to appear: Math. Proc. Cambridge Phil. Soc.MEASURE AND DIMENSION FUNCTIONS:MEASURABILITY AND DENSITIESPertti Mattilaand R. Daniel Mauldin1(Received 3 January 1995)1. IntroductionDuring the past several years, a new type of geometric measure and dimensionhave been introduced, the packing measure and dimension, see [Su], [Tr] and [TT1].These notions are playing an increasingly prevalent role in various aspects of dy-namics and measure theory. Packing measure is a sort of dual of Hausdor� measurein that it is de�ned in terms of packings rather than coverings. However, in contrastto Hausdor� measure, the usual de�nition of packing measure requires two limitingprocedures, �rst the construction of a premeasure and then a second standard limit-ing process to obtain the measure. This makes packing measure somewhat delicateto deal with. The question arises as to whether there is some simpler method forde�ning packing measure and dimension. In this paper, we �nd a basic limitationon this possibility. We do this by determining the descriptive set theoretic com-plexity of the packing functions. Whereas the Hausdor� dimension function on thespace of compact sets is Borel measurable, the packing dimension function is not.On the other hand, we show that the packing measure and dimension functions aremeasurable with respect to the �-algebra generated by the analytic sets. Thus, theusual sorts of measurability properties used in connection with Hausdor� measure,e.g., measures of sections and projections, remain true for packing measure.We now give a somewhat more detailed description of our results and we intro-duce some notation. Throughout this paper (X; d) will be a Polish space, that is, acomplete separable metric space. We equip the space K(X) of non-empty compactsubsets of X with the Hausdor� distance %;%(K;L) = supfdist(x; L); dist(y;K) : x 2 K; y 2 Lg:Then (K(X); %) is a complete separable metric space, see e.g., [R]. We denote byHg and Pg the Hausdor� and packing measures generated by a non-negative non-decreasing function g on the positive reals; their de�nitions will be given later. Weshall study the measurability properties of the functions Hg and Pg on K(X). TheHausdor� measure Hg is rather simple in this respect; it is in Baire's class 2, andin particular a Borel function. The packing measure is much more complicated,1 Research supported by NSF Grant DMS 93038881

2 PERTTI MATTILA AND R. DANIEL MAULDIN1as can be expected from its de�nition. Assuming that g satis�es the doublingcondition; g(2r) � kg(r), we show that Pg is measurable with respect to the �-algebra generated by the analytic sets, but it is not Borel measurable even when Xis the unit interval.From these results on measures we obtain that on K(X) the Hausdor� dimensionis a Borel function and the packing dimension is measurable with respect to the�-algebra generated by the analytic sets. Again, the packing dimension is not aBorel function. We shall actually prove the measurability of the packing dimensionin two ways: via the packing measures and via the box counting dimension (alsoknown as the Minkowski dimension).We shall mainly work with packing measures whose de�nition is based on theradii of balls. In Section 5 we brie y discuss the measurability questions related todiameter-based packing measures.By composing suitable functions, we can use the above results to deduce mea-surability of various functions. For example, we show in Section 6 that if K is acompact subset of a product space T � X, where T is another Polish space, thenthe Hausdor� measures and dimensions of the sections fx 2 X : (t; x) 2 Kg areBorel functions of t, and their packing measures and dimensions are measurablewith respect to the �-algebra generated by analytic sets. Our example in Section 7shows that these latter functions need not be Borel measurable.The last section is independent of the others. There we study the lower densitydg(A; x) = lim infr!0 Pg�A \ B(x; r)�g(2r) ;where g satis�es the doubling condition. For packing measures this is more naturalthan the corresponding upper density because even for compact subsets K of Rnwith positive and �nite Ps measure the upper density may be in�nite everywherein K. For the lower density it was shown in [TT2] and [ST] that if A � Rn andPs(A) < 1, then dg(A; x) = 1 for Ps almost all x 2 A. In Example 8.9 we shallshow that this does not extend to any in�nite-dimensional inner product space.However, we shall prove for g(r) = rs the optimal inequalities 2�s � dg(A; x) � 1for Pg almost x 2 A if A is a subset of an arbitrary Polish space with Pg(A) <1,see Theorem 8.3. We shall also prove in Theorem 8.5 that dg(A; x) does equal onealmost everywhere in A, but almost everywhere with respect toHg and not Pg. Thishas the following consequence: if A � X with Pg(A) <1, then Hg(A) = Pg(A) ifand only if(1.1) limr!0 Hg�A \ B(x; r)�g(2r) = 1 for Pg almost x 2 A:In Rn this has been proved in [TT2] and [ST] (see also [M]) with the help of theaforementioned fact that dg(A; x) = 1 for Pg almost all x 2 A. Still in Rn , forg(r) = rs (1) implies that s must be an integer and A rather regular, i.e. recti�ablewith respect to Pg. This gives for sets A � Rn with g(r) = rs and Pg(A) <1 thatHg(A) = Pg(A) if and only if A is such a recti�able set, see [ST] or [M]. We shallextend this result to separable Hilbert spaces in Theorem 8.8.

MEASURE AND DIMENSION FUNCTIONS: MEASURABILITY AND DENSITIES 3We shall denote by R the set of real numbers, by Rn the Euclidean n-space, byQ the set of rational numbers, and by N the set of positive integers. For A � X,the closure of A will be �A. The �-algebra generated by analytic sets will be denotedby B(A). In a product space, proj1 stands for the projection onto the �rst factor.Finally, we wish to thank Lars Olsen and Jouni Luukkainen for their correctionsand comments during the preparation of this manuscript.2. Hausdorff measure and dimension functionsHere and later g : [0;1)! [0;1) will be a non-decreasing function with g(0) =0. We denote by d(A) the diameter of a set A � X with d(;) = 0. Let 0 < � � 1and A � X. The approximating Hausdor� g-measure Hg;�(A) of A is de�ned byHg;�(A) = inf � 1Xi=1 g(d(Ui)) : A � 1[i=1Ui; each Ui is open with d(Ui) < ��and the Hausdor� g-measure Hg(A) byHg(A) = lim�!0Hg;�(A):We write Hg;� = Hs� and Hg = Hs when 0 < s < 1 and g(r) = rs for r � 0. TheHausdor� dimension dimH A of A is de�ned bydimH A = inffs : Hs(A) = 0g:We shall use the usual notion of Baire's classes for functions between metricspaces. Thus Baire's class 0 consists of all continuous functions and Baire's classn + 1 consists of all pointwise limits of sequences of functions in Baire's class n.In particular, the upper and lower semicontinuous functions belong to the Baire'sclass 1.2.1. Theorem. a) For 0 < � � 1, the functionHg;� : K(X)! [0;1]is upper semicontinuous.b) The functions Hg and dimH : K(X)! [0;1]are of Baire's class 2. Moreover, in general, these functions are not of Baire's class1.Proof. Let V be a countable basis for the topology of X and let fWng1n=1 be anenumeration of all �nite unions of the sets of V. Let c 2 R and K 2 K(X). Usingthe de�nition of Hg;� and the compactness of K, we see that Hg;�(K) < c if andonly if there are �nitely many open sets U1; : : : ; Uk such thatkXi=1 g(d(Ui)) < c; K � k[i=1Ui and d(Ui) < � for i = 1; : : : ; k:

4 PERTTI MATTILA AND R. DANIEL MAULDIN1This holds if and only if there are n1; : : : ; nm 2 N such that(2.1) mXi=1 g(d(Wni)) < c; K � m[i=1Wni and d(Wni) < � for i = 1; : : : ;m:Since the set of K 2 K(X) for which there exist n1; : : : ; nm satisfying (2.1) isclearly open in the exponential or Vietoris topology[K2], we conclude that the setfK 2 K(X) : Hg;�(K) < cg is open, and a) follows. Letting Q+ be the set ofpositive rationals, it then follows that for each �,(2.2) fF 2 K(X) : dimH F � �g = \s>�s2Q+ \n fF : Hs1=n(F ) < 1gis a G�-set. From this we get fF : � < dimH F < �g is a G��-set, for all 0 < � < �.Thus, dimH is a Baire class 2 function. As Hg = limn!1Hg;1=n, part a) completesthe proof of the �rst part of b).The following example shows that, in generalHg is not a Baire class one function.Consider X = [0; 1] � [0; 1] and H1 : K(X) ! [0;+1]. Since H1(F ) = 0 for all�nite F , H1 has value 0 on a dense subset of K(X). On the other hand, it is easyto show that each open set of the formP (U1; : : : ; Un) = fF : F � [Ui andF \ Ui 6= ;; i = 1; : : : ; ng;where each Ui is a nonempty open subset of X contains an element F such thatH1(F ) = 1. Since these sets form a basis for the topology of K(X), H1 has value 1on a dense set. Thus, H1 cannot have a point of continuity, whereas a Baire classone function has a dense set of points of continuity. This also shows that dimH hasvalue 0 on a dense set and also has value 1 on a dense set and is therefore also nota Baire class 1 function.3. Packing and box dimension functionsThe packing dimension can be de�ned either via the upper box counting (i.e.,Minkowski) dimension or via the packing measures. In this section we use the �rstapproach and in the next section the second.We begin with the de�nitions and simple measurability properties of the boxcounting dimensions. Let K 2 K(X). For � > 0 let N�(K) be the smallest numberof open balls of radius � that are needed to cover K. The upper and lower boxcounting dimensions dimBK and dimBK of K are de�ned bydimBK = lim sup�!0 logN�(K)�(� log �)= lim supj!1 logN2�j (K)�(j log 2);and dimBK = lim inf�!0 logN�(K)�(� log �)= lim infj!1 logN2�j (K)�(j log 2);

MEASURE AND DIMENSION FUNCTIONS: MEASURABILITY AND DENSITIES 5where the second equalities are easily seen to hold.Instead of covering with �-balls we can also use packings with �-balls. Let P�(K)be the largest integer k such that there are points x1; : : : ; xk 2 K with d(xi; xj) > �for i 6= j. Then one veri�es easily (or see [F] or [M]) thatdimBK = lim sup�!0 P�(K)=(� log �)= lim supj!1 P2�j (K)=(j log 2);and similarly for dimBK.3.1. Lemma. The functions dimB and dimB : K(X) ! [0;1] are of Baire'sclass 2.Proof. Evidently fK 2 K(X) : N�(K) < cg is open for c 2 R and � > 0, whenceN� is upper semicontinuous. Thus the functions gj : K 7! logN2�j (K)=(j log 2)and hk = infj�k gj are also upper semicontinuous. Hence dimB = limk!1 hk is ofBaire's class 2.For the upper box counting dimension we use the packing function P�. It is easyto verify that it is lower semicontinuous. Hence it follows as above that dimB is ofBaire's class 2.We de�ne the packing dimension dimpA for A � X bydimpA = inf � supi dimBKi : A � 1[i=1Ki; Ki 2 K(X)�:Then the packing dimension has the countable stability property, which the boxcounting dimension lacks:dimp � 1[i=1Ai� = supi dimpAi for Ai � X:This leads to the following lemma. It was also essentially proved by Falconer andHowroyd in [FH].3.2. Lemma. Suppose d 2 R, K 2 K(X) and dimpK > d. Then there is anon-empty compact set M � K such that dimp(M \ V ) � d for all open sets Vwith M \ V 6= ;.Proof. Let M0 = K. De�ne a trans�nite sequence of compact subsets of K byrecursion as follows. For each ordinal �, let M�+1 = fx 2 M� : dimp(M� \ V ) �d; for all neighborhoods V of xg. For each limit ordinal �, let M� = \�<�M�.Since (M�)�2Ord is a descending trans�nite sequence of compact sets, there is acountable ordinal such thatM =M +1. Note that for each ordinal �, dimp(M�nM�+1) � d: We claim that for each countable ordinal �, dimpM� > d: This iscertainly true for � = 0. Suppose this claim holds for all � < �. If � = � + 1,then since M� = M�+1 [ (M� n M�+1); dimpM�+1 = dimpM� > d: If � is acountable limit ordinal, then since M0 = [�<�(M� nM�+1) [M�; again we havedimpM� > d. Set M =M . Then M 6= ;: If there were some open set V such thatdimp(M \ V ) < d and M \ V 6= ;, then M +1 6=M , a contradiction.

6 PERTTI MATTILA AND R. DANIEL MAULDIN13.3. Lemma. Let c 2 R and K 2 K(X). Then dimpK � c if and only if for everyd < c there is a non-empty compact set M � K such that dimB(M \V ) � d for allopen sets V with M \ V 6= ;.Proof. The \only if" part follows immediately from Lemma 3.2. To verify the \if"part suppose that the stated condition holds and dimpK < c, contrary to whatis asserted. Choose d < c for which dimpK < d. Then there are compact setsK1; K2; : : : such that K � S1i=1Ki and dimBKi < d for all i. Let M � K benon-empty, compact and such that dimB(M \ V ) � d; for all open sets V withM \ V 6= ;. Since M = S1i=1(M \ Ki); the Baire category theorem implies thatM \Ki has non-empty interior relative to M , for some i . Thus there is an openset V with ; 6=M \ V �M \ V �M \Ki, whencedimB(M \ V ) � dimB(M \Ki) < d;which is a contradiction and completes the proof of the lemma.3.4. Theorem. For c 2 R the setA = fK 2 K(X) : dimpK � cgis analytic. In particular, the function dimp : K(X) ! [0;1] is measurable withrespect to B(A).Proof. Let fV1; V2; : : :g be a basis for the topology of X. For m, n 2 N de�neBm;n = f(K;M) 2 K(X)� K(X) : M � K and either M \ Vn = ;or dimB(M \ Vn) � c� 1=mg:Then by Lemma 3.3, A = 1\m=1proj1 � 1\n=1Bm;n�:Thus it su�ces to show that each Bm;n is a Borel set.The sets f(K;M) : M � Kg and fM : M \ Vn = ;g are closed. The functionM 7! N�(M \ Vn) is easily seen to be upper semicontinuous for all � > 0, whenceM 7! dimB(M \ Vn) is Borel measurable. Thus every Bm;n is a Borel set.4. Packing measure functionLet A � X and � > 0. We say that f(xi; ri)gni=1 is a �-packing of A if xi 2 A,� � 2ri > 0 and ri + rj < d(xi; xj) for i, j = 1; : : : ; n, i 6= j. Then the closed ballsB(xi; ri) are disjoint. We �rst de�ne the prepacking measures Pg;� and Pg byPg;�(A) = sup� nXi=1 g(2ri) : f(xi; ri)gni=1 is a �-packing of A�;Pg(A) = lim�!0Pg;�(A):If g is continuous we can replace the condition ri + rj < d(xi; xj) by ri + rj �d(xi; xj) without changing the de�nitions of Pg;� and Pg. The following simplelemma is given only for completeness; it will not be needed later on.

MEASURE AND DIMENSION FUNCTIONS: MEASURABILITY AND DENSITIES 74.1. Lemma. The function Pg;� : K(X)! [0;1] is lower semicontinuous and thefunction Pg : K(X)! [0;1] is of Baire's class 2.Proof. For c 2 R, fK 2 K(X) : Pg;�(K) > cg = 1[n=1Gnwhere K 2 Gn if and only if there exists a �-packing f(xi; ri)gni=1 of K such thatPni=1 g(2ri) > c. Each Gn is open, consequently Pg;� is lower semicontinuous.Hence Pg = limn!1 Pg;1=n is of Baire's class 2.Since Pg is not countably subadditive one needs a standard modi�cation to getan outer measure out of it. Thus we de�ne the packing g-measure for A � X byPg(A) = inf � 1Xi=1 Pg(Ai) : A � 1[i=1Ai�:If A is compact, the sets Ai can also be taken compact. Then Pg is a Borel regularouter measure. When g(r) = rs we denote Pg = Ps. The packing dimension, whichwas introduced in the previous section, can also be de�ned in terms of the packingmeasures: dimpA = inffs : Ps(A) = 0g:For these relations, see e.g., [TT1], [F] or [M]. The proofs there are in Rn but theygeneralize without changes.Let us indicate why dealing with packing measure is somewhat delicate. Thiscan be seen by carrying out a straightforward logical analysis of its de�nition. LetE = fK 2 K(X) : Pg(K) < cg:ThenE = proj1 ��(K;K1; K2; : : : ) 2 K(X)1 : K � 1[i=1Ki and 1Xi=1 Pg(Ki) < c��= proj1(C \D);where C = �(K;K1; K2; : : : ) 2 K(X)1 : K � 1[i=1Ki�and D = �(K;K1; K2; : : : ) 2 K(X)1 : 1Xi=1 Pg(Ki) < c�:It follows from Lemma 4.1 that D is a Borel set. But, from its de�nition, the setC is a coanalytic set. Also, it is not a Borel set. (If it were, then consider C \ F ,where F is the Borel set of all (K;K1; K2; : : : ) such that each Ki is a singleton.The projection of this set on its �rst coordinate would be an analytic set. But thisis the set of all countable closed sets, a classic example proven by Hurewicz not

8 PERTTI MATTILA AND R. DANIEL MAULDIN1to be analytic, see [H]). Thus, this analysis only yields that E is a so-called PCA(or P12) set and whether these types of sets are measurable is independent of theZermelo{Fraenkel axioms of set theory, see [J, pp. 528 and 563].In the following Pr(X) stands for the set of all Borel probability measures onX. We equip Pr(X) with the topology of weak convergence. Then it is a completeseparable metrizable space. The support of a measure � 2 Pr(X) is denoted byspt�.4.2. Theorem. Suppose there is k <1 such that g(2r) � kg(r) for r > 0. Thenfor all c 2 R the set fK 2 K(X) : Pg(K) � cgis analytic. In particular, the function Pg : K(X) ! [0;1] is measurable withrespect to B(A).Proof. We may assume c > 0. We shall make use of the theorem of Joyce andPreiss [JP] according to which for any compact K with Pg(K) � c there exists acompact M � K such that c � Pg(M) < 1. De�ning � 2 Pr(X) by �(B) =Pg(M \ B)=Pg(M) we have spt� � K and c�(L) � Pg(L) for all L 2 K(X). Theconverse of this holds trivially: if there exists � 2 Pr(X) such that spt� � K andc�(L) � Pg(L) for all L 2 K(X), then Pg(K) � c. It follows thatfK 2 K(X) : Pg(K) � cg = proj1AwhereA = f(K;�) 2 K(X)� Pr(X) : spt� � K; c�(L) � Pg(L) for L 2 K(X)g:Thus it su�ces to show that A is a Borel set. We haveA = S \ 1\m=1Amwhere S = f(K;�) 2 K(X)� Pr(X) : spt� � Kgand Am = f(K;�) 2 K(X)� Pr(X) : c�(L) � Pg;1=m(L)for compact sets L � Kg:First, the set S is clearly closed. Secondly, (K;�) 2 Am if and only if for everyj 2 N and every compact set L � K there is a (1=m)-packing f(xi; ri)gni=1 of Lsuch that (c� 1=j)�(L) < nXi=1 g(2ri):For a �xed j the set of all such pairs (K;�) 2 K(X)� Pr(X) is open. To see thisnote that its complement consists of those (K;�) for which there exists a compactset L � K such that for every (1=m)-packing f(xi; ri)gni=1 of L we have(c� 1=j)�(L) � nXi=1 g(2ri);and this set is easily seen to be closed. Thus each Am is a G�-set, and so is A.

MEASURE AND DIMENSION FUNCTIONS: MEASURABILITY AND DENSITIES 94.3. Question. Is Pg B(A)-measurable without the doubling condition g(2r) �kg(r)?In the proof we needed the doubling condition since Joyce and Preiss provedtheir result using it.5. Diameter-based packing measure and dimensionThe de�nitions for the upper box counting dimension and packing dimension ofSection 3 can also be given in terms of the diameters of balls instead of their radii.More precisely, for K 2 K(X) let eN�(K) be the smallest number of open balls ofdiameter at most � that are needed to cover K. If we replace N� by eN� in thede�nitions of dimB and dimp, we get the same dimensions. For packing measuresand the dimensions induced by them the situation is di�erent in general metricspaces, and we look at that in some detail. We restrict here to the gauge functionsg(t) = ts.Let A � X and � > 0. De�neeP s� (A) = sup nXi=1 d(Bi)swhere the supremum is taken over all �nite sequences B1; : : : ; Bn of disjoint openballs centered at A and of diameter at most �. As before, we then de�neeP s(A) = lim�!0 eP s� (A);ePs(A) = inf � 1Xi=1 eP s(Ai) : A � 1[i=1Ai�;and also gdimpA = inffs : ePs(A) = 0g:Clearly, ePs � Ps and gdimp � dimp, and it is easy to give examples of metric spaces,even compact subspaces of R, where both of these inequalities are strict.We would now like to modify the method of Section 3 to prove the measurabilityof gdimp. For that we need a formula giving gdimp in terms of a box counting-typedimension. For K 2 K(X) and j = 1; 2; : : : , let pj(K) be the largest number n ofdisjoint open balls B1; : : : ; Bn centered at K with 2�j�1 < d(Bi) � 2�j . Of course,pj(K) may be zero for some j's. SetgdimBK = lim supj!1 log pj(K)j log 2(with log 0 = �1). Then pj is lower semicontinuous and so gdimB is of Baire's class2.

10 PERTTI MATTILA AND R. DANIEL MAULDIN15.1. Lemma. For K 2 K(X),gdimpK = inf � supi gdimBKi : K � 1[i=1Ki; Ki 2 K(X)�:Only very minor modi�cations are needed to carry out the argument of Tricotfrom [T] for the radius packing measures and dimensions, it is given also in [F] and[M]. We leave the details to the reader. Using Lemma 5.1 one can now argue asbefore to prove the following theorem.5.2. Theorem. The function gdimp : K(X)! [0;1] is B(A)-measurable.5.3. Question. Is ePs B(A)-measurable?The problem here is that in Section 4 we used the theorem of Joyce and Preiss[JP] which does not hold for the measures ePs; a counterexample has been con-structed by Joyce [Jo]. 6. Measurability of sectionsWe shall now study the measurability of the Hausdor� and packing measuresand dimensions of sections in a product space. Let T be a Polish space.If E � T �X, and t 2 T , then let Et = fx 2 X : (t; x) 2 Eg.6.1. Theorem. Let B � T �X be a Borel set. If Bt is �-compact for all t 2 T ,then the functions t 7! Hg(Bt) and t 7! dimH Bt, t 2 T , are Borel measurable.Proof. Assume �rst that each Bt is compact. Then the map t 7! Bt is Borelmeasurable (even upper semicontinuous, see [K2, p. 58]) from T into K(X). Thus,the assertions follow from Theorem 2.1.If each Bt is �-compact we use a result of Saint Raymond [S] to �nd Borelsets B1 � B2 � : : : such that B = S1n=1Bn and (Bn)t is compact for all t 2 T ,n = 1; 2; : : : . As Hg(Bt) = limn!1Hg((Bn)t) and dimH Bt = limn!1 dimH(Bn)t,the theorem follows.6.2. Remark. Dellacherie proved in [D] that t 7! Hs(Bt), and hence also t 7!dimH Bt, is B(A)-measurable provided B is an analytic subset of T �X and T andX are compact metric spaces; his proof easily extends to the case where T and Xare complete and separable. We shall now show that these functions need not beBorel measurable even when B is a Borel subset of [0; 1]� [0; 1].6.3. Example. There exists a Borel set B � [0; 1]� [0; 1] such that the functiont 7! dimH Bt, t 2 [0; 1], is not Borel measurable.Proof. Let E = fK� : 0 � � � 1g be a copy of the Baire space (that is, a spacehomeomorphic to the irrationals) in K([0; 1]) consisting of pairwise disjoint elementsand such that dimH K� > 0 for all � 2 [0; 1]. To see that such a set E exists,consider f , a typical continuous function with respect to the Wiener measure. Foralmost all y in the range of f , the level set f�1fyg has Hausdor� dimension 1=2,see [T]. The map y 7! f�1fyg is one-to-one and Borel measurable. Choose a copyof the Baire space D � [0; 1] on which this map is a homeomorphism, see [K1].Then we can take as E the image of D.

MEASURE AND DIMENSION FUNCTIONS: MEASURABILITY AND DENSITIES 11Let C be a coanalytic subset of [0; 1] which is not a Borel set. By [K1] thereexists a continuous map ' of E onto A = [0; 1] n C.De�neB = f(x; y) 2 [0; 1]� [0; 1] : x = '(K) and y 2 K for some K 2 Eg:Then B is a Borel set. To see this let S = f(x; y;K) : x = '(K) andy 2 Kg. Then S is a Borel set and the projection map onto the �rst two co-ordinates is one-to-one and maps S onto B whence S is a Borel set by [K1, p. 487].Also the projection of B into the x-axis is A and for x 2 A, Bx contains someK 2 E, whence dimH Bx > 0. If x 2 C, then Bx = ; and dimH Bx = 0. Thus forthe map g, g(x) = dimH Bx, the set g�1f0g = C is not a Borel set and so g is nota Borel function.We now turn to the measurability of the packing dimension and measure of thesections. Applying Theorems 3.4, 5.2 and 4.2 we can use exactly the same argumentas in 6.1 to prove the following theorem.6.4. Theorem. Let B � T �X be a Borel set such that all the sections Bt, t 2 T ,are compact.(1) The functions t 7! dimpBt and t 7! gdimpBt, t 2 T , are B(A)-measurable.(2) If g is as in Theorem 4.2 the function t 7! Pg(Bt) is B(A)-measurable.7. An exampleIn this section we show that the packing dimension and measure functions neednot be Borel measurable. For this we shall use continued fractions, see e.g., [R].Any z 2 [0; 1] can be written asz = [b1; b2; : : : ] = 1b1 + 1b2 + : : :where bi 2 N . The sequence b1; b2; : : : is �nite if and only if z is rational. We thenwrite [b1; : : : ; bj] = pj=qjwhere the integers pj and qj are as in [R, p. 136]. The following basic relations holdfor them, see (3) and (5) on page 136 of [R]: de�ning p�1 = 1, q�1 = 0, p0 = b0,q0 = 1, we havepj = bjpj�1 + pj�2; qj = bjqj�1 + qj�2 for j = 1; 2; : : : ;(7.1) pjqj�1 � pj�1qj = (�1)j�1 for j = 1; 2; : : : :(7.2)For irrationals z = [b1; b2; : : : ] we use the fact that the even convergents [b1; b2; : : : ;b2j] increase to z and the odd convergents [b1; b2; : : : ; b2j+1] decrease to z. Also,[b1; : : : ; bj; b] is increasing with b if j is odd and decreasing if j is even. For thedenominators qj we shall need the following simple estimates.

12 PERTTI MATTILA AND R. DANIEL MAULDIN17.3. Lemma. b1 � : : : � bj � qj � 2jb1 � : : : � bj for j = 1; 2; : : : .Proof. Since q1 = b1 and q2 = b1b2+1, this holds for j = 1; 2. Suppose the assertedinequalities are valid for j < m. Then by (7.1)b1 � : : : � bm � bm(b1 � : : : � bm�1) + qm�2 � bmqm�1 + qm�2= qm � bm(2m�1b1 � : : : � bm�1) + 2m�2b1 � : : : � bm�2� 2mb1 � : : : � bm:Thus the lemma follows by induction.We denote by I = [0; 1]nQ the set of irrationals in [0; 1]. Let z = [a1; a2; : : : ] 2 I,and de�ne N(z) = f[a1; k1; a2; k2; : : : ] : ki 2 N for i = 1; 2; : : :g:7.4. Lemma. For z 2 I, dimpN(z) � 1=2.Proof. We shall show that N(z) contains a compact setM such that dimB(M \ V )� 1=2 whenever V is an open set with M \ V 6= ;. This implies dimpN(z) � 1=2by Lemma 3.3.Temporarily �x n1; n2; : : : 2 N , ni � 11. LetM = f[a1; k1; a2; k2; : : : ] : 1 � ki � ni; ki 2 N ; for i = 1; 2; : : :g:Then M is a compact subset of N(z).Let J = J(a1; k1; : : : ; am; km) be the smallest interval containing all the pointsof M whose expansions begin with a1; k1; : : : ; am; km. Using the above mentionedmonotonicity properties of [b1; : : : ; bj; b], we see that J contains the interval withleft endpoint A = [a1; k1; : : : ; am; km; am+1; 2] and right endpoint B = [a1; k1; : : : ;am; km; am+1; nm+1]. Let pi and qi, i = 0; 1; : : : ; 2m+ 1, be the partial numeratorsand denominators generated by the sequence a1; k1; : : : ; am; km; am+1 with p0 = 0,q0 = 1. Then by (7.1) A = u=v where u = 2p2m+1 + p2m and v = 2q2m+1 + q2m.Similarly, B = U=V where U = nm+1p2m+1 + p2m and V = nm+1q2m+1 + q2m.Using (7.2) we estimate the length of the interval [A;B]:B � A = (nm+1 � 2)(p2m+1q2m � p2mq2m+1)(nm+1q2m+1 + q2m)(2q2m+1 + q2m)= nm+1 � 2(nm+1q2m+1 + q2m)(2q2m+1 + q2m)> nm+1 � 2(nm+1 + 1) 3q22m+1 � 14q22m+1 ;as nm+1 � 11.Hence, recalling Lemma 7.3, the length of J satis�esH1�J(a1; k1; : : : ; am; km)� � 14q22m+1� 142m+2(a1 � : : : � am+1)2(n1 � : : : � nm)2:= �m:

MEASURE AND DIMENSION FUNCTIONS: MEASURABILITY AND DENSITIES 13We want to show now that two di�erent intervals as above are disjoint. By thebasic monotonicity properties stated at the beginning of this section, it su�ces toconsider J1 = J(a1; k1; : : : ; km�1; am; k) and J2 = J(a1; k1; : : : ; km�1; am; k0) where1 � k < k0 � nm. These two sequences generate the same pi and qi for i < 2m.The interval J1 lies to the left of the interval J2. The right hand endpoint of J1 isat most B1 = [a1; k1; : : : ; am; k; am+1; nm+1 + 1]:So B1 = U1=V1 where by (7.1)U1 = (nm+1 + 1)�(kam+1 + 1) + k) p2m�1 + am+1p2m�2�+ kp2m�1 + p2m�2= �(nm+1 + 1)(kam+1) + k� p2m�1 + �(nm+1 + 1) am+1 + 1� p2m�2= �p2m�1 + �p2m�2;and similarly, with the same � and �,V1 = �q2m�1 + �q2m�2:The left hand endpoint of J2 is at leastA2 = [a1; k1; : : : ; am; k0; am+1; 1] = u2=v2where, as aboveu2 = �0p2m�1 + �0p2m�2; v2 = �0q2m�1 + �0q2m�2with �0 = k0am+1 + 1 + k0; �0 = am+1 + 1:Thus by (7.2) A2 � B1 = u2V1 � U1v2v2V1= (�0� � ��0)(p2m�1q2m�2 � p2m�2q2m�1)v2V1= �0� � ��0v2V1 :Now �0� � ��0 = (k0am+1 + k0 + 1)�(nm+1 + 1) am+1 + 1�� �(nm+1 + 1)(kam+1 + 1) + k�(am+1 + 1)= �(nm+1 + 1) a2m+1 + 1�(k0 � k)+ (nm+1 + 2) am+1(k0 � k)� nm+1 > 0;whence A2 �B1 > 0 so that J1 and J2 are indeed disjoint.

14 PERTTI MATTILA AND R. DANIEL MAULDIN1It follows now that at least n1 : : : nm intervals of length �m are needed to coverM . Thus dimBM � lim supm!1 logN�m(M)� log �m� lim supm!1 log(n1 : : : nm)(2m+ 2) log 4 + 2 log(a1 : : : am) + 2 log(n1 : : : nm) :Choosing the sequence (nm) to grow su�ciently fast, the last upper limit is 1=2and then dimBM � 1=2. The same argument shows that also dimB(M \ V ) � 1=2for all open sets V with M \ V 6= ;. Thus dimpM � 1=2 as required.For A � R2 let Ax = fy 2 R : (x; y) 2 Ag:7.5. Theorem. a) There exists M 2 K(R2) such that fx 2 R : dimp(Mx) > 0g isan analytic non-Borel set.b) The set E = fK 2 K([0; 1]) : dimpK > 0g is an analytic non-Borel set.Indeed, E is a complete P11 set: if A is an analytic subset of a Polish space X,then there is a Borel measurable map h of X into K([0; 1]) such that h�1(E) = A.(Complete sets are discussed in [KL].)Proof. The results in Sections 3 and 4 show that the sets in question are analytic.For the claim that they are not Borel sets a) clearly implies b), since x 7! Mx isBorel measurable. To prove the theorem it su�ces to prove E is P11 complete.To do this we shall make some additional observations to a beautiful technique ofMazurkiewicz and Sierpinski, [MS].Let A be an analytic subset of a Polish space X. Let f be a continuous map ofI = NN onto A. For z = (a1; a2; : : : ) 2 I, let'(z) = (a1; a3; a5; : : : ):Then ' is a continuous map of I onto itself. Let g = f � ' : I ! A, which is alsocontinuous and onto. Moreoverg�1fxg � N(a1; a2; : : : )whenever x 2 A and f(a1; a2; : : : ) = x. Thus by Lemma 5.1dimp g�1fxg � 1=2 for x 2 A:LetM be the closure of f(g(�); �) : � 2 Ig. ThenM is a closed subset of X� [0; 1].If x 2 A, then Mx � g�1fxg, whence dimpMx � 1=2. If x 2 [0; 1] and y 2 Mx,then there is a sequence �i 2 I such that (g(�i); �i) ! (x; y). If y 2 I, thenx = lim g(�i) = g(y) 2 A. Thus for x 2 [0; 1] n A, Mx � Q and so dimpMx = 0.Hence we have shown that h�1(E) = A, where h(x) =Mx.

MEASURE AND DIMENSION FUNCTIONS: MEASURABILITY AND DENSITIES 158. DensitiesIn this section we shall study the lower densitydg(A; x) = lim infr!0 Pg�A \ B(x; r)�g(2r)of packing measures Pg for x 2 X and A � X. We shall always assume that thegauge function g satis�es the doubling condition(8.1) g(2r) � kg(r) for r > 0with some k <1. We also denoteds(A; x) = dg(A; x) when g(r) = rs:If A � Rn and Pg(A) <1, thendg(A; x) = 1 for Ps almost all x 2 Aaccording to [ST, Corollary 7.2]. We shall show in Example 8.9 that this is false evenfor g(r) = rs in any in�nite-dimensional Hilbert space. However, in Theorem 8.3 wederive the optimal inequalities 2�s � ds(A; x) � 1. For this we need a modi�cationof well-known covering lemmas.8.2. Lemma. Let 0 < � < 1=2, let B be a collection of closed balls in X such thatsupfd(B) : B 2 Bg <1;and suppose that A � X is such that every x 2 A is a center of some B 2 B. Thenthere exist B(xi; ri) 2 B, i = 1; 2; : : : , such that the balls B(xi; �ri) are disjoint andA �[i B(xi; ri):Proof. For each x 2 A choose some B(x) = B�x; r(x)� 2 B. SetM = supfr(x) : x 2 Ag:Let 2� < t < 1 and de�neA1 = fx 2 A : tM < r(x) �Mg:Choose a subset B1 � A1 which is maximal with respect to the property:if x; y 2 B1 and x 6= y, then x =2 B(y) or y =2 B(x).Then A1 � Sx2B1 B(x) and the balls B�x; �r(x)�, x 2 B1, are disjoint. Next letA2 = �x 2 A n [x2B1B(x) : t2M < r(x) � tM�and choose a maximal subset B2 � A2 such that for any x; y 2 B2 with x 6= y,x =2 B(y) or y =2 B(x). It follows thatA1 [ A2 � [x2B1[B2B(x)and the balls B�x; �r(x)�, x 2 B1 [B2, are disjoint. Continuing in this manner we�nd the desired balls.

16 PERTTI MATTILA AND R. DANIEL MAULDIN18.3. Theorem. Let A � X and Ps(A) <1. Then2�s � ds(A; x) � 1 for Ps almost all x 2 A:Proof. The right hand inequality follows with essentially the same proof which wasused in [M, 6.10], or one can consult Cutler [Cu] who proved this for Pg withgeneral gauge functions. To prove the left hand inequality, we may assume that Ais a Borel set by the Borel regularity of Ps. Given � > 2 it is enough to show thatds(A; x) � ��s for Ps almost all x 2 A. Let t < ��s and letB � fx 2 A : ds(A; x) < tg � At:Given � > 0 we can apply Lemma 8.2 to �nd disjoint balls B(xi; ri=�) such thatxi 2 B, 2ri < �, Ps�A \ B(xi; ri)� < t(2ri)s andB � 1[i=1B(xi; ri):Then Ps(B) � 1Xi=1 Ps�B \ B(xi; ri)�� t 1Xi=1(2ri)s = t�s 1Xi=1(2ri=�)s � t�sP s� (B):Letting � ! 0 we have Ps(B) � t�sP s(B) for all B � At, which implies Ps(At) �t�sPs(At). As t�s < 1, we obtain Ps(At) = 0 for all t < ��s, and this yields thatds(A; x) � ��s for Ps almost all x 2 A as desired.8.4. Remark. The inequalitiesc � dg(A; x) � 1 for Ps almost all x 2 Ahold for any g satisfying (8.1) if Pg(A) < 1. Here c is a positive constant (notnecessarily optimal) depending on g. This was proved by Cutler in [Cu]. The aboveproof also applies.Next we show that instead of inequalities we have equalityHg almost everywherein A. Recall that we are assuming (8.1).8.5. Theorem. If A � X with Pg(A) <1, thendg(A; x) = 1 for Hg almost all x 2 A.Proof. By the Borel regularity of Pg we may assume that A is a Borel set. If ourclaim is false there are t < 1 and a Borel set C � A such that Hg(C) > 0 and

MEASURE AND DIMENSION FUNCTIONS: MEASURABILITY AND DENSITIES 17dg(A; x) < t for x 2 C (recall Remark 8.4). As Hg � Pg (see [Cu, 2.6], also theproof of this fact in Rn for g(r) = rs in [M, 5.12] generalizes without di�culty), wecan use the Radon{Nikodym theorem to �nd a non-negative Borel function f on Csuch that Hg(B) = ZB f dPgfor Borel sets B � C. Since Hg(C) > 0, there are u > 0 and a Borel set D � Csuch that Hg(D) > 0 and f � u on D which yields(8.2) Hg(B) � uPg(B) for B � D.Let � > 0. By the analogue of Theorem 8.3 for g(r) = rs, recall Remark 8.4,c < dg(D; x) < t for Ps almost all x 2 D. For the upper density with respect tothe Hausdor� measure we have, cf. [Fe, 2.10.18 (3)], also the proof of [M, 6.2] forg(r) = rs generalizes easily,(8.3) lim supr!0 Hg�D \B(x; r)�=g(2r) � 1for Hg almost all x 2 X. From (8.2) we see that (8.3) also holds for Pg almost allx 2 D. Let B � D be a Borel set. By a standard covering theorem, see e.g., [Fe,2.8.6], we can �nd xi 2 B and 0 < ri < �=2 such that the balls Bi = B(xi; ri) aredisjoint, B � n[i=1B(xi; ri) [ 1[i=n+1B(xi; 5ri) for all n = 1; 2; : : : ;cg(2ri) � Pg(D \Bi) � tg(2ri) andHg�D \ B(xi; 5ri)� � 2g(10ri):Thus recalling also (8.1) and (8.2) we havePg(B) � nXi=1 Pg(D \ Bi) + 1Xi=n+1Pg�D \B(xi; 5ri)�� t nXi=1 g(2ri) + u�1 1Xi=n+1Hg�D \B(xi; 5ri)�� tPg;�(B) + 2k3u�1 1Xi=n+1 g(2ri)� tPg;�(B) + 2k3u�1c�1 1Xi=n+1Pg(D \Bi):Here 1Xi=n+1Pg(D \ Bi) � Pg(A) <1:

18 PERTTI MATTILA AND R. DANIEL MAULDIN1Thus letting �rst n!1 and then � ! 0, we obtainPg(B) � tPg(B)for all Borel sets B � D. This implies0 < Hg(D) � Pg(D) � tPg(D) <1which is a contradiction since t < 1.8.6. Corollary. Let g satisfy the doubling condition. Let A � X with Pg(A) <1.Then Hg(A) = Pg(A) if and only iflimr!0 Hg�A \ B(x; r)�g(2r) = 1 for Pg almost all x 2 A.The proof of Theorem 6.12 of [M] applies without any essential changes whenwe have Theorem 8.5 at our disposal.8.7. Remark. An interesting application of Corollary 8.6 in Rn is the followingtheorem of Saint Raymond and Tricot. Let A � Rn with 0 < Ps(A) < 1. ThenHs(A) = Ps(A) if and only if s is an integer and A is Ps-recti�able in the sensethat Ps almost all of A can be covered with countably many Lipschitz images ofRs , see [ST] or [M]. Attempts to extend this to more general metric spaces startingfrom 8.6 boil down to the following: in which metric spaces X is it true that ifA � X with Hs(A) <1, thenlimr!0 Hs�A \ B(x; r)�(2r)s = 1 for Hs almost all x 2 Aif and only if s is an integer and Hs almost all of A can be covered with countablymany Lipschitz images of Rs? Kirchheim proved in [K] that the \if" part is validin any metric space. Chleb��k has given in [C] a proof which shows that the \onlyif" part is valid at least in all in�nite-dimensional inner product spaces. Thuscombining these results with Corollary 8.6 the argument to prove Theorem 17.11 in[M] shows that the theorem of Saint{Raymond and Tricot is valid in Hilbert spaces.8.8. Theorem. Let X be a Hilbert space, s > 0 and A � X with 0 < Ps(A) <1.Then Hs(A) = Ps(A) if and only if s is an integer and A is Ps-recti�able.Now we construct an example in an in�nite-dimensional space to show that thelower bound in Theorem 8.3 is the best possible.8.9. Example. Let X be an in�nite-dimensional Hilbert space and s > 0. Thenthere is a compact set K � X such that 0 < Ps(K) <1 andds(K;x) � 2�s for x 2 K:Proof. Let e1; e2; : : : be an orthonormal basis of X. Let (mk) be a sequence ofpositive integers such that m1 > (1 +p2)s(8.4) mk %1 and 1Xk=1m�1k =1:

MEASURE AND DIMENSION FUNCTIONS: MEASURABILITY AND DENSITIES 19De�ne the positive numbers �k by(8.5) mk�sk = �sk�1 with �0 = 1:Then �k=�k�1 ! 0. For i = 1; : : : ;m1 setx(i) = �1ei and B(i) = U(xi; �1=p2)where U(x; r) denotes the open ball with center x and radius r. Suppose then thatfor some k 2 N the points x(i1; : : : ; ik) have been de�ned. We putx(i1; : : : ; ik; i) = x(i1; : : : ; ik) + �k+1ei; i = 1; : : : ;mk+1:Set B(i1; : : : ; ik) = B�x(i1; : : : ; ik); �k=p2�:Let = Q1i=1f1; : : : ;mig have the product topology. The system fB(i1; : : : ; ik) :1 � ij � mj ; j � k; k 2 Ng of open balls is a Cantor scheme:(8.6)for each k the balls B(i1; : : : ; ik), 1 � ij � mj , 1 � j � k, are pairwise disjoint.(8.7) for each k and 1 � i � mk+1,B(i1; : : : ; ik; i) � B(i1; : : : ; ik):To verify these, recall that mk > (1 +p2)s. Moreover,(8.8) d�B(i1; : : : ; ik; i)� = (p2�k+1)sand(8.9) mk+1Xi=1 d�B(i1; : : : ; ik; i)�s = d�B(i1; : : : ; ik)�s:De�ne K = 1\k=1 [i1:::ikB(i1; : : : ; ik) = 1\k=1 [i1:::ik B(i1; : : : ; ik):Then(8.10) K \ B(i1; : : : ; ik) � B�x(i1; : : : ; ik); 2�k+1�:Let g be the coding map from the coding space onto K. Thus,fg(�)g = 1\k=1B(�jk)

20 PERTTI MATTILA AND R. DANIEL MAULDIN1for � = (i1; i2; : : : ) 2 where �jk = (i1; : : : ; ik). By properties (8.6) and (8.7), gis a homeomorphism of the Cantor space onto K and g maps the cylinder set[i1; : : : ; ik] onto K \B(i1; : : : ; ik). De�ne e� on the cylinder sets in bye�([i1; : : : ; ik]) = (p2�k)s:By properties (8.8) and (8.9), e� satis�es Kolmogorov's consistency conditions. Lete� also denote the extension of e� to a Borel measure on . Let � be the imagemeasure on K of e� via the coding map g. Thus(8.11) ��K \B(i1; : : : ; ik)� = (p2�k)s = d�B(i1; : : : ; ik)�s:We use � to show that Ps(K) > 0. Let A � K be a Borel set and � > 0. Thenthere is a disjoint subfamily �Bj = U(xj ; rj) of the family fB(i1; : : : ; ik)g suchthat d(Bj) < �, Bj \A 6= ;, say xj 2 Bj \A, and A � Sj Bj . In fact, by choosingfor every x 2 B the largest B = B(i1; : : : ; ik) such that x 2 B and d(B) < �, we�nd the sequence (Bj). Using (7) and the fact that �k+1=�k ! 0 we see that forsome r0j and x0j with rj � �1+o(�)� r0j , x0j 2 K we have B0j = B(xj; r0j) � Bj . Thusby (8.11), �(A) �Xj �(Bj) =Xj d(Bj)s � �1 + o(�)�Xj d(B0j)s� �1 + o(�)�P s� (A):Hence �(A) � P s(A) for all Borel sets A � K, which gives0 < �(K) � Ps(K)as desired.We now show that Ps(K) < 1 and prove the lower density estimate. Both ofthese will follow from(8.12) P s�K \ B(�j`)� � (p2�`)s for � 2 , ` = 1; 2; : : : :(In fact, it is easy to show that also the converse inequality in (8.12) holds.) Let� 2 , ` � 1, 0 < � < �`+1 and let D1; : : : ; Dn be disjoint closed balls centered atK \ B(�j`) with(8.13) dj = d(Dj) < �:We want to show that(8.14) Xj dsj � �1 + o(�)�(p2�`)swhich clearly yields (8.12). For each j let kj be the unique integer such that(8.15) 2p2�kj+1 < dj � 2p2�kj ;

MEASURE AND DIMENSION FUNCTIONS: MEASURABILITY AND DENSITIES 21and let Bj = B(i1; : : : ; ikj ) be such that �jkj = (i1; : : : ; ikj ) and K \Bj \Dj 6= ;.Since �k+1=�k ! 0 as k !1, we may assume, in order to prove (8.14), that2p2�kj+1 + 4�kj+2 < dj < 2p2�kj � 4�kj+1:This has the e�ect that K \ Bj � Dj , whence the balls Bj are disjoint. It followsthen from (8.6){(8.9) that(8.16) X d(Bj)s � d�B(�j`)�s = (p2�`)s:Set I = fj : dj � p2�kj + 4�kj+1g andJ = fj : dj > p2�kj + 4�kj+1g:Then by (8.16)Xj2I dsj �Xj2I (p2�kj + 4�kj+1)s(8.17) � �1 + o(�)�Xj2I(p2�kj )s = �1 + o(�)�Xj2I d(Bj)s� �1 + o(�)�(p2�`)s:For each k every ball B(i1; : : : ; ik�1) can contain at most one Dj with kj = kand j 2 J because of the disjointness. Thus there are at most m1; : : : ;mk�1 indicesj 2 J with kj = k. Hence by (8.15) and (8.5)Xj2J;kj=k dsj � m1 : : :mk�1(2p2�k)s = (2p2)s=mk:So by (8.15) and (8.4),(15) Xj2J dsj � (2p2)s X�k+1<�m�1k ! 0:Thus (8.17) and (8.18) yield (8.14) and hence also (8.12).Let � 2 and x = g(�) 2 K. By the construction and (8.12),limk!1 P s�K \ B(x;p2�k � 4�k+1)��2(p2�k � 4�k+1)�s = limk!1 P s�K \ B(�jk)��2(p2�k � 4�k+1)�s � 2�s;which completes the proof since Ps � P s.

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