Tensor products of AC_{∗} charges and AC Radon measures are not always AC_{∗} charges

Zoltán Buczolich^{∗}, Department of Analysis, Eötvös Loránd University, Kecskeméti u. 10-12, Budapest, Hungary, 1053

August 17, 2000

Abstract

In this note we give an example of an AC_{∗} charge, F, on ℝ and an absolutely continuous Radon measure μ on ℝ such that F ⊗ μ is not an AC_{∗} charge on ℝ^{2}.

1 Introduction

In [1] the tensor problem was stated for the tensor product of AC_{∗} charges and the Lebesgue measure. Later W.F. Pfeffer suggested to look at possible generalizations of this problem. In this note we show that a more general version of this problem is false. We also give a version of the tensor theorem which seems to be true and will be the subject of a forthcoming paper of this author.

The generalized version of the tensor problem is the following (for the details of the definitions see the Preliminaries section of this paper or [1]):

Let μ be an absolutely continuous Radon measure in ℝ^{n}, and let F be a charge in ℝ^{m} where m and n are positive integers. (In this paper absolute continuity is always considered with respect to the Lebesgue measure.) Given a bounded BV set B ⊂ ℝ^{m+n}, let

B^{y} = {x ∈ ℝ^{m} : (x, y) ∈ B}

and

(F ⊗ μ)(B) = ∫_{ℝ^{n}} F (B^{y})dμ(y).

It is not difficult to see that F ⊗ μ is a charge and and the question is whether F ⊗ μ is AC_{∗} in E × ℝ^{n} whenever F is AC_{∗} in a locally BV set E ⊂ ℝ^{m}.

∗This author was supported by Grants FKFP 0192/1999 and Hungarian National Foundation for Scientific Research Grant No. T 032042

1991 Mathematics Subject Classification: Primary 26B30 Secondary: 26B05, 46B28.

Keywords: non-absolute integrals, variational measure, tensor product.

This question is of interest because for nonabsolute integrals usually there is no Fubini’s theorem and hence results concerning products are useful.

In this note we give a counterexample in the simplest possible setting. The charge F will be an AC_{∗} charge on ℝ (in fact, it will be supported on [0, 1]) and μ will be an absolutely continuous Radon measure supported on [0, 1]. Their tensor product F ⊗ μ will not be an AC_{∗} charge on ℝ^{2}. More sophisticated similar examples can be obtained in higher dimensions as well.

It is easy to see that if μ has an atomic part then the tensor conjecture is false. Our example shows that even for absolutely continuous measures one can obtain counterexamples. In our construction we give a μ such that its density function (its Radon-Nikodym derivative) ϕ is unbounded. It seems to us that the conjecture on tensor products is true if we assume that ϕ ∈ L^{∞}(ℝ^{n}).

2 Preliminaries

We denote by d(A), and |A| the diameter and the Lebesgue measure of the set A ⊂ ℝ^{m}, respectively.

The open ball centered at x and of radius r will be denoted by B(x, r). Closed intervals in ℝ^{m} are Cartesian products of closed one-dimensional

intervals. Figures in ℝ^{m} are finite unions of closed m-dimensional intervals.

The Lebesgue density points (see for example [2] Corollary 6.2.6) of a measurable set A ⊂ ℝ^{m} form its essential interior, int^{∗}A. The Lebesgue dispersion points (zero-density points) of A form its essential exterior ext^{∗}A. The essential boundary of A equals ∂^{∗}A =^{def}ℝ^{m} \ (int^{∗}A ∪ ext^{∗}A).

The measurable set A ⊂ ℝ^{m} is a BV set if its perimeter ||A|| =^{def}H^{m−1}(∂^{∗}A) is finite (here H^{m−1} denotes the (m − 1)-dimensional Hausdorff measure).

One-dimensional BV sets coincide (modulo sets of zero Lebesgue measure) with finite unions of closed intervals, that is, with one-dimensional figures. Recalling that the zero-dimensional Hausdorff measure is the counting measure if the BV set A ⊂ ℝ is equivalent to ∪_{j=1}^{k}I_{j} (where the I_{j}’s are disjoint closed intervals) then the essential boundary of A coincides with the endpoints of the intervals I_{j} and ||A|| = 2k (the number of the endpoints of these intervals).

A function F defined on m-dimensional BV -sets is an m-dimensional charge if it is

1. additive, that is F (A∪B) = F (A)+F (B) holds for any two nonoverlapping BV sets A and B (recall that A and B are nonoverlapping if int^{∗}A ∩ int^{∗}B = ∅);

2. continuous, that is for every ϵ > 0 there exists ρ > 0 such that |F (A)| < ϵ if |A| < ρ, ||A|| < 1/ϵ and A ⊂ B(0, 1/ϵ).

Assume E ⊂ ℝ^{m}, δ : E → [0, ∞) is a gage on E if {x : δ(x) = 0} is thin, that is, it is of σ-finite (m − 1)-dimensional Hausdorff measure.

A system P = {(A_{i}, x_{i})}_{i=1}^{p} is a partition if the A_{i}’s are nonoverlapping BV sets and x_{i} ∈ A_{i} for i = 1, ..., p. The partition P is anchored in E if x_{i} ∈ E for all i = 1, ..., p. If a gage δ is given on E then P is δ-fine if A_{i} ⊂ B(x_{i}, δ(x_{i})) holds for all i = 1, ..., p.

The regularity, r(A), of a BV -set A ⊂ ℝ^{m} is the number |A|/d(A)||A|| if d(A)||A|| ≠ 0, otherwise r(A) =^{def}0. If η > 0 and r(A) > η we say that A is η-regular. A partition P is η-regular if r(A_{i}) > η for all i = 1, ..., p.

If a one-dimensional BV set A ⊂ ℝ is η-regular then

d(A) ≥ |A| > ηd(A)||A|| and hence 1/η ≥ ||A||,  (1)

that is, A is equivalent to a finite union of less than [1/2η] many closed intervals. Since ||A|| ≥ 2 we also have

|A| > ηd(A)||A|| ≥ 2ηd(A).  (2)

Assume F is a charge defined on BV sets of ℝ^{m}, η > 0, E ⊂ ℝ^{m} and δ : E → [0, ∞) is a gage. Set

V_{η,δ}F (E) = sup_{P} ∑_{i=1}^{p} |F (A_{i})|,

where the supremum is taken for all η-regular, δ-fine partitions P = {(A_{i}, x_{i})}_{i=1}^{p}. The F -variational outer measure of E is defined as

V_{∗}F (E) = sup_{0<η} inf_{δ} V_{η,δ}F (E),

where the infimum is taken for all δ gage functions defined on E. It is not difficult to show that V_{∗}F is a metric outer measure.

We say that F is an AC_{∗} (m-dimensional) charge if V_{∗}F is absolutely continuous with respect to the Lebesgue measure (in ℝ^{m}).

3 The main result

Theorem 1. There exists a charge F defined on BV sets of ℝ and an absolutely continuous Radon measure μ in ℝ, such that F ⊗ μ is not an AC_{∗} charge in ℝ^{2}.

Proof. Let C be the triadic Cantor-set. We denote by C_{k} the union of the 2^{k} many closed intervals of length 3^{−k} which are used in the definition of C at step k. We denote these intervals by G_{1,k}, ..., G_{2^{k},k}. We denote by G′_{1,k}, ..., G′_{2^{k},k} the intervals which are concentric with G_{1,k}, ..., G_{2^{k},k}, respectively, and which are of length

h_{k} = \frac{1}{k^{2}2^{k+1}3^{3k+2}} .

Observe that C_{k+1} ∩ ∪_{j=1}^{2^{k}}G′_{j,k} = ∅.

G_{1,1}  G_{2,1}  C_{1} C_{2} G_{1,2} G′_{1,1} G_{2,2}  G_{3,2} G′_{2,1} G_{4,2}

figure 1.

On G′_{j,k} we set for each j = 1, ..., 2^{k}

ϕ(x) = \frac{1}{h_{k}k^{2}2^{k}} .

Observe that

∫_{∪_{j}G′_{j,k}} ϕ(x)dx = 2^{k} \frac{h_{k}}{h_{k}k^{2}2^{k}} = \frac{1}{k^{2}} .

Set ϕ(x) = 0 for x ∉ ∪_{k=1}^{∞} ∪_{j=1}^{2^{k}} G′_{j,k}.

Put μ(A) = ∫_{A} ϕ(x)dx. Observe that μ is an absolutely continuous Radon measure with support in [0, 1] and

μ(G′_{j,k}) = \frac{1}{k^{2}2^{k}} .  (3)

for each k ∈ ℕ and j = 1, ..., 2^{k}.

To define the charge F we need an auxiliary Cantor type set E. We define this set as the intersection of some auxiliary nested closed sets E_{k} and each E_{k} will be a union of nonoverlapping closed intervals, each of length 3^{−k^{2}}.

Set E_{0} = I_{1,0} = [0, 1], N_{0} = 1.

Assume k ≥ 0, E_{k} is defined and equals the union of nonoverlapping closed intervals I_{j,k}, j = 1, ..., N_{k}, each of length 3^{−k^{2}}.

To define E_{k+1} divide each I_{j,k} into 3^{−k^{2}}/3^{−(k+1)^{2}} = 3^{(k+1)^{2}−k^{2}} many nonoverlapping intervals, each of length 3^{−(k+1)^{2}}, keep the first 3^{(k+1)^{2}−k^{2}} −1 of them, and delete the last one. Repeating this procedure in each I_{j,k} we obtain the closed intervals I_{j′,k+1}, j′ = 1, ..., N_{k+1}. Where N_{k+1} = (3^{(k+1)^{2}−k^{2}} − 1)N_{k}. For j = 1, ..., N_{k} denote by J_{j,k} the last, deleted, subinterval of I_{j,k}. Set E_{k+1} = ∪_{j′=1}^{N_{k+1}}I_{j′,k+1}.

[figure: ...; I_{j′,k+1} intervals; J′_{j,k}; J_{j,k}; I_{j,k}]

figure 2.

By induction we define E_{k} for all k ∈ ℕ and set E = ∩_{k=1}^{∞}E_{k}.

Observe that for k ≥ 1

|E_{k}| = ∏_{j=1}^{k} (1 − 3^{(j−1)^{2}−j^{2}})

and hence |E| > 0.

Assume that k is fixed and we choose a j ∈ {1, ..., N_{k}}. Then the interval I_{j,k} contains a subinterval J_{j,k} of length 3^{−(k+1)^{2}}. Denote by J′_{j,k} the subinterval of J_{j,k} which is concentric with it and of length (1/3)|J_{j,k}|. Observe that dist(J′_{j,k}, E_{k+1}) = (1/3)|J_{j,k}| = 3^{−(k+1)^{2}−1}. Set

κ_{k} = k^{4}2^{k^{2}+1}3^{(k+1)^{2}}.

Divide J′_{j,k} into κ_{k} many equal subintervals, J_{i,j,k}, i = 1, ..., κ_{k}. Then |J_{i,j,k}| = 3^{−(k+1)^{2}−1}/κ_{k} for all i.

For i = 1, ..., κ_{k}/2 on int(J_{2i,j,k}) set f(x) = |J_{j,k}|/|J_{2i,j,k}|; on int(J_{2i−1,j,k}) set f(x) = −|J_{j,k}|/|J_{2i−1,j,k}|. If x ∉ ∪_{k} ∪_{j} ∪_{i}intJ_{i,j,k} then set f(x) = 0.

Observe that if [a, b] is an arbitrary interval and a, b ∉ J_{j,k} then

∫_{[a,b]∩J′_{j,k}} f(x)dx = 0.

If A is a one-dimensional BV set put

F (A) = ∑_{k=0}^{∞} ∑_{j=1}^{N_{k}} ∫_{A∩J_{j,k}} f(x)dx.

By the above observation all but finitely many terms of the double sum in the definition of F equal zero, hence F (A) is well-defined.

Next we need to verify that F is an AC_{∗} charge.

Assume that the BV set A is equivalent (modulo a set of measure zero) to the disjoint closed intervals Z_{1}, ..., Z_{t}. We will denote this property later by A ∼ ∪_{s=1}^{t}Z_{s}.

If no endpoint of an interval Z_{s} belongs to an interval of the form J′_{j,k} then

F (Z_{s}) = ∑_{k=0}^{∞} ∑_{j=1}^{N_{k}} ∫_{Z_{s}∩J_{j,k}} f = 0.  (4)

If at least one of the endpoints of Z_{s} belongs to an interval of the form J_{j_{s},k_{s}} then, without limiting generality, we can assume that the other endpoint of Z_{s} either does not belong to an interval of the form J′_{j,k} ⊂ J_{j,k}, or if it belongs to such an interval then k ≥ k_{s}. It is easy to see that in this case

|F (Z_{s})| ≤ 2|J_{j_{s},k_{s}}|.  (5)

First we show that F is a charge. It is clearly additive. Assume ϵ > 0 is given. Without limiting generality we assume ϵ < 1. We need to choose ρ > 0 such that if the BV set A ⊂ B(0, 1/ϵ), ||A|| < 1/ϵ and |A| < ρ then |F (A)| < ϵ.

Assume A ∼ ∪_{s=1}^{t}Z_{s}. Then ||A|| < 1/ϵ implies t < 1/2ϵ. Choose k_{0} such that

|J_{1,k_{0}}| = |J_{j,k_{0}}| < ϵ^{2}/2,  j = 1, ..., N_{k_{0}}.  (6)

Since f is bounded on H_{k_{0}} =^{def} ∪_{k=1}^{k_{0}−1} ∪_{j=1}^{N_{k}}J_{j,k} we can choose ρ_{1} > 0 such that if the interval Z_{s} ⊂ H_{k_{0}} and |Z_{s}| < ρ_{1} then | ∫_{Z_{s}} f| = |F (Z_{s})| < ϵ^{2}.

If ∫_{Z_{s}∩H_{k_{0}}} f ≠ 0 then there exists k ∈ {1, ..., k_{0} − 1} and j ∈ {1, ..., N_{k}} such that Z_{s}∩J′_{j,k} ≠ ∅. This implies that if we assume that |Z_{s}| ≤ |A| < ρ_{2} = |J_{1,k_{0}}|/3 then Z_{s} ⊂ H_{k_{0}}.

Therefore if ∫_{Z_{s}∩H_{k_{0}}} f ≠ 0 then |Z_{s}| ≤ |A| ≤ ρ =^{def} min(ρ_{1}, ρ_{2}) implies that |F (Z_{s})| < ϵ^{2}.

If ∫_{Z_{s}∩H_{k_{0}}} f = 0, but F (Z_{s}) ≠ 0 then an esimate of the form (5) holds with a k_{s} ≥ k_{0}. Using (6) we obtain

|F (Z_{s})| ≤ 2|J_{j_{s},k_{s}}| ≤ 2|J_{1,k_{0}}| < ϵ^{2}.

This implies

|F (A)| ≤ ∑_{s=1}^{t} |F (Z_{s})| ≤ ϵ^{2}/2ϵ = ϵ/2 < ϵ.

Next we show that F is AC_{∗}.

Since V_{∗}F is a Borel measure, V_{∗}F (ℝ \ [0, 1]) = 0, and one can easily see that V_{∗}F is absolutely continuous on each set J_{j,k}, k = 0, 1, ...; j = 1, ..., N_{k} it is enough to show that it is absolutely continuous on E. Assume S ⊂ E, |S| = 0 and η > 0 is fixed and P = {(A_{i}, x_{i})}_{i=1}^{p} is an η-regular partition, anchored in S ⊂ E. Assume, furthermore, that A_{i} ∼ ∪_{s=1}^{t_{i}}Z_{s,i}. Then (1) implies t_{i} ≤ 1/2η and (2) implies |A_{i}| > 2ηd(A_{i}).

Now, if F (Z_{s,i}) ≠ 0, that is, (4) does not hold then we can use (5) to obtain

|F (Z_{s,i})| ≤ 2|J_{j_{s,i},k_{s,i}}|.

Since x_{i} ∈ A_{i}, x_{i} ∈ S ⊂ E, and E ∩ int(J_{j_{s,i},k_{s,i}}) = ∅ we also have d(A_{i}) ≥ |J_{j_{s,i},k_{s,i}}|/3.

Hence, keeping in mind that ||A_{i}|| ≥ 2, we obtain

|A_{i}| ≥ ηd(A_{i})||A_{i}|| ≥ 2η|J_{j_{s,i},k_{s,i}}|/3 ≥ η|F (Z_{s,i})|/3.

Since t_{i} ≤ 1/2η we obtain

|F (A_{i})| ≤ ∑_{s=1}^{t_{i}} |F (Z_{s,i})| ≤ t_{i} \frac{3|A_{i}|}{η} ≤ \frac{3|A_{i}|}{2η^{2}}.

Cover S by an open set G such that |G| < η^{2}ϵ/3 and for x ∈ H choose δ(x) > 0 satisfying B(x, δ(x)) ⊂ G. If P is δ-fine then ∪_{i=1}^{p}A_{i} ⊂ G and hence

∑_{i=1}^{p} |F (A_{i})| ≤ ∑_{i=1}^{p} \frac{3|A_{i}|}{2η^{2}} < 3|G|/η^{2} < ϵ.

This shows that using this δ we have V_{η,δ}F (S) < ϵ. Using that for each ϵ > 0 one can choose a suitable δ and one can repeat the above argument for all η > 0 we obtain that V_{∗}F (S) = 0. This completes the proof of the fact that F is AC_{∗}.

Finally we show that F ⊗ μ is not AC_{∗}. Using Chapter 7 of [3] one can easily see that the Hausdorff dimension of C × E is 1 + log 2/ log 3, and hence |C × E| = 0 but it is not of σ-finite one-dimensional Hausdorff measure, that is, it is not thin.

We show that V_{∗}F (C × E) ≠ 0.

Assume η is sufficiently small (something like η < 1/20 is suitable). Since C × E is not thin if δ is an arbitrary two dimensional gage function then there exists an x ∈ C × E for which δ(x) > 0. Choose and fix such an x.

Choose k such that 3^{−k^{2}} < δ(x)/2 and assume that G_{x} is the base interval of C_{k^{2}} which contains x and is of length 3^{−k^{2}}, and we denote by G′_{x} the corresponding middle third interval of G_{x}. Recall that G′_{x} is of length h_{k^{2}}, and it was used in the definition of ϕ and μ, furthermore by (3) we have μ(G′_{x}) = k^{−4}2^{−k^{2}}.

[figure: A; J′_{x}; J_{x}; I_{x}; length of I_{x} is 3^{−k^{2}}; G_{x}; G′_{x}; x]

figure 3.

Similarly, assume that I_{x} is the base interval I_{j,k} of length 3^{−k^{2}} which was used in the definition of E_{k} and J_{x} is the corresponding interval J_{j,k} of length 3^{−(k+1)^{2}}. Furthermore, the middle third of J_{x}, J′_{j,k} will be denoted by J′_{x} and the intervals J_{i,j,k}, used in the definition of f will be denoted by J_{x,i} (i = 1, ..., κ_{k}).

Observe that C_{k^{2}+1} ∩G_{x} consists of two intervals, each of length |G_{x}|/3, and G′_{x} ∩ C_{k^{2}+1} = ∅.

Set A = (I_{x} × (C_{k^{2}+1} ∩ G_{x})) ∪ ((∪_{i=1}^{κ_{k}/2}J_{x,2i}) × G′_{x}).

Then ||A|| ≤ 6 · 3^{−k^{2}} + κ_{k}h_{k^{2}} < 7 · 3^{−k^{2}}, d(A) < \sqrt{2} · 3^{−k^{2}}, and |A| > (2/3)3^{−2k^{2}}. This implies that A is η-regular. Since A ⊂ G_{x} × I_{x} ⊂ B(x, δ(x)), the one element partition P = {(A, x)} is η-regular and δ-fine.

On the other hand

(F ⊗ μ)(A) = μ(C_{k^{2}+1} ∩ G_{x})F (I_{x}) + μ(G′_{x})\frac{κ_{k}}{2} ∫_{J_{x,2}} f =

0 + k^{−4}2^{−k^{2}} \frac{κ_{k}}{2} |J_{x,2}|\frac{|J_{x}|}{|J_{x,2}|} = \frac{κ_{k}}{k^{4}2^{k^{2}+1}3^{(k+1)^{2}}} = 1.

This implies V_{η,δ}(F ⊗μ)(C×E) ≥ 1 holds for all gage δ and hence V_{∗}F (C×E) ≥ 1. Thus V_{∗}(F ⊗ μ) is not absolutely continuous.

References

[1] Z. Buczolich, T. de Pauw, and W.F. Pfeffer, Charges, BV functions, and multipliers for generalized Riemann integrals, Indiana Univ. Math. J., Vol. 48, No. 4, (1999), 1471-1512.

[2] D.L. Cohn, Measure Theory, Birkhäuser, Boston 1980.

[3] K. Falconer, Fractal Geometry, John Wiley & Sons, 1990.