NOTES ON UNIVERSALLY NULL SETS C. CARUVANA Here, we summarize some results regarding universally null subsets of Polish spaces and conclude with the fact that, following J. Mycielski, one can produce an analytic subfield of R which is not universally null. Many results come from [7] and [8]. 1. Introduction and Preliminaries For a Polish space X, we will let B(X) denote the class of Borel subsets of X. If X is Polish, then the space of Borel probability measures M(X) on X is Polishable with the weak topology. A continuous measure µ M(X) is a measure giving every finite set measure zero. Let M c (X) be the collection of continuous µ M(X). Then M c (X) is a dense G δ subset of M(X) as long as X has no isolated points. Refer to [10] for more details. Definition 1. Let X be a Polish space. We say that A X is universally null if µ (A) = 0 for every µ M c (X) where µ is the outer measure generated by µ. Definition 2. Let X be a Polish space. For A X, we define the annihilator of A to be N (A) = {µ M(X) : µ (A) = 0}. We say that A is residually null if N (A) is co-meager in M(X). For a more in depth discussion of the theory of residually null sets, refer to [3]. Of immediate interest is that all universally null subsets of any Polish space without isolated points are also residually null. Definition 3. Let X be a topological space. We say that a set A X is a Luzin set if it is uncountable and, for every M X which is meager, A M is countable. Theorem 4 (Mahlo [6], Luzin [5]). which is a Luzin set. Assuming CH, there is a set A R Proof. Let {F α : α < ω 1 } be the collection of all closed nowhere dense subsets of R. For each α < ω 1, inductively choose x α {x β : β < α} {F β : β α}. Then the set {x α : α < ω 1 } is as desired. Date: December 20, 2017. 1
2 C. CARUVANA One can immediately see that the set produced in Theorem 4 is not meager. Definition 5 (Borel [2]). Let X be a metric space with compatible metric ρ. We say that A X has strong measure zero if, given any sequence {ε n > 0 : n ω}, there exists a sequence {A n X : n ω} so that diam ρ (A n ) < ε n and A {A n : n ω}. In a metric space (X, ρ), since diam ρ (A) = diam ρ (cl(a)), we can cover any strong measure zero set with a countable union of closed sets satisfying the requisite properties. Definition 6 (Besicovitch [1]). Let X be a topological space. We say that A X is concentrated on B X provided that, for every open set U so that B U, A \ U is countable. Theorem 7 (Szpilrajn [14]). Let X be a Polish space. A set A X is a Luzin set if and only if A is uncountable and concentrated on every countable dense set of X. Proof. ( ) Suppose A is a Luzin set and D X is countable and dense. Let U be any open set with D U. Then U is dense so X \ U is closed and nowhere dense. It follows that A (X \ U) = A \ U is countable as A is a Luzin set. ( ) Now, suppose A is uncountable and concentrated on every countable dense subset Q of X. We need only check that A meets every meager set on a countable set. It suffices to check that A meets every closed nowhere dense set on a countable set so let F X be closed and nowhere dense. Let U = X \ F and notice that D := Q U is a countable dense set. Since U is open and D U, we have that A\U is countable. That is, A\(X\F ) = A F is countable, finishing the proof. Theorem 8 (Sierpiński [11]). Let X be a Polish space and suppose A X is concentrated on a countable dense set. Then A has strong measure zero. Proof. Let D = {d n : n ω} be dense in X so that A is concentrated on D. Now, let {ε n > 0 : n ω} and, for n ω, pick an open set U 2n that contains d n and satisfies diam(u 2n ) < ε 2n. Define U = {U 2n : n ω} and, since A is concentrated on D and D U, notice that A \ U is countable. Enumerate A \ U = {x n : n ω} and, for n ω, let U 2n+1 be an open set containing x n which satisfies diam(u 2n+1 ) < ε 2n+1. Finally, notice that the family {U n : n ω} covers A and has the property that, for each n ω, diam(u n ) < ε n. Therefore, A has strong measure zero. Lemma 9. Let X be a metric space and µ be a Borel probability measure on X. If µ is continuous, then, for every ε > 0 and x X, there exists a neighborhood U x of x with µ(u x ) < ε.
NOTES ON UNIVERSALLY NULL SETS 3 Lemma 10. Let (X, ρ) be a compact metric space and µ be a continuous Borel probability measure on X. Then, for any ε > 0, there exists a δ > 0 so that, for every closed F X with diam ρ (F ) < δ, µ(f ) < ε. Proof. For each x X, appeal to Lemma 9 to pick r x > 0 so that B ρ (x, 2r x ) satisfies µ(b ρ (x, 2r x )) < ε. As X is compact, pick a finite subset A X so that {B ρ (x, r x ) : x A} covers X. Define δ = min{r x : x A} > 0. Now, let F X be any closed set so that diam ρ (F ) < δ. Fix y F and pick x A so that y B ρ (x, r x ). Now, for any z F, notice that ρ(x, z) ρ(x, y) + ρ(y, z) < r x + δ 2r x. That is, z B ρ (x, 2r x ) and, as z F was arbitrary, we see that F B ρ (x, 2r x ) which provides us with the fact that µ(f ) < ε. Theorem 11 (Szpilrajn [13]). Let X be a Polish space and A X have strong measure zero. Then A is universally null. Proof. Suppose A X is strong measure zero and let µ be any continuous Borel probability on X. Let ε > 0 be arbitrary and define ε n = ε for 2n+2 n ω. By [10, Theorem II.3.2] we can find compact K X so that µ(k) > max {1 ε } 2, 0. Now, notice that ν defined on the Borel sets of K by ν(e) = µ(e) µ(k) is a Borel probability measure on K. So, by Lemma 10, we can pick δ n > 0 so that, whenever F K is closed with diam(f ) < δ n, we have that ν(f ) < ε n. Consider A = A K and notice that A is also strong measure zero. Thus, we can find {A n K : n ω} consisting of closed sets so that diam(a n ) < δ n and A {A n : n ω}. It follows that ν (A ) ν(a n ) ε 2 n+2 = ε 2. Hence, µ (A ) µ(k) = ν (A ) ε 2 = µ (A ) ε 2 µ(k) ε 2. Now, note that 1 ε 2 < µ(k) = µ(x \ K) < ε 2. Finally, we have that µ (A) µ (A \ K) + µ (A ) < ε. As ε was arbitrary, we see that µ (A) = 0, completing the proof. Corollary 12. Every Luzin subset of a Polish space is universally null. Proof. Combine Theorems 7, 8, and 11. Theorem 13. Assuming CH, there exists a set A R which is universally null which fails to have the Baire property.
4 C. CARUVANA Proof. Use Theorem 4 to produce a Luzin set A R and notice that Corollary 12 guarantees that A is universally null. Moreover, A is residually null and, since A is non-meager, we appeal to [3, Theorem 14] to conclude that A cannot have the Baire property. Lemma 14 (Sierpiński-Szpilrajn [12]). Let X be a Polish space, A X, and B A = {B A : B B(X)}. Then the following are equivalent. (i) A is universally null; (ii) any continuous measure on (A, B A ) is identically zero. Proof. ((i) (ii)) Suppose µ is a continuous measure on (A, B A ). Define ν on (X, B(X)) by the rule ν(b) = µ(b A) and observe that ν is a measure on X. Moreover, as µ is continuous, ν is continuous. Hence, ν (A) = 0. From this we see that µ must be identically zero. ((ii) (i)) Let µ be a continuous Borel probability measure on X. Now, define ν on (A, B A ) by the rule ν(e) = inf{µ(b) : B B(X), E B}. To see that ν is a measure on (A, B A ), let {E n : n ω} B A be a pair-wise disjoint family. For each n ω, pick B n B(X) so that E n = A B n and E n B(X) so that E n E n and µ(e n) = ν(e n ). Then let B n = B n E n. Now, E := {E n : n ω} {B n : n ω} =: B and B B(X). For n, m ω with n m, notice that B n B m A = (B n A) (B m A) (B n A) (B m A) = E n E m =. From this, we see that E n B n \ B m which gives µ(b n ) = ν(e n ) µ(b n \ B m ) µ(b n ) = µ(b n \ B m ) + µ(b n B m ). Particularly, µ(b n B m ) = 0 for n m. It follows that ν(e) µ(b) = µ(b n ) = ν(e n ).
NOTES ON UNIVERSALLY NULL SETS 5 For the other inequality, pick G B(X) so that G A = E and µ(g) = ν(e). Then, since E k E B k G B k, n 1 n 1 ν(e k ) µ(g B k ) k=0 k=0 = µ ((G B 0 ) (G B n 1 )) µ(g) = ν(e). Since this inequality holds for arbitary n ω, ν(e n ) ν(e) which establishes that ν is a measure on (A, B A ). As ν is a continuous measure on (A, B A ), by hypothesis, ν is identically zero. This guarantees that µ (A) = 0, finishing the proof. Lemma 15. Let X be Polish and A X. If A is universally null, then any continuous measure on (A) is identically zero. Proof. Let µ be a continuous measure on (A, (A)). Since µ BA is a measure on (A, B A ), µ BA is identically zero. Now, for any E A and B B A with E B, since µ(e) µ(b) = 0, we see that µ is identically zero. Proposition 16. The class of universally null subsets of a Polish space X is invariant under Borel isomorphisms. Proof. Immediate from Lemma 14. Theorem 17 (Szpilrajn). null. The product of universally null sets is universally Proof. Let X and Y be universally null and suppose µ is a continuous Borel probability measure on X Y. Define a measure ν on Y by ν(b) = µ(x B) for each Borel set B Y. Now, since X is universally null, ν({y}) = µ(x {y}) = 0 for every y Y so ν is continuous. It follows that, as Y is also universally null, ν(y ) = µ(x Y ) = 0, finishing the proof. Lemma 18. Suppose G R is co-meager. Then, for any z R, there exist x, y G so that z = x + y. Proof. Let z R and, since G is co-meager, so is z G. G (z G) and pick x G so that y = z x. Now, let y Theorem 19 (Sierpiński). Assuming CH, there exists a universally null set which is not strong measure zero.
6 C. CARUVANA Proof. Let {F α : α < ω 1 } be the collection of all closed nowhere dense subsets of R and {z α : α < ω 1 } be an enumeration of R. Pick x 0, y 0 F 0 so that x 0 + y 0 = z 0. For each 0 < α < ω 1, inductively choose x α, y α {x β : β < α} {y β : β < α} {F β : β α} with the property that x α + y α = z α. Then define X = {x α : α < ω 1 } {y α : α < ω 1 }. Now we argue that X 2 is universally null but not strong measure zero. As X is transparently a Luzin set, it is universally null by Corollary 12. Hence, by Theorem 17, X 2 is universally null in R 2. Now, define p : R 2 R by the rule p(x, y) = x + y. Fix a, b R and let x, y R be so that (x a) 2 + (y b) 2 < ε where ε > 0. Now, (x + y) (a + b) 2 = (x a + y b) 2 = (x a) 2 + 2(x a)(y b) + (y b) 2 < 3 ε 2. From this, we see that p takes ε-balls to 3 ε-balls. Hence, p preserves strong measure zero sets. As p[x 2 ] = R, it can t possibly be the case that X 2 is strong measure zero. Theorem 20 (Grzegorek, Ryll-Nardzewski [4]). There exists a universally measurable set which is not a Borel set modulo a universally null set. Proof. The example here is WO. Since WO Π 1 1, it is universally measurable. Now let B be a Borel set. We will show that WO B is not universally null. Notice that B \ WO Σ 1 1. So if B \ WO were uncountable, it would contain a perfect set ensuring that WO B is not universally null. So B \ WO must be countable. Now, without loss of generality, we can assume that B WO. Since WO is Π 1 1 -complete, WO \ B is Π1 1-complete. Moreover, WO \ B is uncountable. Hence, WO \ B has the perfect set property so WO \ B is not universally null. 2. Subfields of R Theorem 21 (Mycielski [9]). For any meager set Y of irrationals, there exists an uncountable perfect set X so that the field generated by X is disjoint from Y. Corollary 22. There exists an analytic subfield F of R which is not universally null. Incidentally, such an F is meager. Proof. Let Y be a meager set of irrationals, and X be an uncountable perfect set so that the field F generated by X is disjoint from Y by Theorem 21.
NOTES ON UNIVERSALLY NULL SETS 7 Immediately, as F is the field generated by X and X is closed, F is analytic. The fact that F supports a non-degenerate Borel probability measure follows from the fact that F contains an uncountable perfect set. That is, F is not universally null. Now, since F is analytic, it has the Baire property. As F contains the rationals, F is dense. If F were non-meager, it would be open. But then, as an open subgroup of R, F would also be closed yielding F = R, a contradiction. Therefore, F is meager. References [1] Abram Samoilovitch Besicovitch, Concentrated and rarified sets of points, Acta Mathematica 62 (1934), 289 300. [2] Émile Borel, Sur la classification des ensembles de mesure nulle, Bulletin de la Société Mathématique de France 47 (1919), 97 125. [3] C. Caruvana and R. R. Kallman, An Extension of the Baire Property, arxiv:1703.08879 (2017). [4] Edward Grzegorek and Czes law Ryll-Nardzewski, A remark on absolutely measurable sets, Bulletin de l Académie Polonaise. Série des Sciences Mathématiques 28 (1980), 229 232. [5] Nikolai Luzin, Sur un problème de M. Baire, C. R. Hebd. Séances Acad. Sci. 158 (1914), 1258 1261. [6] Paul Mahlo, Über Teilmengen des Kontinuums von dessen Machtigkeit, Sitzungsberichte der Sachsischen Akademie der Wissenschaften zu Leipzig, Mathematisch- Naturwissenschaftliche Klasse 65 (1913), 283 315. [7] Arnold W. Miller, Special subsets of the real line, Handbook of Set-Theoretic Topology (Kenneth Kunen and Jerry E. Vaughan, eds.), North Holland, 1984, pp. 201 233. [8], Special sets of reals, Set Theory of the Reals (Haim Judah, ed.), Israel Mathematical Conference Proceedings, vol. 6, American Math Society, 1993, pp. 415 432. [9] Jan Mycielski, Independent sets in topological algebras, Fundamenta Mathematicae 55 (1964), no. 2, 139 147. [10] K. R. Parthasarathy, Probability Measures on Metric Spaces, AMS Chelsea Publishing Series, Academic Press, 1967. [11] Wac law Sierpiński, Sur un ensemble non dénombrable, dont toute image continue est de mesure nulle, Fundamenta Mathematicae 11 (1928), 302 303. [12] Wac law Sierpiński and Edward Szpilrajn, Remarque sur le problème de la mesure, Fundamenta Mathematicae 26 (1936), 256 261. [13] Edward Szpilrajn, Remarques sur les fonctions complètement additives d ensemble et sur les ensembles jouissant de la propriété de baire, Fundamenta Mathematicae 22 (1934), 303 311. [14], The characteristic function of a sequence of sets and some of its applications, Fundamenta Mathematicae 31 (1938), 47 66.