Talk:Lebesgue measure

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Contents 1 Properties 2 Nice article 3 Borel vs. Lebesgue 3.1 a lebesgue measurable, non Borel set on R 3.2 a non lebesgue measurable set on R 3.2.1 Proof of A is not lebesgue measurable 3.2.2 Proof of A is lebesgue measurable 4 Addition of a minor detail 5 Definition?

[edit] Properties

Many of the properties of te Lebesgue measure as stated on this page are simply properties of measures in general. Items 3 and 4 for example, and perhaps others (I don't know, since I'm just learning about the subject as I browse here!)

Wouldn't it be better to list only those properties that are specific to the Lebesgue measure, or at least to indicate which ones are? The standard definition is already there at measure

Also, a measure is defined on its page as a function on a sigma algebra; if that's really part of the definition, then saying "the Lebesgue measurable sets therefore form a sigma algebra" seems a little redundant and/or confusing.

- Stuart

When the article says, "the Lebesgue measurable sets form a sigma algebra", this is said first without knowing that Lebesgue measure IS a measure. In other words, we're verifying the axioms for the particular case of Lebesgue measure on Euclidean space. This is much the same thing that happens when we verify the group axioms for a set of elements and operation -- it's not redundant, we just have to check that it really is a group. Revolver

Is it clearer now? --AxelBoldt

Thanks. I'm still a little unsure as to which properties listed are common to all measures, and which are special to Lebesgue measure, but that's mainly due to my near-total ignorance of the whole subject. - Stuart

Can any of you give an example of a non-Lebesgue measurable set? JeffBobFrank 03:37, 4 Mar 2004 (UTC)

Sure, it's not too difficult; I'll either put it up here or on a separate article. You basically take a set of representatives of the cosets of Q (the rationals) in R (the reals), (this is where choice comes in) and consider all the translates of this set by rationals. The translates form a countable partition of R, and you use this to get a contradiction. Of course, there are some details I'm leaving out. A proof is given in Royden and a slightly more general result in papa Rudin (that every subset of R of positive Lebesgue measure contains a non-measurable subset). Revolver 17:22, 4 Mar 2004 (UTC)

[edit] Nice article

What a pleasure was to read through this very well-written article! Good work! Oleg Alexandrov (talk) 13:43, 28 October 2005 (UTC)

[edit] Borel vs. Lebesgue

"there are many more Lebesgue-measurable sets than there are Borel measurable sets". cardinal of the class of borel sets (=c=N1=2^N0) < cardinal of the class of Lebesgue measurable set(=N2=2^N1) = cardinal of the powerset of R(=N2=2^N1), but the class of all lebesgue messurable sets is still a strict subset of the power set of R. It is just hard to imagine. Can someone put two examples here at the same time please. Thanks. Jackzhp 17:22, 4 October 2006 (UTC)

The existence follows just from cardinality concerns. The cardinality of the class of Borel sets is the same as the cardinality of the real numbers, but the cardinality of the class of all subsets of the Cantor set is the same as the cardinality of the power set of the real numbers. Now every subset of the Cantor set is Lebesgue measurable of measure zero, because the Cantor set has measure zero. So most subsets of the Cantor set are Lebesgue measurable but not Borel measurable.

I would guess that a particular example of such a set could be constructed, but just like with non-Lebesgue-measurable sets it will not be canonical. CMummert 17:40, 4 October 2006 (UTC)

Please feel free to change the following text.

[edit] a lebesgue measurable, non Borel set on R

Cantor function f is a one to one map, which can map the cantor set C into a Borel set with measure 1. Hence its inverse function exists. Since every lebesgue measurable set with a positive measure contains a non lebesgue measurable set. Denote the non lebesgue measurable subset of f(C) as K, then the range G of the inverse f with domain K is included in C. G is lebesgue measurable, but it is not a Borel set.

f: C <-> f(C)

λ(C)=0, C is null; λ(f(C))=1

K is a non lebesgue measurable set.

f: G <-> K

λ(G)=0, G is negligible, null.

G is not Borel set. If it is a Borel set, K is a Borel set since f is one to one map. Since K is non lebesgue measurable, so K is not a Borel set, neither is G.

Feel free to change all this.

[edit] a non lebesgue measurable set on R

In order for people to understand more easily, let's requre the example set A included in [0,1]. In order to construct the set A, define an equivalent class respect to x: Bx={y: y-x is rational}, then we select a member belong to [0,1] from each equivalent class, then A is not lebesgue measurable.

Bx={y: y-x is rational}

A={ai: ai} is infinite. No equivalent members of relation B.

[edit] Proof of A is not lebesgue measurable

Suppose A is measurable, then A+r (r is a rational number) is also measurable.

rn is any rational number, rn ∈[0,1]

λ([0,1])=1 <= λ(U(A+rn))=sum λ(A) << λ([0,2])=2

such λ(A) can not exist, hence A is not lebesgue measurable. More information can be found: Vitali set

[edit] Proof of A is lebesgue measurable

This is much easy, I can show it here. Although I know it is wrong, I don't know the reason why it is wrong.

Any x belongs to R, its lebesgue measure is zero. All singleton belong to the class of Borel set, so does the union of them, hence A belongs to the class of Borel sets. Lebesgue measure of A equals the sum of the lebesgue measure of all points belong to A. Hence the sum is zero.

A might consist of uncountable many points, so A is not necessary to belongs to the class of Borel sets.

[edit] Addition of a minor detail

I have added one more property of lebesgue integral related with the linear transformation. I was wondering if the property of translation invariance be reatined because it's a special case of a linear transformation. Thanks... Alok Bakshi 06:18, 8 February 2007 (UTC)

I don't see any reason to eliminate it. The two list items could be merged together, but the translation case is important enough to be specifically mentioned. CMummert · talk 14:33, 8 February 2007 (UTC)

[edit] Definition?

Is there not a definition for the Lebesgue measure? Sancho (talk) 05:50, 14 March 2007 (UTC)

Good point. It's not actually defined anywhere in the article. That needs to be fixed.--CSTAR 05:58, 14 March 2007 (UTC)