Is Lebesgue measure the only σ-finite invariant Borel measure?

Márton Elekes^{∗} and Tamás Keleti^{†}

August 31, 2005

Abstract S. Saks and recently R. D. Mauldin asked if every translation invariant σ-finite Borel measure on ℝ^{d} is a constant multiple of Lebesgue measure. The aim of this paper is to investigate the versions of this question, since surprisingly the answer is “yes and no”, depending on what we mean by Borel measure and by constant. According to a folklore result, if the measure is only defined for Borel sets then the answer is affirmative. We show that if the measure is defined on a σ-algebra containing the Borel sets then the answer is negative. However, if we allow the multiplicative constant to be infinity, then the answer is affirmative in this case as well. Moreover, our construction also shows that an isometry invariant σ-finite Borel measure (in the wider sense) on ℝ^{d} can be non-σ-finite when we restrict it to the Borel sets.

Introduction

It is classical that, up to a nonnegative multiplicative constant, Lebesgue measure is the unique locally finite translation invariant Borel measure on ℝ^{d}. R. D. Mauldin [6] asked if we can replace locally finiteness by σ-finiteness. Then he himself gave an affirmative answer in the case when Borel measure means a measure defined on the σ-algebra of Borel sets, and later noticed that this is actually a folklore result, see (in a more general form) e.g. [3, Section 60, Theorem B and Exercise 7]. In fact, the problem already appeared in [8], as an open question posed by Saks. For the sake of completeness we include a proof here. Let λ_{d} denote d-dimensional Lebesgue measure, and B + t = {b + t : b ∈ B}.

Theorem 0.1 Let μ be a σ-finite translation invariant measure defined on the Borel subsets of ℝ^{d}. Then there exists c ∈ [0, ∞) such that μ(B) = cλ_{d}(B) for every Borel set B.

∗Partially supported by Hungarian Scientific Foundation grant no. 37758, 049786 and F 43620.

†Partially supported by Hungarian Scientific Foundation grant no. 049786 and F 43620.

MSC codes: Primary 28C10; Secondary 28A05, 28A10

Key Words: Lebesgue, Borel, measure, unique, translation, isometry, invariant, σ-finite

Proof. First we prove that μ is absolutely continuous with respect to λ_{d}. Let B ⊂ ℝ^{d} be a Borel set with λ_{d}(B) = 0. Define B̃ = {(x, y) ∈ ℝ^{d} × ℝ^{d} : x + y ∈ B}. This set is clearly Borel, and as both λ_{d} and μ are σ-finite measures, we can apply the Fubini theorem to (λ_{d} × μ)(B̃). Note that the x-section B̃_{x} = {y : (x, y) ∈ B̃} = B − x, and similarly B̃^{y} = {x : (x, y) ∈ B̃} = B − y. So by Fubini λ_{d}(B) = 0 implies (λ_{d} × μ)(B̃) = 0. Hence μ(B − x) = 0 for λ_{d}-almost every x, but μ is translation invariant, so μ(B) = 0.

Therefore by the Radon-Nikodým theorem there exists a Borel function f : ℝ^{d} → [0, ∞] such that μ(B) = ∫_{B} fdλ_{d} for every Borel set B. Clearly

μ(B) = μ(B + t) = ∫_{B+t} fdλ_{d} = ∫_{B} f(x − t)dλ_{d}(x)

for every t and every Borel set B. Hence the uniqueness of the Radon-Nikodým derivative implies that for every t for Lebesgue almost every x the equation

f(x − t) = f(x)  (1)

holds.

In order to complete the proof it is clearly sufficient to show that there is a constant c ∈ [0, ∞) such that f(x) = c holds for λ_{d}-almost every x. Suppose on the contrary that there are real numbers r_{1} < r_{2} such that the Borel sets {x : f(x) < r_{1}} and {x : f(x) > r_{2}} are of positive Lebesgue measure. Let d_{1} and d_{2} be Lebesgue density points of the two sets, respectively, and let t = d_{2} − d_{1}. Then d_{2} is the Lebesgue density point of {x : f(x − t) < r_{1}} as well, and so

λ_{d} ({x : f(x − t) < r_{1}} ∩ {x : f(x) > r_{2}}) > 0,

contradicting (1). □

However, in the literature there are at least two different notions that are referred to as Borel measure. The first one is measures defined only for Borel sets (see e.g. [3], [7]), while the second one is measures defined on σ-algebras containing the Borel sets (see e.g. [1], [5]).

In the rest of the paper we investigate the question of Saks and Mauldin in the case of the more general notion. As a spin-off, we also show that σ-finiteness is also sensitive to the definition of Borel measure. This question is related to [2], and was implicitly asked there.

1 The negative result

In this section we prove somewhat more than just a negative answer to our question.

Theorem 1.1 There exists an isometry invariant σ-finite measure μ defined on an isometry invariant σ-algebra A containing the Borel subsets of ℝ^{d} such that, for every Borel set B, if λ_{d}(B) = 0 then μ(B) = 0, while if λ_{d}(B) > 0 then μ(B) = ∞.

Before the proof we need a lemma, which resembles some results proven by various authors, but we were unable to find this version in the literature.

We also need some notation: Isom(ℝ^{d}) is the group of isometries of ℝ^{d}, the symbol |X| denotes the cardinality of a set X, the continuum cardinality is denoted by 2^{ω}, Δ stands for symmetric difference of two sets, and a set P ⊂ ℝ^{d} is perfect if it is nonempty, closed and has no isolated points. Throughout the proof we use the fact that a countable union of sets of cardinality < 2^{ω} is itself of cardinality < 2^{ω} (see e.g. [4, Cor. I.10.41]).

Lemma 1.2 There exists a disjoint decomposition ℝ^{d} = ∪_{n=0}^{∞}A_{n} such that |φ(A_{n})ΔA_{n}| < 2^{ω} for every n ∈ ℕ and every φ ∈ Isom(ℝ^{d}), and such that |A_{n} ∩ P | = 2^{ω} for every n ∈ ℕ and every perfect set P ⊂ ℝ^{d}.

Proof. We say that a set A ⊂ ℝ^{d} is < 2^{ω}-invariant, if |φ(A)ΔA| < 2^{ω} for every φ ∈ Isom(ℝ^{d}). As Isom(ℝ^{d}) is closed under inverses, this is equivalent to |φ(A) \ A| < 2^{ω} for every φ ∈ Isom(ℝ^{d}).

It is enough to construct a sequence A_{n} of disjoint < 2^{ω}-invariant sets such that |A_{n} ∩ P | = 2^{ω} for every n ∈ ℕ and every perfect set P ⊂ ℝ^{d}, since then clearly ℝ^{d} \ ∪_{n=0}^{∞}A_{n} is also < 2^{ω}-invariant, hence we can simply replace A_{0} by A_{0} ∪ (ℝ^{d} \ ∪_{n=0}^{∞}A_{n}).

Now we construct such a sequence by transfinite induction. Let us enumerate Isom(ℝ^{d}) = {φ_{α} : α < 2^{ω}} and define G_{α} to be the group generated by {φ_{β} : β < α}. Note that |G_{α}| < 2^{ω}. For x ∈ ℝ^{d} let G_{α}(x) = {φ(x) : φ ∈ G_{α}}. Let us also enumerate the perfect subsets of ℝ^{d} as {P_{α} : α < 2^{ω}} such that each perfect set P is listed 2^{ω} many times.

Define A_{n}^{0} = ∅ for every n ∈ ℕ. At step α we recursively construct a sequence x_{n}^{α} ∈ P_{α} (n ∈ ℕ) such that for every k ≠ l

[∪_{β<α}A_{k}^{β} ∪ G_{α}(x_{k}^{α})] ∩ [∪_{β<α}A_{l}^{β} ∪ G_{α}(x_{l}^{α})] = ∅.  (2)

To see that this is possible, note first that (2) holds whenever for every n the point x_{n}^{α} is not in the set

∪_{φ∈G_{α}}φ^{−1} (∪_{m≠n} ∪_{β<α} A_{m}^{β} ∪ ∪_{i=0}^{n−1}G_{α}(x_{i}^{α})) ,

which is of cardinality < 2^{ω}. As every perfect set is of cardinality 2^{ω}, this set cannot cover P_{α}, so we can find an x_{n}^{α} with the required property and define A_{n}^{α} = ∪_{β<α}A_{n}^{β} ∪ G_{α}(x_{n}^{α}). Clearly, |A_{n}^{α}| < 2^{ω}. Finally, define A_{n} = ∪_{α<2^{ω}}A_{n}^{α} for every n. These sets are clearly disjoint, they all intersect every perfect set in a set of cardinality 2^{ω}.

Finally, in order to check that the A_{n}’s are < 2^{ω}-invariant, let φ_{α} be given. First note that A_{n}^{α} = ∪_{β≤α}G_{β}(x_{n}^{β}) and A_{n} = ∪_{α<2^{ω}}G_{α}(x_{n}^{α}) for every n. Clearly, for α < β the set G_{β}(x_{n}^{β}) is φ_{α}-invariant (for every n), hence if x ∈ A_{n} is such that φ_{α}(x) ∉ A_{n}, then x ∈ ∪_{β≤α}G_{β}(x_{n}^{β}) = A_{n}^{α}. That is; φ_{α}(A_{n})\A_{n} ⊂ φ_{α}(A_{n}^{α}), so the A_{n}’s are < 2^{ω}-invariant. This completes the proof. □

Proof. (Theorem 1.1) Let A_{n} be the sequence from the previous lemma. Define

A = {[∪_{n=0}^{∞}(A_{n} ∩ B_{n})] ΔH : ∀n B_{n} ⊂ ℝ^{d} Borel, H ⊂ ℝ^{d}, |H| < 2^{ω}}.

Clearly A contains the Borel sets, as B = [∪_{n=0}^{∞}(A_{n} ∩ B)] Δ∅.

In order to check that A is closed under complements note that (XΔH)^{C} = X^{C}ΔH, and therefore ([∪_{n=0}^{∞}(A_{n} ∩ B_{n})] ΔH)^{C} = [∪_{n=0}^{∞}(A_{n} ∩ B_{n})]^{C} ΔH = [∪_{n=0}^{∞}(A_{n} ∩ B_{n}^{C})] ΔH.

In order to show that A is closed under countable unions, we need to show ∪_{k=0}^{∞}(X^{k}ΔH^{k}) ∈ A, where

X^{k} = ∪_{n=0}^{∞}(A_{n} ∩ B_{n}^{k}).  (3)

Using the identity

Z = W ΔW ΔZ  (4)

(note that Δ is associative) we obtain

∪_{k=0}^{∞}(X^{k}ΔH^{k}) = [∪_{k=0}^{∞}X^{k}] Δ [∪_{k=0}^{∞}X^{k}] Δ [∪_{k=0}^{∞}(X^{k}ΔH^{k})] = [∪_{k=0}^{∞}X^{k}] ΔY,

(5)

where Y = [∪_{k=0}^{∞}X^{k}] Δ [∪_{k=0}^{∞}(X^{k}ΔH^{k})]. As

∪_{k=0}^{∞}X^{k} = ∪_{n=0}^{∞} (A_{n} ∩ (∪_{k=0}^{∞}B_{n}^{k}))  (6)

it is sufficient to check that

|Y | < 2^{ω},  (7)

but this is clear, since Y = [∪_{k=0}^{∞}X^{k}] Δ [∪_{k=0}^{∞}(X^{k}ΔH^{k})] ⊂ ∪_{k=0}^{∞}H^{k}, which is of cardinality < 2^{ω}.

To show that A is isometry invariant, let φ ∈ Isom(ℝ^{d}). First note that

φ([∪_{n=0}^{∞}(A_{n} ∩ B_{n})] ΔH) = [∪_{n=0}^{∞}(φ(A_{n}) ∩ φ(B_{n}))] Δφ(H).  (8)

Set

X = ∪_{n=0}^{∞}(φ(A_{n}) ∩ φ(B_{n})) and Y = ∪_{n=0}^{∞}(A_{n} ∩ φ(B_{n})).  (9)

We need to show that XΔφ(H) ∈ A. Using (4) again, write

XΔφ(H) = [Y ΔY ΔX] Δφ(H) = Y Δ [(Y ΔX)Δφ(H)] ,  (10)

where we use again the associativity of Δ. Hence it is enough to show that

|(Y ΔX)Δφ(H)| < 2^{ω},  (11)

which follows from Y ΔX = [∪_{n=0}^{∞}(A_{n} ∩ φ(B_{n}))] Δ [∪_{n=0}^{∞}(φ(A_{n}) ∩ φ(B_{n}))] ⊂ ∪_{n=0}^{∞}(A_{n}Δφ(A_{n})), from |φ(H)| < 2^{ω}, and the < 2^{ω}-invariance of A_{n}.

Let us now define

μ([∪_{n=0}^{∞}(A_{n} ∩ B_{n})] ΔH) = ∑_{n=0}^{∞} λ_{d}(B_{n}).

First we have to show that μ is well-defined. Let [∪_{n=0}^{∞}(A_{n} ∩ B_{n})] ΔH = [∪_{n=0}^{∞}(A_{n} ∩ B′_{n})] ΔH′. We claim that λ_{d}(B_{n}) = λ_{d}(B′_{n}) for every n. Otherwise, without loss of generality, there exists an n_{0} such that λ_{d}(B_{n_{0}}) < λ_{d}(B′_{n_{0}}), hence B′_{n_{0}} \B_{n_{0}} contains a perfect set P (even of positive measure). But |P ∩A_{n_{0}}| = 2^{ω} and |H ∪ H′| < 2^{ω}, hence there exists an x ∈ (P ∩ A_{n_{0}}) \ (H ∪ H′), and then x ∈ [∪_{n=0}^{∞}(A_{n} ∩ B′_{n})] ΔH′ but x ∉ [∪_{n=0}^{∞}(A_{n} ∩ B_{n})] ΔH, a contradiction. (Recall that the A_{n}’s are disjoint.)

In order to prove that μ is σ-additive, let

∪_{k=0}^{∞}(X^{k}ΔH^{k})  (12)

be a disjoint union, where X^{k} is as in (3). First we claim that for every n and every k ≠ k′ we have λ_{d}(B_{n}^{k} ∩ B_{n}^{k′}) = 0. Otherwise, for some n_{0} there exists a perfect set P ⊂ B_{n_{0}}^{k} ∩ B_{n_{0}}^{k′}, and we can find x ∈ (P ∩ A_{n_{0}}) \ (H^{k} ∪ H^{k′}), hence x ∈ [∪_{n=0}^{∞}(A_{n} ∩ B_{n}^{k})] ΔH^{k} and x ∈

[∪_{n=0}^{∞}(A_{n} ∩ B_{n}^{k′})] ΔH^{k′}, but then the union (12) is not disjoint, a contradic-

tion. Therefore λ_{d}(∪_{k=0}^{∞}B_{n}^{k}) = ∑_{k=0}^{∞} λ_{d}(B_{n}^{k}) for every n, so by (5), (6) and (7) we obtain μ(∪_{k=0}^{∞}(X^{k}ΔH^{k})) = ∑_{n=0}^{∞} λ_{d}(∪_{k=0}^{∞}B_{n}^{k}) = ∑_{n=0}^{∞} ∑_{k=0}^{∞} λ_{d}(B_{n}^{k}) = ∑_{k=0}^{∞} ∑_{n=0}^{∞} λ_{d}(B_{n}^{k}) = ∑_{k=0}^{∞} μ(X^{k}ΔH^{k}).

Now we show that μ is isometry invariant. By (8), (9), (10) and (11) we obtain that μ(φ([∪_{n=0}^{∞}(A_{n} ∩ B_{n})] ΔH)) = ∑_{n=0}^{∞} λ_{d}(φ(B_{n})), which clearly equals ∑_{n=0}^{∞} λ_{d}(B_{n}), which is μ([∪_{n=0}^{∞}(A_{n} ∩ B_{n})] ΔH) by definition.

The fact that μ is σ-finite follows from ℝ^{d} = ∪_{n=0}^{∞} ∪_{K=0}^{∞} (A_{n} ∩ [−K, K]^{d}), since μ(A_{n} ∩ [−K, K]^{d}) = λ_{d}([−K, K]^{d}) = (2K)^{d} < ∞ for every n and K.

Finally, for a Borel set B we have μ(B) = μ(∪_{n=0}^{∞}(A_{n} ∩ B)) = ∑_{n=0}^{∞} λ_{d}(B), which is zero if λ_{d}(B) = 0 and ∞ otherwise. □

As an immediate corollary we obtain the following.

Corollary 1.3 There exists an isometry invariant σ-finite measure μ defined on an isometry invariant σ-algebra A containing the Borel subsets of ℝ^{d} such that μ restricted to the Borel sets is not equal to cλ_{d} for every c ∈ [0, ∞).

As ℝ^{d} is not the union of countably many Lebesgue nullsets, the next statement is also a corollary to Theorem 1.1.

Corollary 1.4 There exists an isometry invariant σ-finite measure μ defined on an isometry invariant σ-algebra A containing the Borel subsets of ℝ^{d} such that μ restricted to the Borel sets is not σ-finite.

2 The positive result

The measure μ constructed in the previous section behaves simply on Borel sets; if λ_{d}(B) = 0 then μ(B) = 0, while if λ_{d}(B) > 0 then μ(B) = ∞. So we can say that μ(B) = ∞λ_{d}(B) for every Borel set B. The next theorem shows that this is the only possibility.

Theorem 2.1 Let μ be a σ-finite translation invariant measure defined on a translation invariant σ-algebra containing the Borel subsets of ℝ^{d}. Then there exists c ∈ [0, ∞] such that μ(B) = cλ_{d}(B) for every Borel set B.

Moreover, μ restricted to the Borel sets is σ-finite if and only if c is finite.

The proof of this theorem will be based on the following two lemmas, the second of which is well known.

Lemma 2.2 Let μ be a σ-finite translation invariant measure defined on a translation invariant σ-algebra containing the Borel subsets of ℝ^{d}, and suppose that μ restricted to the Borel sets is not σ-finite. Then for every Borel set B we have either μ(B) = 0 or μ(B) = ∞.

Proof. Let B be a maximal disjoint family of Borel sets of positive finite μmeasure. As μ is σ-finite (on A), B is countable, hence B_{0} = ⋃ B is a Borel set. Define

μ′(B) = μ(B_{0} ∩ B) for every Borel set B.

Note that this measure is only defined for Borel sets. As μ′ is clearly σ-finite, we can apply the Fubini theorem for μ′ × μ and the Borel set B̃_{0}^{C} = {(x, y) ∈ ℝ^{d} × ℝ^{d} : x + y ∈ B_{0}^{C}}, as in the proof of Theorem 0.1. On one hand, (μ′ × μ)(B̃_{0}^{C}) = ∫_{y∈ℝ^{d}} μ′(B_{0}^{C} −y)dμ(y) = ∫_{y∈ℝ^{d}} μ(B_{0}∩(B_{0}^{C} −y))dμ(y). We claim that

μ(B_{0} ∩ (B_{0}^{C} − y)) = 0 for every y, hence (μ′ × μ)(B̃_{0}^{C}) = 0. Indeed, otherwise (using that B_{0} = ⋃ B and B is countable) there is a Borel set B ∈ B such that 0 < μ(B ∩ (B_{0}^{C} − y)) < ∞. But then for D = B ∩ (B_{0}^{C} − y) we obtain that the Borel set D + y is disjoint from B_{0}, hence from all elements of B, and is of positive and finite μ-measure (since μ is translation invariant), contradicting the maximality of B.

On the other hand, 0 = (μ′ × μ)(B̃_{0}^{C}) = ∫_{x∈ℝ^{d}} μ(B_{0}^{C} − x)dμ′(x). As μ restricted to the Borel sets is not σ-finite, μ(B_{0}^{C} − x) = μ(B_{0}^{C}) = ∞ for every x. Therefore we obtain 0 = μ′(ℝ^{d}) = μ(B_{0}), so B = ∅ and we are done. □

Lemma 2.3 Let μ_{1}and μ_{2} be σ-finite translation invariant measures defined on the (not necessarily equal) translation invariant σ-algebras A_{1} and A_{2} containing the Borel subsets of ℝ^{d}, and suppose that μ_{1}(ℝ^{d}), μ_{2}(ℝ^{d}) > 0. Then for every Borel set B, μ_{1}(B) = 0 iff μ_{2}(B) = 0.

Proof. Apply Fubini to μ_{1} × μ_{2} and B̃ = {(x, y) ∈ ℝ^{d} × ℝ^{d} : x + y ∈ B}. □

Proof. (Theorem 2.1) The last statement of the theorem is obvious, as countably many Lebesgue nullsets cannot cover ℝ^{d}.

Now we prove the first statement, namely that the constant c ∈ [0, ∞] exists. If μ restricted to the Borel sets is σ-finite, then we are done by Theorem 0.1. So we can assume that this is not the case. Then applying Lemma 2.2 and Lemma 2.3 with μ_{1} = μ and μ_{2} = λ_{d} the theorem follows. □

References

[1] A. M. Bruckner, J. B. Bruckner, B. S. Thomson: Real Analysis. Prentice Hall, 1997.

[2] M. Elekes, T. Keleti, Borel sets which are null or non-σ-finite for every translation invariant measure, to appear in Adv. Math.

[3] P. R. Halmos: Measure Theory. Springer-Verlag, 1974.

[4] K. Kunen: Set theory. An introduction to independence proofs. North-Holland, 1983.

[5] P. Mattila: Geometry of Sets and Measures in Euclidean Spaces. Cambridge University Press, 1995.

[6] R. D. Mauldin, personal communication, 2004.

[7] W. Rudin: Real and complex analysis. McGraw-Hill, 1987.

[8] E. Szpilrajn: On problems of measure theory (in Russian), Uspehi Mat. Nauk 1 (1946) 179-188. English translation in E. Marczewski, Collected Mathematical Papers, p.276-287, Polish Academy of Sciences 1996.

Márton Elekes Rényi Alfréd Institute P.O. Box 127, H-1364 Budapest, Hungary

Email address: emarci@renyi.hu
URL: http://www.renyi.hu/~emarci

Tamás Keleti Eötvös Loránd University, Department of Analysis Pázmány Péter sétány 1/c, H-1117, Budapest, Hungary

Email address: elek@cs.elte.hu
URL: http://www.cs.elte.hu/anal/keleti