Haar-positive closed subsets of Haar-positive analytic sets
aa r X i v : . [ m a t h . L O ] M a r HAAR-POSITIVE CLOSED SUBSETS OFHAAR-POSITIVE ANALYTIC SETS
M ´ARTON ELEKES, M ´ARK PO ´OR, AND ZOLT ´AN VIDNY ´ANSZKY
Abstract.
We show that every non-Haar-null analytic subset of Z ω contains a non-Haar-null closed subset. Moreover, we also provethat the codes of Haar-null analytic subsets, and, consequently,closed Haar-null sets in the Effros Borel space of Z ω form a ∆ set. It is not hard to see that non-locally-compact Polish groups do notadmit a Haar measure (that is, an invariant σ -finite Borel measure).However, Christensen [4] (and later, independently, Hunt-Sauer-Yorke[8]) generalized the ideal of Haar measure zero sets to every Polishgroup as follows: Definition 0.1.
Let ( G, · ) be a Polish group and S ⊂ G . We saythat S is Haar-null , (in symbols, S ∈ HN ) if there exists a universallymeasurable set U ⊃ S (that is, a set measurable with respect to everyBorel probability measure) and a Borel probability measure µ on G such that for every g, h ∈ G we have µ ( gU h ) = 0. Such a measure µ iscalled a witness measure for S .This notion has found wide application in diverse areas such as func-tional analysis, dynamical systems, group theory, geometric measuretheory, and analysis (see, e.g., [12, 2, 14, 13, 5, 1]). It provides awell-behaved notion of “almost every” (or “prevalent”) element of aPolish group. It is natural to investigate the regularity properties ofHaar-null sets. In particular, one might wonder whether “small setsare contained in nice small sets” and whether “large sets contain nicelarge sets”. Concerning the first question, Solecki [13] has shown a pos-itive statement, namely, that every analytic Haar-null set is containedin a Borel Haar-null set. On the negative side, the first and the thirdauthor [6] proved that, unlike the situation in locally-compact groups, All three authors were supported by the National Research, Development andInnovation Office – NKFIH, grants no. 124749, 129211. The first and third authorswere also supported by the National Research, Development and Innovation Office– NKFIH, grant no. 113047. The third author was also supported by FWF GrantsP29999, P28153, and M2779. in non-locally compact abelian Polish groups there are Borel Haar-nullsets that have no G δ Haar-null supersets.In this paper we address the second question, and answer it positivelyin the case of a concrete non-locally compact Polish group, Z ω , that is,the ω ’th power of the additive group of the integers: Theorem 0.2.
Every analytic non-Haar-null subset of Z ω contains aclosed non-Haar-null subset. Our proof is based on the results of Solecki [13] and Brendle-Hjorth-Spinas [3]. Roughly speaking, a theorem from the former paper allowsus to use witness measures of a very special form, and thus to reducethe understanding of the Haar-null ideal to the understanding of thenon-dominating ideal (see Section 1 for the definitions), while the lattercontains the regularity properties of the latter ideal. The reduction isbased on a coding map and utilizes a compactness argument.We also calculate the exact projective class of the codes of the an-alytic Haar-null subsets of Z ω , as well as the set { C ∈ F ( Z ω ) : C ∈HN } , which turn out to be ∆ .1. Preliminaries and basic facts
We start with the most important definitions and theorems that willbe used in the proof. We will adapt the notation from [9] for descriptiveset theoretic concepts.The following fact is just a trivial consequence of standard results.
Fact 1.1.
Assume that
X, Y are Polish spaces, F ⊂ X × Y is Borel, µ is a Borel probability measure on X , and K ⊂ proj X ( F ) is a compactset with µ ( K ) > . Then there exists a compact set K ⊂ F such that proj X ( K ) ⊂ K and µ (proj X ( K )) > .Proof. Using the Jankov, von-Neumann Uniformization theorem (see,[9, Theorem 18.1]) there exists a measurable function h : K → Y withgraph( h ) ⊂ F . Consequently, by Lusin’s theorem, there exists a com-pact set K ⊂ K with µ ( K ) > h ↾ K is continuous.But then K = graph( h ↾ K ) satisfies the required properties. (cid:3) If b ∈ ω ω let us denote by µ b the natural product probability measureon Q n ∈ ω [0 , b ( n )] ⊂ Z ω . For a Polish space X we will denote by K ( X )and F ( X ) the space of compact subsets of X with the Hausdorff metricand the space of closed subsets of X with the Effros Borel structure,respectively. The following, easy to prove statements will be used: Fact 1.2.
Let X be a Polish space and F ⊂ X be closed. Then AAR-POSITIVE CLOSED SUBSETS OF HAAR-POSITIVE ANALYTIC SETS 3 (1) the map ω ω → P ( Z ω ) (that is, the Polish space of the Borelprobability measures on Z ω ) defined by b µ b (2) the map ω ω × K ( Z ω ) → R defined by ( b, K ) µ b ( K ) (3) the map K ( X × ω ω ) → K ( X ) defined by K proj X ( K ) (4) the set { K ∈ K ( X ) : K ⊂ F } is Borel. For f, g ∈ ω ω we will write f ≤ ∗ g if f ( n ) ≤ g ( n ) holds for each n ∈ ω with finitely many exceptions. Recall that a set S ⊂ ω ω is dominating if it is cofinal in ≤ ∗ . The σ -ideal of non-dominating sets is denoted by N D .The following fact, essentially proved in [3], will play a crucial role.
Fact 1.3.
Assume that A ∈ Σ ( ω ω ) is a dominating set and f : A → ω ω is a Borel function. Then there exists a closed set C ⊂ A such that C
6∈ N D and f ↾ C is continuous.Sketch of the proof. Let W = { w σ , s σ : σ ∈ ω <ω } be a collection withthe following properties: w σ ⊂ ω and dom( s ∅ ) ⊂ ω are finite, s ∅ :dom( s ∅ ) → ω , and for σ = ∅ we have s σ : w σ ↾ lh ( σ ) − → ω and for all i ∈ w σ ↾ lh ( σ ) − s σ ( i ) > σ ( lh ( σ ) − x ∈ ω ω we havethat ω = dom( s ∅ ) ∪ S n w x ↾ n , where the union is disjoint. If T ⊂ ω <ω is a tree, define C T,W = { y ∈ ω ω : s ∅ ⊂ y and ∃ x ∈ [ T ] ∀ n ∈ ω ( y ↾ w x ↾ n = s x ↾ n +1 ) } . It is easy to see that the map φ T,W : [ T ] → C T,W defined by assigningto x ∈ [ T ] the unique y ∈ C T,W with the property that ∀ n ∈ ω ( y ↾ w x ↾ n = s x ↾ n +1 ) is continuous and open. Moreover, one can also checkthat the set C T,W is closed for every T and W .A Laver-tree is a subtree T of ω <ω such that it has a stem s (thatis, a maximal node with ∀ t ∈ T ( t ⊂ s ∨ s ⊂ t )), and for every t s from T the set { n ∈ ω : t ⌢ ( n ) ∈ T } is infinite.In [3] it is shown that A contains a set of the form C ω <ω ,W . Con-sider now the Borel map f ◦ φ ω <ω ,W : ω ω → ω ω . By [7, Example 3.7]there exists a Laver-tree T ⊂ ω <ω such that f ◦ φ ω <ω ,W ↾ [ T ] is con-tinuous. Clearly, φ T,W = φ ω <ω ,W ↾ [ T ], and since this map is open, f ↾ φ T,W ([ T ])(= C T,W ) is continuous. Thus, it is enough to check thatthe set C T,W is dominating.Let z ∈ ω ω be arbitrary. It is not hard to define an x ∈ T such that φ T,W ( x ) ≥ ∗ z inductively: indeed, for every large enough n (namely, if n > lh ( s ), where s is the stem of T ), if x ↾ n is defined, we can find an m so that m > max { z ( i ) : i ∈ w x ↾ n } and x ↾ n ⌢ ( m ) ∈ T . Then foran x obtained this way, it follows from ∀ i ∈ w x ↾ n ( m < s x ↾ n +1 ( i )) that M ´ARTON ELEKES, M ´ARK PO ´OR, AND ZOLT ´AN VIDNY ´ANSZKY whenever n is large enough and i ∈ w x ↾ n then φ T,W ( x )( i ) > z ( i ). Thefact that | w σ | < ℵ implies that φ T,W ( x ) dominates z . (cid:3) We will also make use of another consequence of the results in [3]:
Fact 1.4.
Let A ⊂ ω ω × ω ω be a Σ set. Then the set S N D = { x : A x ∈N D} is ∆ .Proof. Counting the quantifiers in the definition of non-dominating setsgives that the set S N D is Σ . Now by [3], x ∈ S N D holds if and only if ∀ C ∈ F ( ω ω ) ( C ∈ N D ∨ C A x ). As it has been noted by Solecki, itfollows from the construction in [3] that the set { C ∈ F ( ω ω ) : C ∈ N D} is ∆ , thus, a straightforward calculation yields that S N D is Π aswell. (cid:3) A connection between
N D and HN has already been established bySolecki [13]. We will use the following: Lemma 1.5.
Let S ⊂ Z ω be a Haar-null set. There exists a b ∈ ω ω such that for every b ′ ≥ ∗ b we have that µ b ′ is a witness for S ∈ HN . Let us remark first that this statement has been implicitly provedin [13] and used without proof in [2]. We will indicate how to show itusing a slightly different argument from [11].
Sketch of the proof.
It is not hard to see that in order to establish thelemma it suffices to produce a b such that for every b ′ ≥ b the measure µ b ′ is a witness for S ∈ HN . Now, one can check that in the proof of [11,Theorem 3.1] only lower bounds are imposed on the sequence ( N ( n )) n ∈ ω and consequently on the sequence ( a ( n )) n ∈ ω as well. Applying thisobservation and [11, Theorem 3.1] for a Borel Haar-null B ⊃ S , thechoice b = ( a ( n )) n ∈ ω yields the lemma. (cid:3) In this paper solely the group Z ω will be considered, and so we willuse the additive notation for the group operation.For an integer-valued function f : X → Z the notation | f | and cf will be used for the function defined by x
7→ | f ( x ) | and x cf ( x ) for x ∈ X and c ∈ Z . Also we will write f ≤ g if for each x ∈ X we have f ( x ) ≤ g ( x ). 2. Complexity estimation
As a warm up, we calculate the complexity of the codes of Haar-nullanalytic subsets of and the complexity of the closed Haar null subsetsof Z ω in the Effros Borel space. It has been shown by Solecki [13], seealso [10, 15], that the codes for the closed Haar-null subsets, as well asthe set { C ∈ F ( Z ω ) : C ∈ HN } are neither analytic nor co-analytic. AAR-POSITIVE CLOSED SUBSETS OF HAAR-POSITIVE ANALYTIC SETS 5 (In fact, Z ω can be replaced by any non-locally compact Polish groupadmitting a two-sided invariant metric). So the next result is sharp. Theorem 2.1.
The codes for Haar-null analytic subsets of Z ω form a ∆ subset, which is neither analytic nor co-analytic. More precisely, if U ⊂ ω ω × Z ω is a Σ set then the set S = { x ∈ ω ω : U x ∈ HN } is ∆ and there are closed sets U , for which this set is neither analytic norco-analytic. Moreover, the set { C ∈ F ( Z ω ) : C ∈ HN } is also ∆ .Proof. As mentioned above, by Solecki’s results it is enough to provethat S is ∆ . By Fact 1.4 it suffices to define a Σ subset A of ω ω × ω ω with the property that x ∈ S ⇐⇒ A x ∈ N D . Let( x, h ) ∈ A ⇐⇒ ∃ g ∈ Z ω ( µ h ( U x + g ) > . It follows from [9, Theorem 29.27] and Fact 1.2 that the set A is Σ .Let x ∈ ω ω be arbitrary and assume that U x
6∈ HN . We show thatin this case A x = ω ω . Indeed, for any h ∈ ω ω the condition U x
6∈ HN implies the existence of a g ∈ Z ω with µ h ( U x + g ) > U x ∈ HN , and towards a contradiction supposethat A x
6∈ N D . Then, we apply Lemma 1.5 to U x and get a b ∈ ω ω .Pick an h ∈ A x such that h ≥ ∗ b . Then on the one hand µ h shouldwitness that U x is Haar-null, on the other hand µ h ( U x + g ) > g ∈ Z ω , a contradiction.To see that the above argument implies that the set { C ∈ F ( Z ω ) : C ∈ HN } is ∆ , just fix a Borel isomorphism ι between ω ω and F ( Z ω ).It is straightforward to check that the set { ( C, h ) ∈ F ( Z ω ) × Z ω : h ∈ C } is Borel and so is the set B = { ( ι ( C ) , h ) ∈ F ( Z ω ) × Z ω : h ∈ C } .Now, using the first part of the statement, the set { x ∈ ω ω : ι − ( x ) ∈HN } = { x ∈ ω ω : B x ∈ HN } is ∆ , and, consequently, its pullbackunder ι , that is, the set { C ∈ F ( Z ω ) : C ∈ HN } is ∆ as well. (cid:3) The main result
In this section we prove Theorem 0.2. Let us start with an easyobservation.
Lemma 3.1.
Let f ∈ ω ω be arbitrary. Then the set H ( f ) = { g ∈ Z ω : ∃ ∞ n ∈ ω | g ( n ) | ≤ f ( n ) } is Haar-null in Z ω .Proof. Let f ′ ( n ) = 2 n ( f ( n ) + 1). We will show that the measure µ f ′ witnesses that H ( f ) is Haar-null. Let h ∈ Z ω be arbitrary. Clearly, H ( f ) + h = \ k ∈ ω [ n ≥ k { g + h : | g ( n ) | ≤ f ( n ) } , M ´ARTON ELEKES, M ´ARK PO ´OR, AND ZOLT ´AN VIDNY ´ANSZKY and for every n ∈ ω we have µ f ′ ( { g + h : | g ( n ) | ≤ f ( n ) } ) ≤ f ( n ) f ′ ( n ) ≤ n , Thus, using P n ∈ ω n < ∞ and the Borel-Cantelli lemma, we get that µ f ′ ( H ( f ) + h ) = 0. (cid:3) Now we are ready for the proof of the main result. Our strategywill be somewhat similar to the idea of the proof of Theorem 2.1, justsignificantly more sophisticated. To a given analytic set A
6∈ HN wewill assign a Borel set D that encodes the witnesses for A
6∈ HN ,i.e., codes for possible witness measures µ and compact sets K , andtranslations t ∈ Z ω with K + t ⊂ A and µ ( K ) >
0. The codingwill be constructed so that it ensures that D is dominating. Usingthe results of Brendle, Hjorth, and Spinas, we will chose a dominatingclosed subset of D with some additional properties, and from it a non-Haar-null subset of A will be reconstructed. A compactness argumentwill yield that this set is in fact closed. Proof of Theorem 0.2.
Let A ∈ Σ ( Z ω ) be a non-Haar null set and let F ∈ F ( Z ω × ω ω ) with proj Z ω ( F ) = A . Fix a Borel bijection ψ : 2 ω →K ( Z ω × ω ω ) and define a Borel partial mapping φ : ω ω × Z ω × Z ω →K ( Z ω × ω ω ) as follows: let ( b, t, c ) ∈ dom( φ ) iff the conjunction of thefollowing holds:(1) c − t ∈ ω .(2) 2 b ≤ ∗ | t | .(3) proj Z ω ( ψ ( c − t )) ⊂ Q n ∈ ω [0 , b ( n )].(4) µ b (proj Z ω ( ψ ( c − t ))) > p for the ( Z ω × ω ω ) × Z ω → Z ω × ω ω mappingthat is the translation of the first coordinate, i.e., ( r, x ) + p t = ( r + t, x ).Define φ for ( b, t, c ) ∈ dom( φ ) by letting φ ( b, t, c ) = ψ ( c − t ) + p t, in other words, φ ( b, t, c ) = ψ ( c − t )+ p t is the compact subset of Z ω × ω ω defined by { ( r, x ) + p t : ( r, x ) ∈ ψ ( c − t ) } .Finally, we will need a homeomorphism bij : ω ω → ω ω × Z ω × Z ω . Inorder to be precise, let us fix a concrete one by letting for every n ∈ ω • bij( f )(0)( n ) = f (3 n ) , • for i ∈ { , } define bij( f )( i )( n ) = ( f (3 n + i ) / , if f (3 n + i ) is even, − ( f (3 n + i ) + 1) / , if f (3 n + i ) is odd. AAR-POSITIVE CLOSED SUBSETS OF HAAR-POSITIVE ANALYTIC SETS 7
Lemma 3.2.
Let D = { f ∈ ω ω : φ (bij( f )) ⊂ F } . Then D is dominat-ing.Proof. Assume otherwise, and let f ∈ ω ω witness this fact. Withoutloss of generality, we can assume that f is constant on the sets of theform { n, n + 1 , n + 2 } , and it attains only positive values. Definean element f ′ ∈ Z ω by f ′ ( n ) = 2( f (3 n ) + 1).Using Lemma 3.1 and A
6∈ HN we get that A \ H (3 f ′ )
6∈ HN .This yields that there exist a compact set K ⊂ Q n [0 , f ′ ( n )] with µ f ′ ( K ) > t ∈ Z ω such that K + t ⊂ A \ H (3 f ′ ). Now,using Fact 1.1 for the sets F + p ( − t ), K and µ f ′ we get a compact set K ⊂ F + p ( − t ) (or, equivalently, K + p t ⊂ F ) with µ f ′ (proj Z ω ( K )) > Z ω ( K ) ⊂ K .Consider now the function g = bij − ( f ′ , t, t + ψ − ( K )) . We claimthat g ≥ ∗ f and g ∈ D , contradicting our initial assumption andthus finishing the proof. In order to see g ≥ ∗ f , notice that, as ∅ 6 = K + t ⊂ ( Q n [0 , f ′ ( n )] + t ) ∩ ( A \ H (3 f ′ )), necessarily f ′ + | t | ≥ ∗ f ′ ,so | t | ≥ ∗ f ′ . Then, it is straightforward to check from the definitionof bij that g ≥ ∗ f holds.Checking g ∈ D is just tracing back the definitions: clearly, bij( g ) =( f ′ , t, t + ψ − ( K )) ∈ dom( φ ) holds, as we have already seen (1), (2),and for (3), (4) note that ψ ( t + ψ − ( K ) − t ) = K and proj Z ω ( K ) ⊂ K .Finally, φ (( f ′ , t, t + ψ − ( K ))) = K + p t ⊂ F . (cid:3) Using Fact 1.2 we get that dom( φ ) is a Borel set, and as + p iscontinuous, φ is a Borel map. Moreover, using (4) from Fact 1.2, D must be Borel as well. Since non-dominating sets form a σ -ideal, bypassing to a dominating Borel subset of D , we can assume that thereis an n ∈ ω and a sequence ( β, τ, γ ) ∈ ω n × Z n × Z n such that foreach f ∈ D if bij( f ) = ( b, t, c ) then 2 b ( k ) ≤ | t ( k ) | for each k ≥ n andfor each k < n we have β ( k ) = b ( k ) , τ ( k ) = t ( k ) , γ ( k ) = c ( k ).Now Fact 1.3 implies the existence of a closed dominating set C ⊂ D such that φ ◦ bij ↾ C is continuous. We claim that the set C ′ =proj Z ω ( S x ∈ C φ (bij( x ))) is closed and non-Haar null, which finishes theproof, as it is clearly a subset of A .First, we show that the set is closed. Let r n ∈ C ′ with r n → r and assume that r n ∈ proj Z ω ( φ ( b n , t n , c n )), where bij − ( b n , t n , c n ) ∈ C .Then, r n ∈ proj Z ω ( φ ( b n , t n , c n )) = proj Z ω ( ψ ( c n − t n ) + p t n ) and by( b n , t n , c n ) ∈ dom( φ ) we get that proj Z ω ( ψ ( c n − t n )) ⊂ Q k [0 , b n ( k )]and by our assumptions on D we have 2 b n ( k ) ≤ | t n ( k ) | for k ≥ n .Then | r n ( k ) | ≥ | t n ( k ) | − | b n ( k ) | ≥ | t n ( k ) | / | r n ( k ) | + 1 ≥ max {| b n ( k ) | , | t n ( k ) | , | c n ( k ) |} . By our assumptions on the n -long ini-tial segments of the elements of bij( D ), and the convergence of r n we M ´ARTON ELEKES, M ´ARK PO ´OR, AND ZOLT ´AN VIDNY ´ANSZKY get that the sequence ( b n , t n , c n ) n ∈ ω must contain a convergent subse-quence, and, as bij is a homeomorphism, bij − ( b n , t n , c n ) contains such asubsequence as well. If f ∈ ω ω is its limit then of course f ∈ C , and thecontinuity of proj Z ω ◦ φ ◦ bij ↾ C yields that r ∈ proj Z ω ( φ (bij( f ))) ⊂ C ′ holds.Second, assume that C ′ is Haar-null. By Lemma 1.5 there exists a b ∈ ω ω such that for each b ′ ≥ ∗ b the measure µ b ′ witnesses C ′ ∈ HN . Sincethe set C is dominating, there exists an f ∈ C such that f (3 n ) ≥ b ( n )holds for every large enough n . Then if bij( f ) = ( b ′ , t ′ , c ′ ), by definition b ′ ( n ) = f (3 n ), so µ b ′ must witness that C ′ is Haar-null. On the otherhand, proj Z ω ( φ ( b ′ , t ′ , c ′ )) = proj Z ω ( ψ ( c ′ − t ′ ) + p t ′ ) = proj Z ω ( ψ ( c ′ − t ′ )) + t ′ ⊂ C ′ , so µ b ′ must witness that the set proj Z ω ( ψ ( c ′ − t ′ )) isHaar-null as well. But, ( b ′ , t ′ , c ′ ) ∈ dom( φ ) holds, so by (4) we have µ b ′ (proj Z ω ( ψ ( c ′ − t ′ ))) >
0, a contradiction. (cid:3)
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AAR-POSITIVE CLOSED SUBSETS OF HAAR-POSITIVE ANALYTIC SETS 9 [15] S. Todorˇcevi´c and Z. Vidny´anszky. A complexity problem for borel graphs. submitted, https: // arxiv. org/ abs/ 1710. 05079 . Alfr´ed R´enyi Institute of Mathematics, Hungarian Academy of Sci-ences, PO Box 127, 1364 Budapest, Hungary and E¨otv¨os Lor´and Uni-versity, Institute of Mathematics, P´azm´any P´eter s. 1/c, 1117 Bu-dapest, Hungary
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