Andrzej Roslanowski and Saharon Shelah- Measured Creatures

download Andrzej Roslanowski and Saharon Shelah- Measured Creatures

of 35

Transcript of Andrzej Roslanowski and Saharon Shelah- Measured Creatures

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    1/35

    arXiv:math/0

    010070v3[math.L

    O]23Mar2003

    MEASURED CREATURES

    ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

    Abstract. We prove that two basic questions on outer measure are undecid-

    able. First we show that consistently

    every sup-measurable function f : R2 R is measurable.The interest in sup-measurable functions comes from differential equations and

    the question for which functions f : R2 R the Cauchy problem

    y = f(x, y), y(x0) = y0

    has a unique almost-everywhere solution in the class ACl(R) of locally abso-lutely continuous functions on R.

    Next we prove that consistently

    every function f : R R is continuous on some set of positive outerLebesgue measure.

    This says that in a strong sense the family of continuous functions (from thereals to the reals) is dense in the space of arbitrary such functions.

    For the proofs we discover and investigate a new family of nicely definableforcing notions (so indirectly we deal with nice ideals of subsets of the reals the two classical ones being the ideal of null sets and the ideal of meagre ones).

    Concerning the method, i.e., the development of a family of forcing notions(equivalently, nice ideals), the point is that whereas there are many such ob-

    jects close to the Cohen forcing (corresponding to the ideal of meagre sets),little has been known on the existence of relatives of the random real forcing(corresponding to the ideal of null sets), and we look exactly at such forcing

    notions.

    0. Introduction

    The present paper deals with two, as it occurs closely related, problems concern-ing real functions. The first one is the question if it is possible that all superpositionmeasurable functions are measurable.

    Definition 0.1. A function f : R2 R is superpositionmeasurable (in short:supmeasurable) if for every Lebesgue measurable function g : R R the super-position

    fg : R R : x f(x, g(x))

    is Lebesgue measurable.

    The interest in sup-measurable functions comes from differential equations and

    the question for which functions f : R2 R the Cauchy problemy = f(x, y), y(x0) = y0

    1991 Mathematics Subject Classification. Primary 03E35; Secondary 03E75, 28A20, 54H05.The first author thanks the Hebrew University of Jerusalem for support during his visits to

    Jerusalem and the KBN (Polish Committee of Scientific Research) for partial support through

    grant 2P03A03114.The research of the second author was partially supported by the Israel Science Foundation.

    Publication 736.

    1

    http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3http://arxiv.org/abs/math/0010070v3
  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    2/35

    2 ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

    has a unique almost-everywhere solution in the class ACl(R) of locally absolutelycontinuous functions on R. For the detailed discussion of this area we refer thereader to Balcerzak [2], Balcerzak and Ciesielski [3] and Kharazishvili [17]. Grandeand Lipinski [14] proved that, under CH, there is a non-measurable function whichis sup-measurable. Later the assumption of CH was weakened (see Balcerzak [2,Thm 2.1]), however the question if one can build a non-measurable sup-measurablefunction in ZFC remained open (it was formulated in Balcerzak [ 2, Problem 1.10]and Ciesielski [7, Problem 5], and implicitly in Kharazishvili [17, Remark 4]). Inthe third section we will answer this question by showing that, consistently, everysup-measurable function is Lebesgue measurable.

    Next we deal with von Weizsackers problem. It has enjoyed considerable popu-larity, and it has origins in measure theory and topology. In [25], von Weizsackernoted that if

    () non(N)def= min{|X| : X R has positive outer Lebesgue measure } = c,

    then

    () there is a function f : [0, 1] [0, 1] such that the graph of f is of (twodimensional) outer measure 1 but for every Borel function g : [0, 1] [0, 1]the set {x [0, 1] : f(x) = g(x)} is of measure zero.

    Then he showed that () implies

    () there is a countably generated algebra A containing Borel([0, 1]) suchthat the Lebesgue measure can be extended to A, but there is no extremalextension to A.

    So it was natural to ask if the statement in () can be proved in ZFC (i.e., withoutassuming ()). A way to formulate this question was to ask

    ()vW Is it consistent to suppose that for every function f : R R there is aBorel measurable function g : R R such that the set {x R : f(x) =g(x)} is not Lebesgue negligible ?

    One can arrive to question ()vW also from the topological side. In [6], Blumbergproved that if X is a separable complete metric space and f : X R, then thereexists a dense (but possibly countable) subset D of X such that the restrictionf D is continuous. This result has been generalized in many ways: by consideringfunctions on other topological spaces, or by aiming at getting a large set onwhich the function is continuous. For example, in the second direction, we mayrestrict ourselves to X = R and ask if above we may request that the set D isuncountable. That was answered by Abraham, Rubin and Shelah who showed in[1] that, consistently, every real function is continuous on an uncountable set. Thenext natural step is to ask if we can demand that the set D is of positive outermeasure, and this is von Weizsackers question ()vW. It appears in Fremlins list

    of problems as [9, Problem AR(a)] and in Ciesielski [7, Problem 1].We will answer question ()vW in affirmative in the fourth section. The respec-

    tive model is built by a small modification of the iteration used to deal with thesup-measurability problem (and, as a matter of fact, it may serve for both pur-poses). We do not know if () fails in our model (and the question if () isconsistent remains open).

    Let us note that the close relation of the two problems solved here is not verysurprising. Some connections were noticed already in Balcerzak and Ciesielski [3].

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    3/35

    MEASURED CREATURES 3

    Also, among others, these connections motivated the following strengthening of()vW:

    ()+vW Is it consistent that for every subset Y ofR of positive outer measure and

    every function f : Y R, there exists a set X Y of positive outermeasure such that f Y is continuous?

    However, as Fremlin points out, the answer to ()+vW is NO:

    Proposition 0.2 (Fremlin [10]). There are a set Y R of positive outer measureand a function f : Y R such that f X is not continuous for any X Y ofpositive outer measure.

    Proof. Recall that a Hausdorff space Z is universally negligible if there is no Borelprobability measure on Z that vanishes at singletons. By Grzegorek [15], there isa universally negligible set Z R of cardinality non(N) (see also [11, Volume IV,439E(c)]). Pick a non-null set Y R of size non(N) and fix a bijection f : Y Z.

    IfX Y is such that f X is continuous, then we may transport Borel measureson X to Z, and therefore X is universally negligible and thus Lebesgue negligible.(See also [11, Volume IV, 439C(f)].)

    The notion of sup-measurability has its category version (defined naturally byreplacing Lebesgue measurability by Baire property). It was investigated inE.Grande and Z.Grande [13], Balcerzak [2], and Ciesielski and Shelah [8]. Thelatter paper presents a model in which every Baire-sup-measurable function has theBaire property. Also von Weizsacker problem has its category counterpart whichwas answered in Shelah [23]. What is somewhat surprising, is that the models of[8] and [23] seem to be totally unrelated (while for the measure case presented herethe connection is striking). Moreover, neither the forcing used in [8] (based on theoracle-cc method of Shelah [24, Chapter IV]), nor the one applied in Shelah [23],

    are parallel to the method presented here.

    The present paper is a part of authors program to investigate the family offorcing notions with norms on possibilities. One of the points is that we knowmany forcing notions in the neighbourhood of the Cohen forcing notion (see, e.g.,Roslanowski and Shelah [20], [19]), but we have not known any relatives of therandom real forcing. In the present paper we further develop the theory of forcingnotions with norms on possibilities introducing measured creatures. This enrich-ment of the method of norms on possibilities creates a bridge between the forcingsof [21] and the random real forcing (including the latter in our framework), and wecome with bounding friends of the random forcing. Though they are not ccc,they do make random not so lonely.

    Our presentation is self-contained, and though we use the notation of [ 21], thetwo basic definitions we need from there are stated in somewhat restricted formbelow (in 0.3, 0.4). The general construction of forcing notions using measure (tree)creatures is presented in the first section, and only in the following section we definethe particular example that works for us. The forcing notion Qmt4 (K

    , , F)(defined in section 2) is the basic ingredient of our construction. The requiredmodels are obtained by CS iterations ofQmt4 (K

    , , F); in the fourth section wealso add in the iteration random reals (on a stationary set of coordinates).

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    4/35

    4 ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

    Let us point out that measured creatures presented here have their ccc relativewhich appeared in [19, 2.1] (and more general constructions will be presented in[18, Chapter 2]).

    Notation: Most of our notation is standard and compatible with that of classi-cal textbooks on Set Theory (like Bartoszynski and Judah [4]). However in forcingwe keep the convention that a stronger condition is the larger one.

    (1) R0 stands for the set of non-negative reals. For a real number r and a setA, the function with domain A and the constant value r will be denotedrA.

    (2) For two sequences , we write whenever is a proper initial segmentof , and when either or = . The length of a sequence isdenoted by lh().

    (3) A tree is a family T of finite sequences such that for some root(T) T wehave

    ( T)(root(T) ) and root(T) T T.

    (4) For a tree T, the family of all branches through T is denoted by [T], andwe let

    max(T)def= { T : there is no T such that }.

    If is a node in the tree T then

    succT() = { T : & lh() = lh() + 1} andT[] = { T : }.

    A set F T is a front of T if

    ( [T])(k )( k F).

    (5) The Cantor space 2 (the spaces of all functions from to 2) and the spacei

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    5/35

    MEASURED CREATURES 5

    such that nor R0, dis H(1) (i.e., dis is hereditarily countable), andfor some sequence

    i

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    6/35

    6 ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

    Basic Notation: In this section, H stands for a function with domain andsuch that (m )(|H(m)| 2). Moreover we demand H H(1) (i.e., H ishereditarily countable).

    Definition 1.1. (1) A measured (tree) creature for H is a pair (t, Ft) such thatt LTCR[H] and

    Ft : [0, 1]pos(t) [0, 1].

    (2) We say that (K, , F) is a measured tree creating triple for H if(a) (K, ) is a local treecreating pair for H,(b) F is a function with domain K, F : t Ft, such that (t, Ft) is a

    measured (tree) creature (for each t K).

    (3) If (K, , F) is as above, t K, X pos(t), and r : X [0, 1]X,then we define Ft(r : X) as Ft(r : pos(t)), where

    r

    = r if X,

    0 if pos(t) \ X.We think of Ft as a kind of averaging function. At the first reading the reader

    may think that pos(t) is finite and

    Ft(r : pos(t)) =

    {r : pos(t)}

    |pos(t)|.

    For this particular function, our construction results in the random real forcing.However in general our averaging function does not have to be additive (as long asit has the properties stated in 1.2 below), and the result is not the random forcing(and this is one of the points of our construction). Also having Ft depend on tallows us to cheat: if we do not like the results of our averaging we may passto a tree creature s (t) (dropping the norm a little) with possible better for usaveraging function Fs.

    Regarding the requirements of 1.2, note that they are meant to provide us withsome features of the Lebesgue measure, without imposing additivity on the aver-aging functions Ft (specifically see 1.2()).

    Definition 1.2. A measured tree creating triple (K, , F) is nice if for every t K:

    () if r : pos(t), r : pos(t) [0, 1], r r for all pos(t), then

    Ft(r : pos(t)) Ft(r : pos(t)),

    () if nor[t] > 1, {} = dom(val[t]), r , r0 , r1 [0, 1] (for pos(t)) are such

    that r0 + r1 r and Ft(r : pos(t)) 2

    2lh() , then there are realnumbers c0, c1 and tree creatures s0, s1 (t) such that

    c0 + c1 = (1 22lh() )Ft(r : pos(t))

    and() if < 2, c > 0, then nor[s] nor[t] 1, pos(s) { pos(t) :

    r > 0}, and

    Fs(r : pos(s)) c,

    () if b [0, 1] and r [0, 1] (for pos(t)), then

    Ft(b r : pos(t)) = b Ft(r : pos(t)),

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    7/35

    MEASURED CREATURES 7

    () if r : pos(t) [0, 1], > 0, then there are r > r (for pos(t))such that for each r : pos(t) [0, 1] satisfying r r

    < r

    (for

    pos(t)) we have

    Ft(r : pos(t)) < Ft(r : pos(t)) + .

    From now on (till the end of this section), let (K, , F) be a fixed strongly finitaryand nice measured tree creating triple for H. Note that then the condition (c)4 ofDefinition 0.4 is equivalent to

    (c)5 (k )(n)( Tp)(lh() n nor[t] k).

    Proposition 1.3. Lett K. Then:

    () If r = 0 for pos(t), then Ft(r : pos(t)) = 0.() If r : pos(t) [0, 1], > 0, then there are r < r (for pos(t))

    such that for each r : pos(t) [0, 1] satisfying r < r

    r (for

    pos(t)) we have

    Ft(r : pos(t)) < Ft(r : pos(t)).

    Proof. () Follows from 1.2() (take b = 0).() If Ft(r : pos(t)) , then any r < r (for pos(t)) work. So assume

    Ft(r : pos(t)) > and let b =Ft(r :pos(t))/2Ft(r :pos(t))

    . Then 0 < b < 1. For

    pos(t) put

    r =

    1 if r = 0,b r otherwise.

    We are going to show that these r s are as required. To this end suppose thatr : pos(t) [0, 1] is such that r

    < r

    r (for all pos(t)). Then also

    b r r (for pos(t)) and by 1.2(, ) we get

    Ft(r : pos(t)) Ft(b r : pos(t)) = b Ft(r : pos(t)) =

    Ft(r : pos(t)) /2 > Ft(r : pos(t)) .

    Definition 1.4. Let p = tp : Tp Qtree (K, ).

    (1) For a front A Tp of Tp, we let T[p,A] = { Tp : ( A)( )}.(2) Let A be a front of Tp and let f : A [0, 1]. By downward induction on

    T[p,A] we define a mapping fp,A : T[p,A] [0, 1] as follows:

    if A then fp,A() = f(),

    if fp,A() has been defined for all pos(tp), T[p,A] \ A, then

    we put fp,A() = Ftp(fp,A() : pos(t

    p)).

    (3) For Tp we define

    F

    p () = inf{fp[],A() : A is a front of (T

    p

    )[]

    and f = 1A},

    and we let F(p) = Fp (root(p)).(4) For e {, 4} we let

    Qmte (K, , F) = {p Qtreee (K, ) :

    F(p) > 0}.

    It is equipped with the partial order inherited from Qtreee (K, ).

    Proposition 1.5. Assume p Qtree (K, ) and A is a front of Tp.

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    8/35

    8 ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

    (1) If f0, f1 : A [0, 1] are such that f0() f1() for all A, then

    ( T[p,A])(f0p,A() f1p,A()).

    (2) If f0 : A [0, 1], b [0, 1], and f1() = b f0() (for A), then

    ( T[p,A])(f1p,A() = b f0p,A()).

    (3) If A is a front of Tp above A (that is, ( A)( A)( )) and

    T[p,A], then 1Ap,A()

    1Ap,A().

    Definition 1.6. Let p Qtree (K, , F).

    (1) A function : Tp [0, 1] is a semiFmeasure on p if

    ( Tp)

    () Ftp(() : pos(tp))

    .

    (2) If above the equality holds for each Tp, then is called an Fmeasure.

    Proposition 1.7. Letp Qtree (K, ).

    (1) If : Tp

    [0, 1] is a semiFmeasure on p, then for each Tp

    wehave () Fp ().

    (2) If there is a semiFmeasure on p such that (root(p)) > 0, then p Qmt (K, , F).

    (3) If p Qmt (K, , F), then the mapping Fp () : T

    p [0, 1] is anFmeasure on p.

    Lemma 1.8. Assume p Qmt (K, , F) and 0 < < 1. Then there is Tp such

    that Fp () 1 .

    Proof. Assume toward contradiction that Fp () < 1 for all Tp. Choose

    inductively fronts Ak of Tp such that

    A0 = {root(p)},

    ( Ak+1)( Ak)(

    ),

    1Ak+1p,Ak+1

    () < 1 for all Ak.

    Note that then (by 1.5(1,2)) for each k < we have

    1Ak+1p,Ak+1

    (root(p)) (1 )k+1.

    Since the right hand side of the inequality above approaches 0 (as k ), we getan immediate contradiction with the demand F(p) > 0.

    Definition 1.9. A condition p Qmt (K, , F) is called normal if for every Tp

    we have Fp () > 0. We say that p is special if for every Tp we have Fp ()

    22lh()+1

    .

    Proposition 1.10. (1) Special conditions are dense inQmt4 (K, , F). (So also

    normal conditions are dense.)(2) If p is normal, and A is a front of Tp, then F(p) = fp,A(root(p)), where

    f() = Fp () (for A).

    Proof. 1) Let p Qmt4 (K, , F); clearly we may assume that nor[tp] > 1 for all

    Tp. Also we may assume that F(p) > 3/4 (remember 1.8) and lh(root(p)) > 4.

    Fix Tp such that Fp () 22lh() for a moment. Let 1 < a < 2. For each

    pos(tp) pick a front A of (Tp)[] such that

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    9/35

    MEASURED CREATURES 9

    if Fp () < 22lh()+1 , then

    1Ap[],A

    () < 22lh()+1

    ,

    if Fp () 22lh()+1 , then

    1Ap[],A

    () < aFp ().

    Let X0 = { pos(tp) : Fp () < 2

    2lh()+1}, X1 = pos(tp) \ X0, r = 1Ap[],A

    (),

    and

    r =

    r if X,0 if X1.

    Apply 1.2() for tp, r0 , r

    1 , r (note that Ftp(r : r pos(t

    p))

    Fp () 2

    2lh())

    to pick sa0, sa1 (t

    p) and c

    a0 , c

    a1 such that

    ca0 + ca1 = (1 2

    2lh())Ftp (r : pos(tp)),

    and

    ()a if < 2, ca > 0, then nor[sa ] nor[t

    p] 1, pos(s

    a ) X, and

    Fsa

    (r : pos(sa )) c

    a .

    Note that, if ca0 > 0, then ca0 Fsa0 (r : pos(sa )) 2

    2lh()+1 , and thus

    ca1 (1 22lh())Ftp (r : pos(t

    p)) 2

    2lh()+1 > 0.

    Also, letting r = min{aFp (), 1},

    Fsa1 (r : pos(sa1 )) Fsa1 (r

    : pos(s

    a1 )) a Fsa1 (

    F

    p () : pos(sa1 )).

    Together

    ()a (1 22lh()

    )Ftp(r : pos(tp)) 2

    2lh()+1 a Fsa1 (Fp () : pos(s

    a1 )).

    Since (K, ) is strongly finitary, considering a 1 (and using 1.2()), we find

    s (tp) such that nor[s] nor[tp]1 and

    Fp () 2

    2lh()+1 for all pos(s),and

    Fp () = Ftp (Fp () : pos(tp)) Fs (

    F

    p () : pos(s)) + 22lh()+1

    1 22lh().

    Note that also, as 22lh()

    Fp (),

    Fp ()(1 22lh()) 22

    lh()+1

    Fp ()(1 212lh()),

    so

    () Fp () (1 212lh()) Fs(

    Fp () : pos(s)).

    Now, starting with root(p), build a tree S and a system q = s : S suchthat succS() = pos(s). It should be clear that in this way we will get a conditionin Qtree4 (K, ) (stronger than p). Why is q in Q

    mt4 (K, , F)? Let k

    > lh(root(q)),A = { S : lh() = k} and f = 1A. Using (), we may show by downwardinduction that for every T[q, A] we have

    fq,A() Fp ()

    k1k=lh()

    (1 212k

    ) Fp () (1 222lh())

    22lh()

    (1 222lh()

    ) 22lh()+1

    .

    Now we may easily conclude that q Qmt4 (K, , F) is special.

    2) Let A be a front of Tp, p normal (so, in particular, Fp () > 0 for A). Fixa > 1 for a moment.

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    10/35

    10 ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

    For each A pick a front A of (Tp)[] such that 1Ap,A

    () < a Fp (). Let

    B = AA and f() = Fp () for A. By downward induction one can show

    that for all T[p,A] we have 1Bp,B() a fp,A(). Then, in particular, we have

    F(p) 1Bp,B(root(p)) a fp,A(root(p)),

    and hence (letting a 1) F(p) fp,A(root(p)). The reverse inequality is eveneasier (remember 1.5(1)).

    Lemma 1.11. Let p Qmt4 (K, ) be a normal condition such that F(p) > 12 ,

    nor[tp] > 2 for all Tp, and let k0 = lh(root(p)) > 4, 0 < 2(1+k0). Suppose

    that B is an antichain of Tp, and that for each B we are given a normalcondition q p[] such that

    root(q) = and F(q) 1 .

    Then at least one of the following conditions holds.

    (i) There is a normal condition q Qmt4 (K, , F) such that

    q p, root(q) = root(p), and Tq B = .

    (ii) There is a normal condition q Qmt4 (K, , F) such that q p, root(q) = root(p), F(q) (1 2k0)F(p), and Tq B is a front of Tq, and q[] = q for Tq B, and if Tq, B, then nor[tq] nor[t

    p] 2.

    Proof. Let e = 212

    (for < ); note that (e)2 = 2e+1.Fix k > lh(root(p)) for a while. Let A be a front of Tp such that

    { B : lh() k} A and ( A)( / B lh() = k).

    By downward induction, for each T[p,A], we define r0 , r1 [0, 1] and s

    0 , s

    1

    (tp) such that

    () If A B, then r0 = 0, r1 =

    F(q).() If A \ B, then r0 =

    Fp (), r

    1 = 0.

    () If T[p,A] \ A, lh() = m, then:

    if Fp () (1 ) k1=m

    (1 3e) < em, then r0 = r1 = 0,

    else r0 + r1

    Fp () (1 )

    k1=m

    (1 3e).

    Clauses (), () define r0 , r1 for A; s

    0 , s

    1 are not defined then (or are arbitrary).

    Suppose T[p,A] \ A, lh() = k 1. If Fp () (1 ) (1 3ek1) < ek1,then we let r0 = r

    1 = 0 (and s

    0, s

    1 are not defined). So assume now that

    Fp () (1 ) (1 3ek1) ek1.

    Then also (as r0 + r1

    Fp () (1 ))

    Ftp(r0 + r

    1 : pos(t

    p))

    Fp () (1 )

    Fp () (1 ) (1 3ek1) ek1 > 22k1 ,

    and we may apply 1.2() to pick r0, r1 and s

    0, s

    1 (t

    p) such that

    (i) r0 + r1 (1 ek1) Ftp (r

    0 + r

    1 : pos(t

    p))

    Fp () (1 ) (1 3ek1),

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    11/35

    MEASURED CREATURES 11

    (ii) if r > 0, < 2, then nor[s] nor[t

    p] 1, pos(s

    ) { pos(t

    p) : r

    >

    0}, and Fs(r : pos(s

    )) r

    .

    Suppose now that T[p,A] \ A, lh() = m 1 < k 1, and r0 , r1 have beendefined for all pos(tp) (and they satisfy clause ()). If

    Fp () (1 ) k1

    =m1

    (1 3e) < em1,

    then we let r0 = r1 = 0 (and s

    0, s

    1 are not defined). So assume

    Fp () (1 ) k1

    =m1

    (1 3e) em1.

    Then for pos(tp) we let

    r =

    r0 + r1 if r0 + r1 > 0,em otherwise,

    and we note that

    Ftp (r : pos(t

    p))

    Fp () (1 )

    k1=m

    (1 3e) em1 > 22m1.

    Applying 1.2() choose t0, t1 (tp) and c0, c1 such that c0+c1 (1em1)Ftp (r :

    pos(tp)) and

    if c0 > 0, then pos(t0) { pos(tp) : r0 + r

    1 = 0}, nor[t

    0] nor[tp] 1and Ft0(r

    : pos(t

    0)) c0, if c1 > 0, then pos(t1) { pos(tp) : r

    0 + r

    1 > 0}, nor[t

    1] nor[tp] 1

    and Ft1(r : pos(t1

    )) c1.Now look at the definition of r . If c0 > 0, then Ft0(r

    : pos(t

    0)) em, soc0 em (em1)2. Therefore

    Ft1(r : pos(t

    1)) c1 (1 em1) Fp () (1 ) k1=m

    (1 3e) em

    (1 em1)Fp ()(1 ) k1=m

    (1 3e) em1Fp ()(1 ) k1=m

    (1 3e) =

    Fp ()(1 ) k1=m

    (1 3e) (1 2em1) em1 (1 2em1) > 22m1

    .

    Hence we may apply 1.2() again and get r0, r1 and s

    0, s

    1 (t

    1) (tp) suchthat

    r0 + r1 (1 em1) Ft1(r : pos(t1))

    Fp ()(1 ) k1=m

    (1 3e) (1 2em1) (1 em1)

    Fp () (1 ) k1

    =m1(1 3e),

    and if r > 0, < 2, then pos(s) { pos(t

    p) : r

    > 0}, nor[s

    ] nor[t

    p] 2

    and Fs(r : pos(s

    )) r

    .

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    12/35

    12 ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

    Note that (as k0 > 4)

    +

    k1=k0

    3e

    1

    2k0+1 + 6

    =k0

    1

    22

    3

    2k0+2 .

    Therefore,

    Fp (root(p)) k1=k0

    (1 3e) (1 ) Fp (root(p)) (1 ( +k1=k0

    3e))

    Fp (root(p)) (1 3

    2k0+2) > 12

    2932 > ek0 .

    Hence also (by ())

    Fp (root(p))(1 3

    2k0+2) Fp (root(p))

    k1=k0

    (1 3e) (1 ) r0root(p) + r

    1root(p).

    Now, ifrroot(p) > 0, < 2, then we build inductively a finite tree Sk T[p,A] as fol-

    lows. We declare that root(Sk ) = root(p), s,kroot(p) = sroot(p), and succSk (root(p)) =

    pos(s,kroot(p)). If we have decided that Sk , / A (and r

    > 0), then we also

    declare s,k = s, succSk

    () = pos(s,k ) (note r

    > 0 for pos(s

    ,k )).

    Then, ifSk0 is defined, Sk0 B = , and, if S

    k1 is defined, S

    k1 A B. Also, if we

    extend Sk0 using p[] (for Sk0 A), then we get a condition q

    k0 p such that

    F(qk0 ) r0root(p)

    def= r0,k. Likewise, if we extend Sk1 using q (for S

    k1 A),

    then we get a condition qk1 p such that F(qk1 ) r

    1root(p)

    def= r1,k.

    If for some k > lh(root(p)) we have r1,k (1 2k0)F(p), then we use therespective condition qk1 to witness the demand (ii) of the lemma. So assume thatfor each k > lh(root(p)) we have r1,k < (1 2k0)F(p), and thus

    r0,k > (1 32k0+2

    )F(p) (1 12k0

    )F(p) = 12k0

    F(p) > 0.

    Apply the Konig Lemma to find an infinite set I \ (k0 + 1) such that for allk, k, k I, k < k < k , we have

    ( Sk

    0 )(lh() k Sk

    0 & s0,k

    = s0,k

    ).

    Then Sq = { : (k I)( Sk0 )}, sq = s

    0,k (for sufficiently large k I)

    determine a condition q witnessing the first assertion of the lemma.

    Lemma 1.12. Assume that is aQmt4 (K, , F)name for an ordinal, n m < and p Qmt4 (K, , F) is a normal condition such that

    F(p) > 12 , and nor[tp] >

    n + 2 for Tp. Letk0 = lh(root(p)) > 2. Then there is a normal conditionq Qmt4 (K, , F) such that

    (a) q p, root(q) = root(p), F(q) (1 2k0)F(p), and(b) ( Tq)(nor[tq] n), and(c) there is a front A of Tq such that for every A:

    the condition q[] forces a value to , Fq () >

    12 , lh() > k0,

    if Tq, then nor[tq] m.

    Proof. Let B consist of all Tp such that

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    13/35

    MEASURED CREATURES 13

    () lh() > k0 and there is a normal condition q Qmt4 (K, , F) stronger thanp[] and such that root(q) = , F(q) (1 2(2+k0)), ( Tq)(nor[tq]

    m), and for some front A of T

    q

    , for every A:() Fq () > 1/2 and the condition q[] decides the value of ,

    and() no initial segment of has the property stated in () above.

    Note that B is an antichain of Tp, and B Tp

    = for every condition p psuch that root(p) = root(p) (by 1.8). For each B fix a condition q witnessingclause () (for ). Now apply 1.11: case (i) there is not possible by what we statedabove, so we get a condition q as described in 1.11(ii). It should be clear that it isas required here.

    Theorem 1.13. Suppose that p Qmt4 (K, , F), and n are Qmt4 (K, , F)names

    for ordinals (n < ). Then there are a condition q p and fronts An of Tq (forn < ) such that for each n < and An, the condition q[] decides the value

    of n.

    Proof. We may assume that p is normal, k0 = lh(root(p)) > 2, F(p) >12 , and

    nor[tp] > 3 for Tp. We build inductively a sequence qn, An : n < such that

    (1) qn Qmt4 (K, , F) is a normal condition, root(qn) = root(p), qn qn+1,q0 = p,

    (2) An Tqn+1 is a front of Tqn+1, ( An)( An+1)( ),(3) if An, then Fqn+1() >

    12 , and for each T

    qn+1 such that we

    have nor[tqn+1 ] n + 4,

    (4) if root(p) An, then tqn+1 = t

    qn+2 ,

    (5) for each An, the condition (qn+1)[] decides the value of n,

    (6) F(qn+1) k0+n

    =k0

    (1 2) F(p).

    The construction can be carried out by 1.12 (q1, A0 are obtained by applying 1.12to p and 0; ifqn+1, An have been defined, then we apply 1.12 to n+1 and (qn+1)[]

    for An; remember 1.5). Next define q = tq : Tq so that root(q) = root(p),

    each An is a front of Tq, and if root(p) An then tq = tqn+1 . It is

    straightforward to check that q is as required in 1.13.

    Corollary 1.14. Let (K, , F) be a strongly finitary nice measured tree creatingtriple. Then the forcing notionQmt4 (K, , F) is proper and

    bounding.

    Proposition 1.15. If (K, , F) is a strongly finitary nice measured tree creatingtriple, then the forcing notionQmt4 (K, , F) is explicitly 0snep (see [22, Defini-tions 1.5, 1.8].

    Proof. Should be clear at the moment; compare the proof of [22, Lemma 3.1].

    2. The Forcing

    In this section we define a nice, strongly finitary measured tree creating triple(K, , F), and we show several technical properties of it and of the forcing notionQmt4 (K

    , , F). This forcing will be used in the next two sections to show ourmain results 3.2 and 4.14.

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    14/35

    14 ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

    Let k < . Fix a function k : such that

    k(0) = 2k+4 and k(i + 1) > 2

    2k+3 + k(i) +22k+7

    log2(1 + 222k+7).

    Let Nk = 21+log2(k(k+1)) (where r is the integer part of the real number r),

    and let H(k) = 2Nk .

    Let K consist of tree creatures t LTCR[H] such that

    dis[t] = (kt, t, nt, gt, Pt), where nt kt < , t i 1, andg

    is a partial function fromNktto 2 such that g gt and |g \ gt| 2kt+3. Furthermore, suppose that r [0, 1](for pos(t)) are such that

    22kt+3

    2|g|Nkt

    {r : pos(t) & g

    (kt)}def= a.

    Then there is s (t) such that

    () nor[s] = nor[t] 1, g gs,

    () Fs (r : pos(s)) a (1 22kt+3),

    () ifh is a partial function fromNks to 2 such thatgs h and|h\gs| 2ks+3,

    then {r : pos(s) & h (ks)}

    2Nks|h|

    is in the interval [F

    s (r : pos(s)), F

    s (r : pos(s)) (1 + 22k+3

    )].Proof. Let k = kt, n = nt.

    We try to choose inductively partial functions g from Nk to 2 such that

    (a) g = g0 g1 . . ., |g \ g| 2k+3,

    (b) 2|g|Nk

    {r : pos(t) & g (k)} a (1 + 22

    2k+7

    ).

    Note that in (b), the left hand side expression is not more than 1, so if the inequality

    holds, then (as a 22k+3

    )

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    15/35

    MEASURED CREATURES 15

    () 2k+3

    log2(1 + 222k+7)

    .

    Consequently, in the procedure described above, we are stuck at some 0 satisfying(). Let

    gs = g0 , ns = n 1, ks = k, s = t, Ps = {f Pt : g0 f}.

    So this defines s, but we have to check that s K. For this note that

    |gs| |g| + 0 2

    k+3 k(k n) + 2k+3 +

    22k+6

    log2(1 + 222k+7)

    k(k ns).

    (So indeed s K, and plainly s (t).) Also note that

    2|gs|Nk

    {r : pos(s)} a (1 + 222k+7)0

    def= a a.

    Now, suppose that u Nk \ dom(gs), |u| 2k+3. Let h : u 2. We cannot use

    gs

    h as g0+1, so the condition (b)0+1 fails for it. Therefore

    bhdef= 2|gs|+|h|Nk

    {r : pos(t) & gsh (k)} 22k+3.

    For < 2 let a = 2|gt|Nk

    {r : pos(t)}.

    First, we consider the case when both a0 and a1 are not smaller than 22k+3 .

    Then we may apply 2.1 and get s0, s1 (t) such that nor[s] = nor[t] 1,pos(s) { pos(t) : r > 0} and

    cdef= Fs(r

    : pos(s)) a (1 2

    2k+3).

    Then

    c0 + c1 (a0 + a1) (1 22k+3) Ft (r : pos(t)) (1 2

    2k),

    and we are done.

    So suppose now that a < 22k+3

    . Then

    a1 2|gt|Nk

    {r : pos(t)} 2

    2k+3 22k

    22k+3

    22k+3

    ,

    and using 2.1 we find s1 (t) such that nor[s1] = nor[t] 1, pos(s1) { pos(t) : r1 > 0}, and

    c1def= Fs1(r

    1 : pos(s1)) a1 (1 2

    2k+3)

    (Ft (r : pos(t)) 22k+3) (1 22

    k+3

    )

    Ft (r : pos(t)) (1 22k) + 22

    k

    (22k

    22k+3

    ) 22k+3

    + 22k+4

    =

    Ft (r : pos(t)) (1 22k) + 22

    k+1

    292k

    22k+3

    + 22k+4

    Ft (r : pos(t)) (1 22k).

    The following lemma and the proposition are, as a matter of fact, included in2.5, 2.6. However, we decided that 2.3 and 2.4 could be a good warm-up, and alsowe will use their proofs later.

    Lemma 2.3. Assume that:

    (i) t K, nor[t] > 1, k = kt, [0, 1],

    (ii) r : pos(t) [0, 1], a = Ft (r : pos(t)), a 262k ,

    (iii) Y is a finite set,(iv) for pos(t), u is a function from Y to [0, 1] such that

    r |Y| {u(y) : y Y},(v) for y Y we let

    u(y) = sup{b : there is s (t) such that nor[s] nor[t] 1 andb Fs (u(y) : pos(s))}.

    Then

    a (1 22k

    )

    {u(y) : y Y}

    |Y|.

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    17/35

    MEASURED CREATURES 17

    Proof. Let k = kt, N = Nkt , g = gt.First note that

    a = Ft (r : pos(t)) 2|g|N

    {r : pos(t)}

    2|g|N 1 1|Y|

    pos(t)

    yY

    u(y) =1

    1|Y|

    yY

    2|g|N

    pos(t)

    u(y)

    .

    Let Cdef= {y Y : 2|g|N

    pos(t)

    u(y) 22k+3

    }. For each y C we may use 2.1

    to pick sy (t) such that nor[sy] nor[t] 1 and

    Fsy (u(y) : pos(sy)) 2|g|N

    pos(t)

    u(y) (1 22k+3).

    Hence,

    a 1 |Y\C||Y| 2

    2k+3 + 1 1|Y|

    yC

    Fs (u(y):pos(sy))

    122k+3

    1 2

    2k+3 + 1 1|Y|

    1

    122k+3yC

    u(y).

    Consequently,

    (a 22k+3

    )(1 22k+3

    )

    yC

    u(y)

    |Y|

    yY

    u(y)

    |Y|,

    and hence

    a(1 22k

    )

    yY

    u(y)

    |Y|.

    Proposition 2.4. The forcing notionQmt4 (K, , F) preserves outer (Lebesgue)

    measure.

    Proof. Assume that A i 2 for

    all Tq

    , and F

    (q) >1

    2 ,() for some

    j

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    18/35

    18 ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

    Fix i < for a moment, and let Yi = {y

    j 262k . Also note that

    () u(y) defined as in 2.3(v) is u(y).

    [Why? First suppose that u(y) < u(y). By the definition ofu we may find q q

    such that root(q) = , nor[tq

    ] nor[tq ] 1 for T

    q , and q y T, andF

    (q) > u(y). Note that F

    q () u(y) for all pos(tq

    ), and thus

    u(y) < F

    (q) = Ftq

    (F

    q () : pos(tq

    )) F

    tq

    (u(y) : pos(tq

    )).

    By the definition of u(y), the last expression is u(y), a contradiction. Nowsuppose u(y) > u(y). Take s (tq) such that nor[s] nor[t

    q] 1 and

    Fs (u(y) : pos(s)) > u(y); clearly we may request that u(y) > 0 for pos(s). Let z < u(y) (for pos(s)) be positive numbers such that ifz r u(y) for pos(s), then Fs (r : pos(s)) > u(y) (compare1.3). Pick conditions q such that

    F

    (q) > z , q as in definition of u(y), and

    let q be such that root(q) = , tq

    = s, and (q)[] = q for pos(s). Then

    F

    (q) > u(y) giving an easy contradiction.]Thus we get

    7

    8 F

    q () (1 22ki1) |Yi|

    {u(y) : y Yi},

    as required.

    Now suppose k0 k = lh() < ki 1, and we have proved the assertion of theclaim for all pos(tq). We again apply 2.3, this time to =

    78

    ki1=k+1

    (1 22

    ),

    and tq, u , r = F

    q () (for pos(tq)) and Yi. We note that

    78

    ki1=k+1

    (1 22

    ) Ftq (r : pos(tq)) =

    78

    ki1=k+1

    (1 22

    ) F

    q ()

    78 (1 2

    12k+1) 22k+1

    262k

    ,

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    19/35

    MEASURED CREATURES 19

    so the assumptions of 2.3 are satisfied. Therefore we may conclude that

    7

    8

    ki1=k+1

    (1 22

    ) F

    q () (1 22k

    ) |Yi|

    {u(y) : y Yi},

    as needed.

    Applying 2.4.1 to = root(q) we get

    7

    8ki1=k0

    (1 22

    ) F

    (q)

    uroot(q)(y) : y Yi

    |Yi|

    ,

    and hence 34 F

    (q) |Yi|

    {uroot(q)(y) : y Yi}. Then necessarily

    1

    4|Yi| |{y Yi : uroot(p)(y)

    1

    4F

    (q)}|

    (remember

    F

    (q) >

    1

    2 ). Let Zi = {y Yi : uroot(p)(y)

    1

    4

    F

    (q)}.Look at the set {x

    i18

    F

    (q), and ( Tqi)(nor[tqi ] nor[t

    q] 1), and

    qi x (n + i) T.

    By Konig Lemma (remember (K, ) is strongly finitary) we find an infinite setI such that for each i < j0 < j1 from I we have

    Tqj0 k 1, [0, 1],

    (ii) r : pos(t) [0, 1], a = Ft (r : pos(t)), a 262k ,

    (iii) Y is a finite set, Y = Y Nk,(iv) for pos(t), u is a function from Y to [0, 1] such that

    r |Y|

    {u(y) : y Y},

    (v) for y = (y0, y1) Y Nk and < 2 we let

    u(y, ) = sup{b : there is s (t) such that nor[s] nor[t] 1 and( pos(s))((k)(y1) = ) and b Fs (u(y) : pos(s))}.

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    20/35

    20 ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

    Then

    a (1 22k

    ) 1

    2 |Y| {u(y, ) : y Y & < 2}.Proof. Let k = kt, N = Nk, g = gt. Note that

    a = Ft (r : pos(t)) 2|g|N

    {r : pos(t)}

    2|g|N 1 1|Y|

    pos(t)

    1,

    a(1 22k

    ) 1

    |Y 2|

    (y,)Y2

    u(y, ).

    Let W be the canonical Qmt4 (K, , F)name for the generic real (so W is a

    name for a function ini

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    21/35

    MEASURED CREATURES 21

    Proposition 2.6. Suppose that A i

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    22/35

    22 ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

    Proof of the claim. The proof, by downward induction on , is similar to that of2.4.1, but this time we use 2.5.

    First note that if k = ki, then our assertion is exactly what is stated in ().

    So suppose that T[p,Fi], lh() < ki, and that we have proved our claim for

    all pos(tp). We are going to apply 2.5 to t = tp, =

    78

    ki+11=k+1

    (1 22

    ),

    Y = {y (ki \ {k}) : y Yi } 2[k + 1, ki) (and Y = Y Nk being interpreted

    as Yi 2[k + 1, ki)), and r = F

    p (), and u(y, ) = u(y, ) (for pos(t

    p),

    (y, ) Xi), so we have to check the assumptions there. Note that (as p is special)

    Ft (r : pos(t)) = F

    p () 7

    8

    ki+11=k+1

    (1 22

    ) 22k+1

    > 262k

    (so the demand in 2.5(ii) is satisfied). Also, by the inductive hypothesis,

    |Y

    Nk| r

    {u(y, ) : (y, ) X

    i

    }

    (so 2.5(iv) holds). Finally note that if (y, ) Yi 2[k + 1, ki), < 2, and

    : [k, ki) 2 is such that (k) = , [k + 1, ki) = , then u(y,,) definedby 2.5(v) is u(y,

    ).So, by 2.5, we may conclude that

    78

    ki+11=k+1

    (1 22

    ) F

    p () (1 22k) 2 |Yi | 2

    kik1 {u(y,

    ) : (y, ) Xi},

    as needed.

    In particular, it follows from 2.6.1 that

    78

    ki+11=k0

    (1 22

    ) F

    (p)

    {uroot(p)(y, ) : (y, ) Yi 2[k0, ki)}|Yi | 2

    kik0,

    and hence

    3

    4 F

    (p)

    {uroot(p)(y, ) : (y, ) Y

    i 2

    [k0, ki)}

    |Yi | 2kik0

    .

    Let :j

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    23/35

    MEASURED CREATURES 23

    is a Borel set of positive (Lebesgue) measure, so we may choose (x0, x1) A suchthat for infinitely many i < we have (x0ki, x1ki) Z

    +i . For each such i pick a

    condition qi

    p such that

    root(qi) = root(p), F

    (qi) >18

    F

    (p), and ( Tqi)(nor[tqi ] nor[t

    p] 1), and

    qi x0 ki T and

    j [k0, ki)

    W(j)

    x0(j)

    = x1(j)

    .

    By Konig Lemma, we may find a condition q Qmt4 (K, , F) stronger than p,

    and an infinite set I such that

    () if i < j are from I, then i + 1 < j and

    Tqj

    k

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    24/35

    24 ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

    if B C is Borel, then B has measure 0 if and only if its image [B] hasmeasure zero.

    Now note that the set [A] has outer measure 1 and inner measure 0 (ink

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    25/35

    MEASURED CREATURES 25

    Remark 3.3. Note that for the iteration P, Q : < 2 to work for the proofof 3.2 we do not need that all iterands are Qmt4 (K

    , , F). It is enough thatfor some stationary set S { <

    2: cf() =

    1}, for every S, we have

    P Q = Qmt4 (K

    , , F), and that the forcings used in the iteration are suchthat each P2/P+1 preserves non-nullity of sets from V

    P+1 . So, in particular, wemay use in the iteration also other (s)nep forcing notions preserving old reals arenot null. This will be used in the next section, where we will add the randomforcing here-and-there.

    4. Possibly every real function is continuous on a non-null set

    The aim of this section is to show that a slight modification of the iteration fromthe previous section results in a model in which every function f : R R agreeswith a continuous function on a set of positive outer measure. Let us start with areduction of the problem that exposes relevance of the tools developed earlier.

    Proposition 4.1. Assume:(a) the condition ()3sup of 3.1 holds true,

    (b) for every function f : 2 2 there are functions f1, f2 and a set Asuch that

    A 2 and f1 : A 2 is such that the set

    {(x, f1(x)) : x A} 2 2

    has positive outer measure, f2 : 2 2 2 is Borel, and (x A)(f(x) = f2(x, f1(x))).

    Then for every function f : R R there is a continuous function g : R Rsuch that the set {x R : f(x) = g(x)} has positive outer measure.

    Proof. Assume f : R R. Let : R 2

    be a Borel isomorphism preservingnull sets (see, e.g., [16, Thm 17.41]), and let f = f 1. Let f1, f2, A begiven by the assumption (b) for f. Put A = {(x, f1(x)) : x A} 2 2. Weknow that A is a non-null set (and consequently it is non-measurable), so applying()3sup we may pick a Borel function g0 : 2

    2 such that the set

    Bdef= {x A : f1(x) = g0(x)}

    has positive outer measure, and so does 1[B]. Let g1 : R R be defined by

    g1(x) = 1

    f2

    (x), g0((x))

    .

    Clearly g1 is Borel and for each x 1[B] we have g1(x) = f(x). Finally, usingLusins theorem (see, e.g., [16, Thm 17.12]) we may pick a continuous function

    g :R

    R

    such that the set {x 1

    [B] : g1(x) = g(x)} is not null (just take gso that it agrees with g1 on a set of large enough measure).

    The iteration of 3.2 will be changed by adding random reals on a stationaryset. So just for uniformity of our notation we represent the random real forcingas Qmt4 (K

    r, r, Fr). Let Hr(i) = 2 (for i < ). Let Kr consist of tree creaturest LTCR[Hr] such that

    dis[t] = (kt, t, Pt), where kt < , t i

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    26/35

    26 ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

    nor[t] = kt, val[t] = {t, : t ikt

    Hr(i) & (kt) Pt}.

    The operation r is trivial:

    r(t) = {s Kr : s = t & Ps Pt}.

    For t Kr and a sequence r : pos(t) [0, 1] we let

    Frt (r : pos(t)) =

    {r : pos(t)}

    2.

    It is easy to check that (Kr, r, Fr) is a (nice) measured tree creating triple forHr, and that the forcing notion Qmt4 (K

    r, r, Fr) is (equivalent to) the random realforcing.

    Like in 3.2, we start with universe V satisfying CH. Let Z { < 2 : cf() =1} be a stationary set such that { < 2 : cf() = 1} \ Z is stationary as well.Let Q = P

    , Q

    : <

    2 be countable support iteration such that

    if Z, then P Q = Qmt4 (K

    r, r, Fr),

    if 2 \ Z, then P Q = Qmt4 (K

    , , F).

    We are going to show that

    P2 every real function is continuous on a non-null set ,

    and for this we will show that the assumptions of 4.1 are satisfied in VP2 . First notethat P2 ()

    3sup (see 3.3; remember 3.1). To show that, in V

    P2 , the assumption(b) of 4.1 holds, we need to analyze conditions and continuous reading of names inthe iteration.

    Definition 4.2. Let (K, , F) be a measured tree creating triple for H (say, either(K, , F) defined in the second section, or (Kr, r, Fr) defined above).

    (1) A finite candidate for (K, , F) (or just for (K, )) is a system s = s : S\ max(S) such that

    S n

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    27/35

    MEASURED CREATURES 27

    such that() w is a finite non-empty set of ordinals below 2, w = {0, . . . , n}

    (the increasing enumeration);let xi be r if i Z, and xi be if i 2 \ Z,

    () k : w , c = c0 , . . . , cn, Y = Y0 , . . . , Yn (we treat c, Y asfunctions with domain w),

    () c0 FC(Kx0 , x0), ht(c0) = k(0), Y0 = {s : s max(c0)},

    and for 0 < i n:() ci : Yi1 FC(K

    xi , xi) is such that ht(ci()) = k(i) for each Yi1 ,Yi = {

    i : = 0 , . . . , i1 Yi1 & i max(ci())}.(We think of elements ofYi as functions from {0, . . . , i} with valuesbeing sequences in appropriate

    j is such that wt and k < is such that k is thepredecessor of in w

    t, then for every 0 , . . . , 1 Yt

    1 wehave

    i kt(i) : i < & i wt Ytk and

    ct(i kt(i) : i < & i wt) end ct

    (0 , . . . , 1).

    (3) For an ordinal < 2 and a finite pre-template t we define the restriction

    t = t of t in a natural way: wt

    = wt , kt

    = kt wt

    , ct

    = ct wt

    and Yt

    = Yt wt

    . (Note that t t.)(4) We say that finite pre-templates t, t are isomorphic if |wt| = |wt

    |, and ifh : wt wt

    is the order preserving isomorphism, then h[wt Z] = wt

    Z, and kt = kt

    h, ct = ct

    h, and Yt = Yt

    h.We also may say that h is an isomorphism from t to t.

    Definition 4.4. By induction on n = |wt| 1 we define

    (a) when a condition p P2 obeys a pre-template t, and(b) if wt = {0, . . . , n}, = 0 , . . . , n Y

    t, and p P2 obeys t, then

    we define a condition p[t,] P2 stronger that p.

    First consider the case when n = 0. Let t be a pre-template such that wt = {0}and let p P2. We say that p obeys t if

    p 0 P0 p(0) end extends the candidate ct

    0 .

    If p obeys t as above, and = 0 Yt0 , then p

    [t,] is defined as follows:

    p[t,] (2 \ {0}) = p (2 \ {0}), and p[t,] 0 P0 p

    [t,](0) = (p(0))[0 ] .

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    28/35

    28 ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

    Now, suppose that wt = {0, . . . , n} (the increasing enumeration; n > 0), andthat we have dealt with n 1 already. We say that a condition p P2 obeys t if

    p obeys t n, and for every = 0 , . . . , n1 Ytn1 , the condition p

    [tn,] n forces

    (in Pn) that p(n) endextends the candidate ctn().

    In that case we also define p[t,] for = 0 , . . . , n Ytn :

    p[t,] 2 \ {n} = p[tn,n] 2 \ {n}, p[t,] n Pn p

    [t,](n) = (p(n))[n ] .

    Definition 4.5. (1) A weak template is a increasing sequence t = tn : n < of finite pre-templates such that

    ( n

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    29/35

    MEASURED CREATURES 29

    proper+bounding in [24, Ch. VI] or [12]. (Of course, we use here Lemma1.12.

    Note that there are weak templates t such that no condition p P2 obeys t there could be a problem with norms and/or measures! From all weak templateswe will select only those which correspond to conditions in P2 (and they will becalled just templates; see 4.9 below).

    Definition 4.8. (1) A cover for a condition p Qtree (K, ) is a condition

    q Qtree (K, ) defined so that root(p) = root(q), q p and:

    if Tp, k = lh(), then nor[tq] = nor[tp], gtq = gtp , and

    Ptq = {f H(k) : gtq f},

    if / Tp, k = lh(), then gtq = , Ptq = H(k) and nor[tq ] = k.

    (2) Let p Qtree (K, ), and let q be the cover of p, and assume that Tq is

    a perfect tree. The covering mapping for p is the natural homeomorphism

    hp : [Tq] 2, defined as follows. First we define a mapping hp : Tq 2< : we let hp(root(Tq)) = . Suppose that hp() has been defined, Tq, and say hp() 2n, n < . We note that |pos(tq)| is a power of 2,

    and thus we may pick k > n such that |pos(tq)| = |2[n, k)|. Now, hp maps

    pos(tq) onto { 2k : hp(tq) } (preserving some fixed well-ordering of

    H(1)). Finally we let hp() =n

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    30/35

    30 ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

    condition t(z), and let h : [Tq] 2 be the covering mapping for t(z) (see

    4.8). Put t+1(z) = t(z)

    h(z).

    If + 1 < t, then we also define t

    +1(z

    ) as a condition in Qtree (K

    x+1,

    x+1)

    such that

    if n < , wtn = {0 , . . . , m} (the increasing enumeration), and = + 1, m, then t+1(z

    ) end extends ctn(z0 k

    tn(0), . . . , z1

    ktn(1)).

    Suppose now that t is a limit ordinal, and that we have defined Zt ,

    t and

    t for < . We put

    Zt = {z : < : ( < )(z : < Zt

    )}

    (again, above, like before and later, we ignore the first term whenever consid-

    ering elements of Zt). The mapping t : Z

    t (2

    ) is such that t(z) =

    t

    (z ) (for z Z

    t

    ). Also if, additionally, < t, then for z = z : < Z

    t

    we let t(z) be an element ofQ

    tree (K

    x , x) such that

    if n < , wtn = {0 , . . . , m} (the increasing enumeration), and = , m, then t(z) end extends c

    tn

    (z0 ktn(0), . . . , z1 k

    tn(1)).

    Definition 4.9. Let t be a weak template, and wt = : < t be the increasingenumeration. Also for < t let x be r if Z, and if / Z. We say that tis a template if for every < t and z Z

    t we have

    t(z) Qmt4 (K

    x , x , Fx ).

    Proposition 4.10. (1) Assume that p P2. Then there are a conditionq P2 and a template t such that q p obeys t, t < 1, and

    for some enumerationn

    : n < of t we have:() for every n < and z Ztn ,

    F(tn(z))

    1 2n10

    ,

    where F is suitably Fr or F.[If a template t satisfies () for an enumeration = n : n < of t,then we we will say that t behaves well for .]

    (2) For every template t, there is a condition p P2 which obeys t.

    For a countable ordinal , the space (2) is equipped with the product measuremLeb of countably many copies of 2. We will use the same notation mLeb for thismeasure in various products (and related spaces), hoping that no real confusion iscaused.

    Lemma 4.11. Let < 1. Suppose that C (2) is a closed set of positiveLebesgue measure. Then there is a closed set C C of positive Lebesgue measuresuch that for each < :

    ()C for every y (2), the set

    (C)ydef= {y (2)[, ) : yy C}

    is either empty or has positive Lebesgue measure (in (2)[, )).

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    31/35

    MEASURED CREATURES 31

    Proof. For a set X (2), < , and y (2) we let

    (X)ydef= {y (2)[, ) : yy X}.

    We may assume that (otherwise the lemma is easier and actually includedin this case). Fix an enumeration = {n : n < } such that 0 = 0, and let

    e0 = 24 mLeb(C), en+1 = 2

    6(n+2)2 (en)n+2.

    We are going to define inductively a decreasing sequence Cn : n < of closed(non-empty) subsets of C such that C0 = C1 = C and

    () for each m < n and y (2)m we have

    either (Cn)y = or mLeb((Cn)y) em

    1

    n=m+2

    4

    .

    (Note that () implies mLeb(Cn) e0 (1 n=2

    4).)

    Suppose that Cn has been defined already, n 1. Let { : } enumeratethe set

    {m : m n & m n}

    in the increasing order. By downward induction on 0 < we chose open sets

    U (2) . So, the set U (2) is such that (remember = n):

    (y (2)n \ U)

    mLeb((Cn)y) en

    ,

    mLeb

    Cn (U (2)[n, ))

    < en.

    Now suppose that U, . . . , U +1 have been chosen already so that

    mLebCn (Uk (2)[k, )) <

    23n+3

    en1 k

    en

    for each k { + 1, . . . , }. Let

    U = U+1 (2)[+1, ) . . . U (2

    )[, ).

    Note that (by our assumptions)

    mLeb(U Cn) < ( )

    23n+3en1

    1 en.

    Let = m and A = {y (2)m : mLeb

    (Cn U)y

    > em22n+2

    }. Note that

    mLeb(A) em

    22n+2< mLeb(Cn U) < (

    ) 23n+3

    en1

    1 en,

    and hence

    mLeb(A) k0 for a moment. Let A = { T : lh() = k} (so it is a

    front of T). For each T[q, A], by downward induction, we define s (t)and a real a [0, 1] such that

    () a k1

    =lh()

    1 22

    +3

    Fq ().

    If A, then we let a = 1 (and s is not defined).Suppose that a has been defined for all pos(t) so that () holds. Then

    Ft(a : pos(t)) k1

    =lh()+1

    1 22

    +3

    Ft (Fq () : pos(t)) =

    k1=lh()+1

    1 22

    +3

    Fq () 22lh()+3

    (remember our requests on q). Consequently we may apply 2.1 (for t = t, r = aand g = gt) to pick s

    (t) such that

    ()nor

    [s] =nor

    [t] 1, and() a

    def= Fs (a : pos(s))

    1 22

    lh()+3

    Ft (a : pos(t))

    k1=lh()

    1 22

    +3

    Fq ().

    This completes the choice of ss and as. Now we build a system sk :

    Sk \ max(Sk) such that Sk T[q, A] is a finite tree, root(Sk) = root(T), sk = sand succSk() = pos(s

    k) for Sk \ max(Sk).

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    33/35

    MEASURED CREATURES 33

    Next, applying Konig Lemma, we pick an infinite set I and a system p =tp

    : Tp Q(K

    , ) such that root(Tp

    ) = root(T) and

    Tp

    & k1, k2 I & lh() < k1 < k2 tp

    = sk2 .

    It follows from our construction that necessarily p Qmt4 (K, , F), and it is a

    condition stronger than p, and [Tp

    ] [T] = C.

    Theorem 4.13. In VP2 , the condition (b) of 4.1 holds.

    Proof. For < 2 let x be a Pname for the generic real added at stage (so itis a member of 2 if Z, and a member of

    k

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    34/35

    34 ANDRZEJ ROSLANOWSKI AND SAHARON SHELAH

    wt

    and let z = x , and z = z : < . Note that the conditions p

    and

    q are compatible. Also, as x is (a name for) a random real over VP , we have

    q P+1 the setB

    def= {y (2)[

    + 1, t) :

    t

    +1( zx)()

    y D}

    is null .

    Using Lemma 4.11, we may pick (a P+1name for) a closed set C (2)[ + 1, t)

    such that the condition q forces (in P+1):

    C {t

    t(z) [ + 1, t) : z

    x z Zt

    t},

    C B = , and the condition ()

    Cof 4.11 holds true for every [ + 1, t).

    (For the first demand remember that t

    is well behaving, so the set on the right-hand side has positive Lebesgue measure.) But now, using 4.12, we may inductively

    build a condition q

    P2 stronger than both q and p

    (and with the supportincluded in ( + 1) wt

    ) such that

    q P2 t

    t(x : w

    t

    ) [ + 1, t) / B ,

    getting an immediate contradiction.

    Pick any Z and let z = z : < be as defined in the proof of 4.13.1

    above. Let E be a Pname for the set

    {(r0, r1) 2 2 : zr0 Z

    t+1 and

    t+1( z

    r0)1(r1) rng(

    t

    t)}.

    So E is (a name for) a closed subset of 2 2. Let f2 be a name for a Borelfunction from 2 2 to 2 such that

    if (r0, r1) E, and t+1( zr0)1(r1) = tt(z : <

    t),

    then for each n <

    f2(r0, r1) n = n(z ktn() : wtn)

    (remember (i)). It should be clear that f2 is (a name for) a continuous functionand

    p P2 (x A)(f(x) = f2(x, f1(x))) ,

    finishing the proof.

    Corollary 4.14. It is consistent that

    every sup-measurable function is Lebesgue measurable, and for every function f : R R there is a continuous function g : R R

    such that the set {x R : f(x) = g(x)} has positive outer measure.

    References

    [1] Uri Abraham, Matatyahu Rubin, and Saharon Shelah. On the consistency of some partition

    theorems for continuous colorings, and the structure of1-dense real order types. Annals ofPure and Applied Logic, 29:123206, 1985.

    [2] Marek Balcerzak. Some remarks on sup-measurability. Real Anal. Exchange, 17:597607,

    1991/92.[3] Marek Balcerzak and Krzysztof Ciesielski. On the sup-measurable functions problem. Real

    Anal. Exchange, 23:787797, 1997/98.

  • 8/3/2019 Andrzej Roslanowski and Saharon Shelah- Measured Creatures

    35/35

    MEASURED CREATURES 35

    [4] Tomek Bartoszynski and Haim Judah. Set Theory: On the Structure of the Real Line. A KPeters, Wellesley, Massachusetts, 1995.

    [5] James E. Baumgartner. Iterated forcing. In A. Mathias, editor, Surveys in Set Theory, vol-

    ume 87 of London Mathematical Society Lecture Notes, pages 159, Cambridge, Britain,1978.

    [6] Heinrich Blumberg. New properties of all real functions. Transactions of the American Math-ematical Society, 24:113128, 1922.

    [7] Krzysztof Ciesielski. Set theoretic real analysis. Journal of Applied Analysis, 3:143190, 1997.[8] Krzysztof Ciesielski and Saharon Shelah. Category analog of sub-measurability problem.

    Journal of Applied Analysis, 6:159172, 2000. math.LO/9905147.[9] David Fremlin. Problem list. Circulated notes;

    available from http://www.essex.ac.uk/maths/staff/fremlin/measur.htm .

    [10] David H. Fremlin. E-mail message to A.Roslanowski.Message-Id: , 16 October 2000.

    [11] David H. Fremlin. Measure Theory. Torres Fremlin.See http://www.essex.ac.uk/maths/staff/fremlin/mt.htm .

    [12] Martin Goldstern. Tools for your forcing construction. In Set Theory of the Reals, volume 6

    of Israel Mathematical Conference Proceedings, pages 305360.

    [13] Eulalia Grande and Zbigniew Grande. Quelques remarques sur la superposition F(x, f(x)).Fundamenta Mathematicae, 121:199211, 1984.

    [14] Zbigniew Grande and Jan S. Lipinski. Un exemple dune fonction sup-mesurable qui nestpas mesurable. Colloquium Mathematicum, 39:7779, 1978.

    [15] Edward Grzegorek. On some results of Darst and Sierpinski concerning universal null anduniversally measurable sets. Bull. Acad. Polon. Ser. Sci. Math., 29:15, 1981.

    [16] Alexander S. Kechris. Classical Descriptive Set Theory, volume 156 of Graduate Texts in

    Mathematics. SpringerVerlag New York, Inc., 1994.[17] Aleksander B. Kharazishvili. Sup-measurable and weakly sup-measurable mappings in the

    theory of ordinary differential equations. Journal of Applied Analysis, 3:211223, 1997.[18] Andrzej Roslanowski and Saharon Shelah. Norms on possibilities III: strange subsets of the

    real line. in preparation.

    [19] Andrzej Roslanowski and Saharon Shelah. Sweet & Sour and other flavours of ccc forcingnotions. Archive for Mathematical Logic, accepted. math.LO/9909115.

    [20] Andrzej Roslanowski and Saharon Shelah. Norms on possibilities II: more ccc ideals on 2 .

    Journal of Applied Analysis, 3:103127, 1997. math.LO/9703222.[21] Andrzej Roslanowski and Saharon Shelah. Norms on possibilities I: forcing with

    trees and creatures. Memoirs of the American Mathematical Society, 141(671), 1999.math.LO/9807172.

    [22] Saharon Shelah. Non-elementary proper forcing notions. Journal of Applied Analysis, ac-cepted. math.LO/9712283.

    [23] Saharon Shelah. Possibly every real function is continuous on a nonmeagre set. Pub-

    lications de LInstitute Mathematique - Beograd, Nouvelle Serie, 57(71):4760, 1995.math.LO/9511220.

    [24] Saharon Shelah. Proper and improper forcing. Perspectives in Mathematical Logic. Springer,1998.

    [25] Heinrich von Weizsacker. Remark on extremal measure extensions. In Measure theory, Ober-

    wolfach 1979, volume 794 of Lecture Notes in Mathematics, pages 7980. Springer, Berlin,1980.

    Department of Mathematics, University of Nebraska at Omaha, Omaha, NE 68182-

    0243, USA, and Mathematical Institute of Wroclaw University, 50384 Wroclaw, PolandE-mail address: [email protected]

    URL: http://www.unomaha.edu/aroslano

    Institute of Mathematics, The Hebrew University of Jerusalem, 91904 Jerusalem,

    Israel, and Department of Mathematics, Rutgers University, New Brunswick, NJ 08854,

    USA

    E-mail address: [email protected]

    URL: http://www.math.rutgers.edu/shelah

    http://arxiv.org/abs/math/9905147http://www.essex.ac.uk/maths/staff/fremlin/measur.htmhttp://www.essex.ac.uk/maths/staff/fremlin/measur.htmhttp://www.essex.ac.uk/maths/staff/fremlin/mt.htmhttp://www.essex.ac.uk/maths/staff/fremlin/mt.htmhttp://arxiv.org/abs/math/9909115http://arxiv.org/abs/math/9703222http://arxiv.org/abs/math/9807172http://arxiv.org/abs/math/9712283http://arxiv.org/abs/math/9511220http://arxiv.org/abs/math/9511220http://arxiv.org/abs/math/9712283http://arxiv.org/abs/math/9807172http://arxiv.org/abs/math/9703222http://arxiv.org/abs/math/9909115http://www.essex.ac.uk/maths/staff/fremlin/mt.htmhttp://www.essex.ac.uk/maths/staff/fremlin/measur.htmhttp://arxiv.org/abs/math/9905147