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In Kanamori's book, "The Higher Infinite" p. 376, he defines a minimal cover of some $A \subseteq \omega^\omega$, to be any $B \subseteq \omega^\omega$, such that $A\subseteq B$ and that $B$ is Lebesgue measurable and if $Z \subseteq B-A$ is Lebesgue measurable, then $m_L(Z) = 0$. And he claims that picking some $B$ with $A\subseteq B$ and $m_L(B)$ minimal, does the job. [Here $m_L$ denotes the Lebesgue measure.]

Now here is my problem. The whole premise of this chapter is that we don't want to use choice to do these things. But any way I try to construct such a $B$, I inevitably use some form of choice. The best I can do is $\mathsf{AC}_\omega(\omega^\omega)$. Is there some choice-free way to do this?

A sketch of a proof with $\mathsf{AC}_\omega(\omega^\omega)$: Let $x = \inf\{m_L(B): A\subseteq B \text{ and } B \text{ is Lebesgue measurable}\}$. By $\mathsf{AC}_\omega(\omega^\omega)$, let $\langle B_n: n<\omega\rangle$ be a sequence such that $A\subseteq B_n$ and $m_L(B_n) \rightarrow x$ as $n \rightarrow \infty$. Now $B = \bigcap_n B_n$ is the desired minimal cover. $\square$

Asaf Karagila
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Shervin Sorouri
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    Measure theory without at the very least $\sf AC_\omega(\Bbb R)$ is nothing short of terrifying. – Asaf Karagila Aug 12 '20 at 17:48
  • @AsafKaragila, I guess so. :) And since we are working with $\mathsf{AD}$ and we already have $\mathsf{AC}_\omega(\omega^\omega)$, I don't think there is any critical issue in the text. I was just wondering if I was missing something trivial. Because the book says in the beginning of this chapter that any use of $\mathsf{AC}$ will be explicitly mentioned. – Shervin Sorouri Aug 12 '20 at 17:50
  • Well. How do you define the Lebesgue (or even Borel) measure in the Feferman–Levy model? – Asaf Karagila Aug 12 '20 at 17:52
  • @AsafKaragila, I don't think defining the measures on the Borel and Lebesgue sigma algebras needs choice, does it? I don't know. I'll need to review my analysis books. – Shervin Sorouri Aug 12 '20 at 17:56
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    Okay, another question, what is the Borel $\sigma$-algebra? – Asaf Karagila Aug 12 '20 at 17:57
  • Well, if we want to go with the classical route, it is the intersection of all the sigma algebras containing the open subsets. But I can see how the recursive definition might become troublesome. – Shervin Sorouri Aug 12 '20 at 17:58
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    @ShervinSorouri In ZF alone that could be the full powerset of the reals. See e.g. here. – Noah Schweber Aug 12 '20 at 17:59
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    So in the Feferman–Levy model, where every set of reals is a countable union of countable sets, how do you define a measure on the Borel sets which is both countably additive and non-trivial? – Asaf Karagila Aug 12 '20 at 17:59
  • @NoahSchweber, yes. But that still wouldn't be a problem, would it? We still can go on and do our analysis I guess. – Shervin Sorouri Aug 12 '20 at 18:00
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    @ShervinSorouri Well it means for example that measure isn't countably additive anymore. – Noah Schweber Aug 12 '20 at 18:01
  • @AsafKaragila, Oh, I get it now. That's a grim situation. :) Thanks for the clarification. – Shervin Sorouri Aug 12 '20 at 18:01
  • @NoahSchweber, Thanks for the great link! – Shervin Sorouri Aug 12 '20 at 18:04
  • Just to clarify, now that you're aware of the horrors of measure theory without countable choice, do you want to clarify your question (or delete it altogether)? – Asaf Karagila Aug 13 '20 at 08:41
  • @AsafKaragila, since my issue is resolved and maybe in the future this might be a question for someone else, and because of the effort put into the comments, I would prefer if you could write an answer and I would accept it, so it would be off of the unanswered list. If not, I can post an answer myself. Or if maybe none of the options are okay, I can delete it. – Shervin Sorouri Aug 14 '20 at 19:54
  • I'd normally opt for one of us (you, @Noah, me) to write an answer. But since the comments sort of clarify that the idea is somehow problematic to begin with, I'd actually go with deleting. But it's up to you. – Asaf Karagila Aug 14 '20 at 20:13
  • @AsafKaragila, since I think there is something to be learned here(in the comments)(at least for someone like me), I am reluctant to delete it. Since you have already notified Noah Schweber, if you guys won't want to post an answer, I will try to post one based on the comments, until tomorrow night. I hope this is okay with everyone. – Shervin Sorouri Aug 14 '20 at 20:38

2 Answers2

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This is in fact a theorem of ZF. The Caratheodory construction of Lebesgue measure works in ZF, though without choice it need not be $\sigma$-additive. Every Borel codable set of reals is measurable but not necessarily every Borel set. See the paper cited below.

Let $U_i$ enumerate the basic open sets of $\omega^{\omega}.$ By (finite) subadditivity of $\lambda^*$ and adjusting the proof of the Lebesgue density theorem, we have that for any $X$ with $\lambda^*(X)>0$ and $\epsilon >0,$ there is $U_i$ such that $\lambda^*(X \cap U_i) > (1-\epsilon) \lambda(U_i).$

Define $S_n$ recursively by having $i \in S_n$ if $\lambda^*(A \cap U_i \setminus \bigcup_{j<i, j \in S_n} U_j)>\frac{n-1}{n} \lambda(U_i).$ Let $V_n = \bigcup_{i \in S_n} U_i$ and $V = \bigcap_{n<\omega} V_n.$

By our density lemma, each $A \setminus V_n$ is null. Since $\lambda^*$ is additive among subsets of disjoint measurable sets, we have

$$\lambda(V_n) - \frac{1}{n} \le \frac{n-1}{n}\lambda(V_n) \le \sum_{i \in S_n} \lambda^*\left (A \cap U_i \setminus \bigcup_{j<i, j \in S_n} U_j \right ) \le \lambda^*(A) \le \lambda^*(A \cup V_n) =\lambda(V_n).$$

We compute $$\lambda^*(A \setminus V) \le \inf_{n<\omega} \left ( \lambda^*\left (\bigcap_{i<n} V_n \setminus V \right ) + \lambda^*\left (A \setminus \bigcap_{i<n} V_n \right ) \right ) =0.$$

Therefore, $B := A \cup V$ is measurable, with

$$\lambda^*(A) \le \lambda(B) = \lambda(V) \le \inf_{n<\omega} \lambda(V_n) \le \inf_{n<\omega} \left (\lambda^*(A) + \frac{1}{n} \right) = \lambda^*(A).$$ So $B$ is as desired.

Foreman, Matthew; Wehrung, Friedrich, The Hahn-Banach theorem implies the existence of a non-Lebesgue measurable set, Fundam. Math. 138, No. 1, 13-19 (1991). ZBL0792.28005.

Elliot Glazer
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  • This is very cool, thanks! So in light of the other answer, the usual notions of real analysis seem to malfunction in the absence of choice. I'm wondering to what extent do the usual theorems and "definitions" go through in this setting; like the one you showed above. Would you know if there a systematic account of this available? – Shervin Sorouri Apr 24 '22 at 22:16
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    Fremlin's Measure Theory Vol. 5 goes into the measure theory of Borel-codable sets. Most of concrete measure theory has choiceless analogues in this setting. There isn't much written about what works and what doesn't in choiceless measure theory in more general settings. I'm working on filling that gap. See also this post: https://mathoverflow.net/a/393162/109573 – Elliot Glazer Apr 25 '22 at 03:26
  • Oh I see, very nice. Thanks. – Shervin Sorouri Apr 25 '22 at 20:32
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This answer is purely a compilation of the ideas in the comments, given by Asaf Karagila and Noah Schweber. I have also made this a "Community Wiki", so I won't gain any reputation from any upvotes.

The thing is that in choiceless setting, most of our definitions in measure theory either don't make sense, or have different non-equivalent forms. To see an example look here. And also in certain cases we fail to have a measure on the Borel sets. For example in the Feferman–Levy model every set of reals is a countable union of countable sets and so it is not possible to have a measure on the Borel sets which is both countably additive and non-trivial. This is why some level of choice is required to even get started. And so the use of $\mathsf{AC}_\omega(\omega^\omega)$ in the above proof is justified.

Shervin Sorouri
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