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(I'm not sure if this is entirely suitable here so feel free to close it if it's not.) The statement "there is a Lebesgue measure on $\mathbb{R}$($2^\omega$)" means: there is a total $\sigma$-additive monotone (wrt set inclusion) function $\mu$ identical with the usual Lebesgue measure on Cantor space (aka interval [0,1]) so Lebesgue measure is total. It is well known over ZFC Lebesgue measure is not total.

Some typical consequences of ZFC+ "there is a measure on $\mathbb{R}$" would be negation of CH (as there exists an $\omega_1$-scale). Of course the measure here is necessarily not Lebesgue. I'm wondering if there are references about nice consequences of ZF+ "there is a Lebesgue measure on $\mathbb{R}$" in areas other than set theory (or in a more hideous language? Analysis?)? Thanks in advance!

Jing Zhang
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  • You want to look for "real-valued measurability" and "Ulam's problem". There is a vast literature on this. Some consequences are naturally set-theoretic (e.g., the assumption can be restated in terms of generic elementary embeddings), but there are consequences on analysis (to Fubini-type theorems, for instance). David Fremlin has an excellent survey on the topic. – Andrés E. Caicedo Jan 22 '15 at 06:47
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    For the title it looks like you're asking about consequences of "all sets of reals are Lebesgue measurable". But from the body of the question is sounds more like you're asking about consequences of "there is a total $\sigma$-additive (but not necessarily translation-invariant?) extension of Lebesgue measure. Which is it? For the latter, you don't have to drop the axiom of choice, it could happen in ZFC. – bof Jan 22 '15 at 06:58
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    @Zhang Jing I'm not sure what an $\omega_1$-scale is, but I assume by CH you mean the continuum hypothesis and not the axiom of choice. The theory is ZF+DC+CH+"All sets of reals are Lebesgue measurable" is consistent relative to the existence of an inaccessible cardinal, and the theory ZF+DC+CH+"there is a measure on $\mathbb{R}$'' is consistent relative to ZF (though such a measure is necessarily not translation invariant). So, under either interpretation of your question raised by bof in the previous comment, the negation of CH is not a consequence of the theory in question. – Jesse Elliott Jan 22 '15 at 07:53
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    @JesseElliott: I think what Zhang was trying to refer to is the following theorem: Lemma 10. 16 and Corollary 10.17 in Jech's book. On the other hand, I am not really sure why we are dropping AC. There can be such measures relative to real valued measurable cardinal assumptions. See my former common mistake here! – Burak Jan 22 '15 at 14:35
  • @bof,@Jesse: the negation of CH is indeed a consequence of ZFC+exists a total measure. What I'm really interested is the extension of Lebesgue measure. I'll edit the question. – Jing Zhang Jan 22 '15 at 14:41
  • @AndresCaicedo: thanks! Which survey by Fremlin do you refer to? – Jing Zhang Jan 22 '15 at 15:13
  • Your definition of Lebesgue measure in the first paragraph of the question is incorrect. Anyway, you probably want to assume a modicum of choice to avoid pathological situations (such as $\mathbb R$ being a countable union of countable sets). Otherwise, you need some additional care about how the measure is defined. (Fremlin also addresses this, in his treatise on measure theory.) – Andrés E. Caicedo Jan 22 '15 at 15:42
  • Sorry Cantor space and R confusion. I remedied this. Isn't that such measure simply does not exist if R is a countable union of countable sets, by $\sigma$-additivity? – Jing Zhang Jan 22 '15 at 16:18
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    Your description in the first paragraph is now ambiguous: The existence of a total measure extending Lebesgue measure is not the same as Lebesgue measure being total. – Andrés E. Caicedo Jan 22 '15 at 16:31
  • I meant Lebesgue measure is total otherwise choice is not necessarily ruled out. Sorry for the ambiguity. – Jing Zhang Jan 22 '15 at 16:37
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    Which again brings us to the question of how precisely is Lebesgue measure being defined. The thing is this: You can define the length of an interval $[a,b]$ or $(a,b)$ or ... as $b-a$, and then the outer measure of a set as the infimum over all countable coverings of the set by open intervals of the sums of the lengths of these intervals. You can then define inner measure (using compact sets) and then the measurable sets are those where inner and outer measures coincide. Or you can define measurable sets directly following Caratheodory. (Do these definitions coincide?) Anyway, (Cont.) – Andrés E. Caicedo Jan 22 '15 at 17:02
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    you cannot prove in $\mathsf{ZF}$ that the measure of an interval is its length. Do you want this equality as an additional requirement? Or do you have a different definition in mind? – Andrés E. Caicedo Jan 22 '15 at 17:04
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    @JingZhang: I am getting confused by what you have written: "I meant Lebesgue measure is total otherwise choice is not necessarily ruled out".

    You want to rule out choice and want the usual Lebesgue measure to be total? In this case, ~CH does not necessarily follow (and you have to be careful when you state what CH is). It can be that all sets of reals are Lebesgue measurable and have the perfect set property. The latter implies CH in the sense that every uncountable set of reals has the size of continuum.

     – Burak Jan 22 '15 at 17:24
  • @Burak: I mean ~CH is a consequence of ZFC + there is a measure on \mathbb{R}, and this measure is necessarily not Lebesgue. You are right over ZF they are not really orthogonal. – Jing Zhang Jan 22 '15 at 17:39
  • @AndresCaicedo: I did have in mind that the length of an interval equals its measure. – Jing Zhang Jan 22 '15 at 17:52
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    One nice one, per http://journals.cambridge.org/download.php?file=%2FJAZ%2FJAZ1_19_03%2FS1446788700031505a.pdf&code=94bdeb599e2a7f0d9a01ec5fab104dcd : if all sets of reals are Lebesgue measurable, then the additive groups of $\mathbb{R}$ and $\mathbb{C}$ are (no longer) isomorphic. – Steven Stadnicki Jan 22 '15 at 19:19
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    Here is another one: if all sets of reals are measurable, then the real line can be partition into strictly more parts then elements. See http://boolesrings.org/asafk/2014/anti-anti-banach-tarski-arguments/ – alexod Feb 12 '15 at 06:22

2 Answers2

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There are some useful tables in the back of Gregory Moore's book Zermelo's Axiom of Choice showing the deductive relations between several principles that lie between the Axiom of Choice and the existence of a non-measurable set. Some quick examples I see from looking at the tables and taking the constrapositive are that if there are no non-measurable subsets of $\mathbb{R}$, then the Hahn-Banach Theorem fails, not every Boolean algebra has a probability measure, not every infinite set has a non-principal ultrafilter, and not every vector space has a basis.

Are these the kind of consequences you mean?

Also, the result Stadnicki mentioned in the comments is pretty cool, and I haven't seen it before. (the additive groups of $\mathbb{R}$ and $\mathbb{C}$ are no longer isomorphic.)

Michael Cotton
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  • This is a Dover book too, so it's really easy to find. Worth having, in my opinion, as it's a pleasant read. But none of the proofs of these results will be in there. It's mostly just a history book. – Michael Cotton Feb 12 '15 at 04:31
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One can multiply examples, of course. For the amusement of anyone interested, here are three more consequences of there being no nonmeasurable sets (also of the Banach-Tarski decomposition failing).

  1. There are no full finitely-additive conditional probabilities defined for all pairs of subsets of $[0,1]$ (with the second subset non-empty).

  2. There are partial orders that don't extend to total orders (and even ones that don't extend to total preorders while keeping strict inequalities).

These sound negative, like the one about Hahn-Banach failing. Here's a positive-sounding one:

  1. If $G$ is a totally ordered abelian group and $\mu$ is a finitely-additive $G$-valued measure (with the obvious definitions, including non-negativity) defined on all subsets of $[0,1]$, then $\mu(A)=0$ for some non-empty $A$.

My proofs use a modification of the proof that Hahn-Banach implies Banach-Tarski. (By the way, sorry, it's a philosophy and not a math paper that I'm referencing.)

Asaf Karagila
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Alexander Pruss
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