and so x + s(Q) is the samewhether we do addition or addition modulo one. But E \ (−x + Z) is justthe translate of (x + E) \ Z via −x and so we are done.■Theorem 4 There exists E ⊆R which is analytic (hence measurable andhaving the property of Baire) such that E′ = {(x, y) : x + y ∈E} is not ineither the σ-algebra generated by rectangles of the form A×B with A arbitraryand B measureable, nor is it in the σ-algebra generated by rectangles of theform A × B with A arbitrary and B having the property of Baire.proof:7

Suppose for contradiction that E′ = σ⟨An × Bn : n ∈ω⟩where the Anare arbitrary and the Bn are measurable. Let Z be a measure zero Borelset and ˆBn be Borel such that Bn∆ˆBn ⊆Z for every n ∈ω.

Suppose thatE \ (x + Z) is not Borel. By translating this set by −x we must have that(−x+E)\Z is not Borel.

Define ˜Bn as follows. If −x ∈An let ˜Bn = ˆBn andif −x /∈An let ˜Bn = ∅.

Define C = σ⟨˜Bn : n ∈ω⟩, i.e., C has exactly thesame Borel code as E′ = σ⟨An×Bn : n ∈ω⟩except at the base we substitute˜Bn for An × Bn. Since each ˜Bn is Borel the set C is a Borel set.

Now for anyy /∈Z we have that1. y ∈(−x + E) iff2.

x + y ∈E iff3. (x, y) ∈E′ iff4.

(x, y) ∈σ⟨An × Bn : n ∈ω⟩iff5. (x, y) ∈σ⟨An × ˆBn : n ∈ω⟩iff6.

y ∈σ⟨˜Bn : n ∈ω⟩= C.(4) and (5) are equivalent because y /∈Z. (5) and (6) are proved equivalentby an easy induction on the Borel code σ.Consequently, for every y /∈Z we have that y ∈(−x + E) iffy ∈C.

Butthis means that (x + E) \ Z = C \ Z which contradicts the assumption that(−x + E) \ Z is not Borel (both C and Z are Borel). A similar proof worksfor the property of Baire case.■There is other work on measurable rectangles which does not seem tobe directly related to this.For example, Eggleston [3] proves that everysubset of the plane of positive measure contains a rectangle X × Y with Xuncountable (in fact perfect) and Y of positive measure.

Martin [9] gives ametamathematical proof of this result.Erdos and Stone [4] show that there exists Borel sets A and B such thatthe set A + B = {x + y : x ∈A, y ∈B} is not Borel.Friedman and Shelah (see Burke [2] or Steprans [15]) proved that in theCohen real model for any Fσ subset E of the plane, if E contains a rectangleof positive outer measure, then E contains a rectangle of positive measure.8

One corollary of this is that it is consistent that there is a subset of the planeof full measure which does not contain any rectangle A×B with both A andB having positive outer measure. To see this let E ⊆R be any meager Fσset with full measure.

Then E′ = {(x, y) : x+y ∈E} is a subset of the planeof full measure which is meager. It cannot contain a rectangle of positivemeasure A × B since by the classical theorem of Steinhaus the set A + Bwould contain an interval and hence E would not be meager.References[1] T.Bartoszynski and H.Judah, Jumping with random reals, Annals ofPure and Applied Logic, 48 (1990), 197-213.

[2] M.Burke, A theorem of Friedman on rectangle inclusion and its conse-quences, to appear. [3] H.Eggleston, Two measure properties of Cartesian product sets, Quar-terly Journal of Mathematics Oxford, 5 (1954), 108-115.

[4] P.Erdos and A.H.Stone, On the sum of two Borel sets, Proceedings ofthe American Mathematical Society, 25 (1969), 304-306. [5] K.Kunen, Inaccessible properties of cardinals, Doctoral Dissertation,Stanford University, 1968.

[6] K.Kuratowski, Topology, Vol I, Academic Press, New York 1966. [7] R.Mansfield, The solution of one of Ulam’s problems concerning analyticrectangles, in Axiomatic Set Theory, Proceedings of the Sympo-sium in Pure Mathematics, Vol.

8 part I, American MathematicalSociety, (1971) 241-245. [8] R.Mansfield, The solution of one of Ulam’s problems concerning analyticrectangles II, Proceedings of the American Mathematical Society, 26(1970), 539-540.

[9] D.A.Martin, The use of set-theoretic hypotheses in the study of measureand topology, in General Topology and Modern Analysis editedby Rao, McAuley, Academic Press (1981), 417-430.9

[10] A.W.Miller, On the length of Borel hierarchies, Annals of MathematicalLogic, 16 (1979), 233-267. [11] A.W.Miller, Generic Souslin sets, Pacific Journal of Mathematics, 97(1981), 171-181.

[12] B.V.Rao, On discrete Borel spaces and projective sets, Bulletin of theAmerican Mathematical Society, 75 (1969), 614-617. [13] B.V.Rao, Remarks on analytic sets, Fundamenta Mathematicae, 66(1970), 237-239.

[14] F.Rothberger, A remark on the existence of a denumerable base for afamily of functions, Canadian Journal of Mathematics, 4 (1952), 117-119. [15] J.Steprans, Combinatorial consequences of adding Cohen reals, to ap-pear.10

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