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Postscript to Shelah & Fremlin p90D.H.Fremlin

University of Essex, Colchester

25 September 1991

1. The statement () of Shelah & Fremlin p90, 2G, can be strengthened, as follows. Write () forthe statement

there is a closed negligible set Q [0, 1] such that (L)(Q1[D]) LD for every D [0, 1].

Proposition IfP is a partially ordered set as in Shelah & Fremlin p90, then 11P P ().

proof (a) Take X and P and and as in Shelah & Fremlin p90, 1. Then if D X and > 0, thereis a closed set F X, with F D , such that

11P P F R1[D],

writing F for the P-name for a closed subset of X corresponding to F.PPP Choose k0 such that 2

k0 14, and a closed set F0 such that F0 =

(D F0) D 12 . Set

F = {x : k k

0,

(D {w : wk = x

k}) > 2

k+1

{w : wk = x

k}}.Then F is closed, and

F0 \ F {x : x F0, k k

0, (F0 {w : wk = xk}) 2k+1{w : wk = xk}}

has measure at most 2k

0+1 12 , so F

D .Now suppose that is a P-name such that

11P P F.Set r = 1, Lk = nk for every k N and follow the argument of Lemma 1R of Shelah & Fremlin p90down to the end of part (c), but insisting at the beginning that k0 k0.

Observe thatp3 P s,

where s = Hi(vi )i (1 2

k1+1){x : s x};

but asp3 P F, s ,

we must have(D {x : s x}) > 2k1+1{x : s x},

and there is a z D such that s z and z / Jk for every k k1.Now continue the argument as in (e)-(i) of the proof of Lemma 1R to get p5 p3 such that p5 P (, z)

R. QQQ

(b) Now we find that11P P D X, (R1[D]) D.

PPP Let 0 be a P-name for a subset of X, and , (ground-model) rationals such that

11P P 0 > > .

Take < such that whenever is a P-name for a closed subset of X and

11P P > 1 then there is a P-name for a member of 0. Now taking to be a P-name for the subset of Xconsisting of those members of 0 which can be represented by P-names, we see that

11P P .

Using part (a) in VP , we have a P-name for a closed subset of X such that

11P P & 11P P () R1[]

,

expressing P as an iteration P P as in the proof of Theorem 1S of Shelah & Fremlin p90. But thismust mean that

11P P (R1[0])

,and as was arbitrary this proves the claim. QQQ

(c) This proves the result for (X, ) rather than for ([0, 1], L). But as remarked in 2G of Shelah &Fremlin p90, we have a continuous inverse-measure-preserving f : X [0, 1] such that f1[D] =

L

[D]1

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for every D [0, 1]; so that setting Q = {(f(x), f(y)) : (x, y) R} we obtain the result for Lebesguemeasure, as stated.

2. In fact wse can go a little further: in VP, Q has the property that

whenever D [0, 1] and E is a measurable set such that (E\ D) = 0, then (E\ Q1[D]) = 0.

To see this, follow the arguments above, observing that it is enough to consider closed E, and that the setF of part (a) of the proof can be taken to be a subset of E.

ReferenceShelah S. & Fremlin D.H. [p90] Pointwise compact and stable sets of measurable functions, to appear

in J. Symbolic Logic.