aa r X i v : . [ m a t h . M G ] D ec CORRECTION OF METRICS
P. B. ZATITSKIY AND F. V. PETROV
Metric triple (
X, ρ, µ ) is a space with structures of metric ρ andBorel probabilistic measure µ . The studying of them was initiatedby M. L. Gromov in [2]. Namely, it was proved that metric triplesare uniquely parametrized by measures on the set of distance matrices( ρ ( x i , x j )) i,j , generated by random and independent choice of points x , x , · · · ∈ X . A. M. Vershik gave another formulation of this fact andproposed its new proof, based on the Ergodic theorem, and initiated thestudying of variable metrics on the measurable space. In particular, in[3, 4] he suggested to use dynamics of metrics for studying the entropyand other invariants of dynamical systems. It requires the preliminarystudying of the set of matrics and “almost metrics” on the standardmeasurable space. The present note is devoted to correction of almostsemimetrics to real metrics on the set of full measure. We are goingto use the obtained theorems in further work on application of metricsdynamics to ergodic theory.In this article we will consider only Lebesgue spaces with continuousmeasure. Definition 1.
Let ( X, µ ) be Lebesgue space with probabilistic measure, ρ ( x, y ) be measurable non-negative function on ( X × X, µ × µ ) such that ρ ( x, y ) = ρ ( y, x ) and ρ ( x, z ) ≤ ρ ( x, y ) + ρ ( y, z ) for almost all x , y , z in X . We say that ρ is almost-metric on X .We say that almost-metric ρ is essentially separable, if for any ε > one may cover X by a countable set of measurable subsets of essentialdiameter less then ε (essential diameter of the set A ⊂ X is defined asthe essential supremum of the function ρ ( x, y ) , restricted to A × A ).Two almost-metrics which coincide on the set of full ( µ × µ ) -measureas functions of 2 variables, are called equivalent. Sometimes we also saythat one of two equivalent metrics is a correction of another (usuallycorrected metric is in a sense better then the initial one).Metric (or semimetric) on the Lebesgue space is called admissible,if the corresponding (semi-)metric space is separable on the set of fullmeasure. Date : 1 Oct. 2011.
Key words and phrases.
Lebesgue space, semimetric.This research is supported by the Chebyshev Laboratory (Department of Math-ematics and Mechanics, St. Petersburg State University) under RF Governmentgrant 11.G34.31.0026, and RFBR grants 11-01-12092-ofi-m and 11-01-00677-a.
We say that a semimetric ρ on the space with Borel probabilisticmeasure has a finite ε -entropy H ε ( ρ ) < ∞ , if there exist a finite numberof balls of radius ε , which cover the set of measure at least − ε .Closed ball of radius r centered in x is denoted by B ( x, r ) . The following lemma lists several equivalent definitions of admissiblemetric.
Lemma 1.
Let ( X, ρ ) be a semimetric space and µ be Borel measureon it. Then the following statements are equivalent: For any ε > the semimetric ρ has finite ε -entropy. There exists a set of full measure µ such that restriction of ρ on thisset is separable For µ -almost all x all balls (in semimetric ρ ) centered in x havepositive measure.Proof. Let’s prove implication from 1) to 3). Consider the sets T n = { x ∈ X : µ ( B ( x, n )) = 0 } . It suffices to prove that µ ( T n ) = 0 for all n . A priori we do not even know that T n is measurable, but we donot need it. Choose any ε ∈ (0 , n ). Condition 1) says that X maybe partitioned onto disjoint sets X , X , . . . , X k such that µ ( X ) < ε and diam(X j ) < ε for all j ∈ { , . . . , k } . Without loss of generalitysets X j have positive measure (if µ ( X j ) = 0, remove it and replace X to X ∪ X j ). Then for any point x / ∈ X the ball B ( x, ε ) containsone of the sets X j , hence has strictly positive measure. It follows that T n ⊂ X . So we have proved that for arbitrarily small ε > T n is contained in a set of measure less then ε . It follows that T n ismeasurable and have zero measure. The union of T n ’s taken by allpositive integers n has zero measure aswell. This proves 3).Let’ prove that 3) implies 2). Consider the set Y of points such thatall balls centered in those points have positive measure. Then Y is aset of full measure and it suffice to prove that for any ε > ε -net. Assume that for some ε > ε -net, then by Zorn’s lemma Y has a more then countablesubset with mutual distances at least ε . But then the balls of radius ε/ { x n } ∞ n =1 bedense in a set X ′ of full measure. For fixed ε > B n = B ( x n , ε ). The union of them has full measure. But then forsome finite N the finite union N S n =1 B n has measure more then 1 − ε , asdesired. (cid:3) Theorem 1. Let ( X, µ ) be a Lebesgue space, ρ be almost metric on X . Then ρ may be corrected to everywhere finite semimetric on X . ORRECTION OF METRICS 3 If ρ was essentially separable, then corrected semimetric may bechosen so that ( X, ρ ) is separable. In other words, the corrected semi-metric may be chosen admissible.Proof.
1) Note that for almost all x the function ρ ( x, · ) is measur-able and inequality ρ ( x, y ) + ρ ( x, z ) ≥ ρ ( y, z ) holds for almost allpairs ( y, z ). Fix such point x . At first, we replace the measure µ to some quivalent measure so that the function f ( t ) = ρ ( x , t ) becomessummable. For example, we may take A n = f − ([ n − , n )) and put˜ µ ( B ) = c P ∞ n =1 − n µ ( B ∩ A n ) for any measurable B and the constant c is chosen so that ˜ µ ( X ) = 1. Note that now the function ρ ( y, z ) becomessummable as a function of two variables by triangle inequality.Now we identify Lebesgue space and the unit circle S = R / Z equippedbyt he Lebesgue probabilistic measure. Define the new metric ˜ ρ (pos-sibly infinite somewhere) by equality˜ ρ ( x, y ) = lim sup T → +0 T − Z T Z T ρ ( x + t, y + s ) dtds (1)Note that by Lebesgue theorem on differentiation of integral the limitexists and equals ρ ( x, y ) almost surely (i.e. for almost all pairs ( x, y )).This function ˜ ρ is obviously symmetric. Let’s prove triangle inequalityfor it. Note that for almost all ( s, t, τ ) ∈ [0 , T ] one has ρ ( y + s, z + t ) ≤ ρ ( y + s, x + τ ) + ρ ( x + τ, z + t ) . Let’s integrate it by [0 , T ] , divide by T and take an upper limit for T → +0. Using inequality lim sup( F + G ) ≤ lim sup F + lim sup G weget a triangle inequality for ˜ ρ . If ˜ ρ does not vanish in some points ondiagonal, replace those values to 0. Note that the function ˜ ρ is finitealmost everywhere (since equivalnet function ρ is almost everywherefinite). Hence one may choose a point x so that ˜ ρ ( x , x ) < ∞ foralmost all x ∈ S , i.e. for all x in the set S of full measure. Nowchange the semimetric ˜ ρ outside S × S by the rules ˜ ρ ( x, y ) := ˜ ρ ( x , y )for x / ∈ S , y ∈ S , ˜ ρ ( x, y ) := ˜ ρ ( x, x ) for y / ∈ S , x ∈ S and ˜ ρ ( x, y ) := 0for x, y / ∈ S . This semimetric is almost everywhere finite.2) Using the statement of p. 1), we may suppose that the semimetric ρ is defined on whole X and satisfies triangle inequality everywhere.Let’s prove that there exists a set Y of full measure so that restrictionof ρ on Y is a separable semimetric space. The correction outside Y is done as descibed above in the end of proof of 1), and separabilityis saved after such correction. It suffices for any fixed positive integer n to find a countable union of balls of radius 1 /n which has a fullmeasure, then define Y as the intersection of such sets. Let’s cover X by a countable number of measurable sets of diameter at most 1 /n .Consider one such set A . For almost all points x ∈ A the distancesfrom x to almost all points of A do not exceed 1 /n . Hence A mod 0 iscovered by a ball of radius 1 /n , and we are done. (cid:3) P. B. ZATITSKIY AND F. V. PETROV
Usually when one says about the space with metric and measure, sheconsiders the metric structure as “main”, and restricts measure to sat-isfy some properties in terms of metric (Borel, regular measures). Wefollow A. M. Vershik’s approach, considering the metric as measurablefunction. Note that if (
X, ρ ) is separable semimetric space, and µ is aBorel measure on X , then ρ as a function on X × X is measurable w.r.t.the measure µ × µ (since it is continuous and hence Borel measurable).The follwoing theorem shows that in separable case those two condi-tions (“measure is Borel” and “metric is measurable”) are equivalent. Theorem 2.
Let ( X, µ ) be the Lebesgue space, ρ be admissible semi-metric on X . Then the measure µ is Borel w.r.t. topology of metric ρ .Proof. For any rational r > { ( x, y ) : ρ ( x, y ) < r } is mea-surable, hence almost all its sections (balls of radius r , hereafter ballsmean open balls). Hence foralmost all ponts x ∈ X all balls with ra-tional radius center in x are measurable. In this case all balls centeredin x are measurable. Denote by X the set of such points, let X bethe complement of X ( µ ( X ) = 0). Consider a countable dense subset X ′ ⊂ X . Let’s prove that any ball is measurable, from separabil-ity it follows that measure is Borel. Consider the ball B = B ( x , r )centered in x with raidus r . For any point x ∈ X ′ ∩ B considerthe ball B ( x, r − ρ ( x, x )). It lies in B and is measurable. Let’sprove that the union of such balls contains X ∩ B . Then B con-tains their union U and is contained in U ∪ X , hence B is measurableand B = U mod 0. Take any point x ∈ X ∩ B . Let’s find a point y ∈ X ′ such that ρ ( x, y ) < ( r − ρ ( x , x )) /
2. Then by triangle in-equality ρ ( x , y ) < ( r + ρ ( x , x )) /
2, so y ∈ B and moreover the ball B ( y, r − ρ ( x , y )) contains a point x , as we wish. (cid:3) If the metric is not separable, the conclusion of this theorem mayfail. Indeed, let A be a non-measurable subset of X , x ∈ X \ A be apoint, define a metric ρ ( x, y ) = , if x = y ;1 , if x = x , y ∈ A or y = x , x ∈ A ;2 , otherwise.It equals 2 almost everywhere and is so measurable, but the ball B ( x , / ρ on the space X . Consider its Borel sigma-algebra B . Is ittrue that ρ , as a function on X × X , is measurable w.r.t. sigma-algebra B × B ? We do not know the answer. ORRECTION OF METRICS 5
Now let’s prove that two equivalent admissible metrics coincide onthe set of full measure. Of course, this statement fails without admis-sibility condition (for example, for metrics with distances values 1 and2.)
Theorem 3.
Let two admissible metrics ρ , ρ coincide almost every-where on X × X w.r.t. measure µ × µ . Then there exists a subset X ′ ⊂ X of full measure such that ρ and ρ coincide on X ′ × X ′ .Proof. Note that for almost all x ∈ X , we have ρ ( x, y ) = ρ ( x, y ) foralmost all y ∈ X . Removing the set of zero measure from X , we maysuppose that it holds for all x . Now for any x ∈ X and any positive r the balls { y ∈ X : ρ ( x, y ) < r } and { y ∈ X : ρ ( x, y ) < r } haveequal measure. Using 1 we may find a subset X ′ of full measure suchthat for any x ∈ X ′ any ball { y ∈ X : ρ ( x, y ) < r } has a positivemeasure. Then the same holds for ρ . Let’s prove that ρ and ρ coincide on X ′ × X ′ Take x , x ∈ X ′ . Take any r > y such that ρ ( x , y ) < r one has ρ ( x , y ) = ρ ( x , y )and ρ ( x , y ) = ρ ( x , y ). Since the set of such y ’s (it is a ball) haspositive measure, we may find at least one point y in it. Then ρ ( x , x ) ≤ ρ ( x , y )+ ρ ( y, x ) = ρ ( x , y )+ ρ ( y, x ) ≤ r + ρ ( x , x ) . Since it holds for any r >
0, we conclude that ρ ( x , x ) ≤ ρ ( x , x ).Opposite inequality is analogous. So, ρ and ρ coincide on X ′ × X ′ asdesired. (cid:3) The natural question to ask is the following: which structures onmeasure spaces (except semimetric space structure) admit correctiontheorems like Theorem 1. May one correct almost-group to the group,almost-vector space to vector-space and so on? We do not know non-trivial examples, in which the answer is negative (the trivial examplesinclude, say, the observation that almost-metric may be corrected onlyto semimetric, not to metric). We present another positive result
Theorem 4.
Let semimetric ρ be defined on a Lebesgue space X , andlet it satisfy ultrametric inequality ρ ( x, z ) ≤ max( ρ ( x, y ) , ρ ( y, z )) foralmost all triples ( x, y, z ) ∈ X . Then there exists an ultrametric on X , which coincides with ρ on almost all pairs and satisfies ultrametricinequality for all triples.Proof. For any n consider the correction of almost-metric ρ n given bythe formula( ρ n ( x, y )) n = lim sup T → +0 T − Z T Z T ρ n ( x + t, y + s ) dtds By power means inequalities the sequence ρ n ( x, y ) increases by n for fixed x, y , so it has (finite or infnite) limit ˜ ρ ( x ; y ). By Lebesgue’sintegrals differentiation theorem this limit is almost everywhere finite P. B. ZATITSKIY AND F. V. PETROV and equal to ρ ( x, y ). The function ˜ ρ n is a semimetric for all n (thisis so for ρ m instead ˜ ρ with m ≥ n , and one may pass to limit in thecorresponding triangle inequality). The infinite values may be avoidedon the same way as in the proof of Theorem 1 for usual metrics. (cid:3) We also formulate the following general statement, which includesTheorems 1, 4 as partial cases.
Conjecture 1.
Given positive integers k ≤ n and measurable real-valued function f ( x , x , . . . , x k ) . For almost all y , . . . , y n the vector { f ( y i , . . . , y i k ) } ≤ i j ≤ n of dimension n k belongs to the given closed sub-set in the space of dimension n k . Then there exists a function ˜ f , equiv-alent to f , for which this condition holds for all y , . . . , y n . The questions of this paper arised in the program in ergodic theory,initiated by A. M. Vershik. We are also grateful to him for supportand numerous helpful discussions.
References [1] P. R. Halmos, J. von Neumann,
Operator methods in classical mechanics, II. — Ann. Math. , No. 2, 332–350 (1942).[2] M. Gromov, Metric Structures for Riemannian and Non-Riemannian Spaces.Birkhauser, 1998.[3] The universal Uryson space, Gromov’s metric triples, and random metrics onthe series of natural numbers. Uspekhi Mat. Nauk , No. 5, 57–64 (1998).English translation: Russian Math. Surveys , No. 5, 921–928 (1998).[4] A. Vershik, Scaling entropy and automorphisms with purely point spectrum , arXiv:1008.4946v4 . Zatitskiy P. B., Petrov F. V. Correction of metrics.We prove that a symmetric nonnegative function of two variables on aLebesgue space that satisfies the triangle inequality for almost all triplesof points is equivalent to some semimetric. Some other properties ofmetric triples (spaces with structures of a measure space and a metricspace) are discussed.
St. Petersburg Department, of V.A.Steklov Institute of Mathe-matics, of the Russian Academy of Sciences, Fontanka 27,, 191023, St.Petersburg, Russia, P. L. Chebyshev Laboratory, of St. PetersburgState University, 14th Line 29B, Vasilyevsky Island,, 199178, St. Pe-tersburg, Russia
E-mail address : [email protected] St. Petersburg Department, of V.A.Steklov Institute of Mathe-matics, of the Russian Academy of Sciences, Fontanka 27,, 191023, St.Petersburg, Russia
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