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First, yes, I know that the Lebesgue outer measure isn't necessarily countably subadditive without the principle of countable choices, and thus isn't an outer measure. Nonetheless, I eagerly anticipate all the answers patiently explaining this to me.

So anyway, to be more precise: given a subset of the real numbers whose Lebesgue submeasure (i.e., a function defined identically to the Lebesgue outer measure, which demonstrably isn't an outer measure in some models of ZF) is infinite, and which fulfills the Carathéodory criterion with respect to the same, is it possible to show in ZF that there exists a subset with arbitrarily large finite Lebesgue submeasure (preferably one that fulfills the Carathéodory criterion, but at all)? I'm pretty sure I can get arbitrarily close to a set of finite submeasure, but I can't seem to extend this to the infinite case. If not, why not?

user361424
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  • Where do you think choice gets involved? – Asaf Karagila Jan 04 '18 at 10:14
  • There's a model of ZF in which the real numbers can be expressed as a countable union of countable sets. In that model, the Lebesgue outer (sub)measure couldn't be countably subadditive, since ZF alone is enough to show that it's zero on a countable set, so what you've got is $ 0 + 0 + 0 + \cdots < +\infty $. Moreover, ZF alone is enough to show that countable sets and the real line both meet the Carathéodory criterion, so the usual definition of Lebesgue measure isn't subadditive either. – user361424 Jan 04 '18 at 10:31
  • Okay... So you obviously need some countable choice here. But since you already know the countable unions of countable sets issue, what more do you want to know? – Asaf Karagila Jan 04 '18 at 10:45
  • I don't need countable choice. I'm interested in what I can and can't show of the submeasure without countable choice. – user361424 Jan 04 '18 at 10:49
  • Or, more clearly: I know that the ordinary proof of this fact won't work without countable choice, but I'm wondering if there might be another way to come to it, or if there's a good reason (not necessarily a rigorous proof, but more substantial than "the ordinary proof doesn't work") that there couldn't be. – user361424 Jan 04 '18 at 18:51
  • And the argument "without countable choice it is consistent to have a counterexample" is not enough to convince you that countable choice is really needed? This is what's confusing me. The answer I would give you to your question is exactly this example, but you already know it. – Asaf Karagila Jan 05 '18 at 00:26
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    What's the counterexample? The counterexample I gave to subadditivity, the reals expressed as a countable union of countable sets, certainly isn't a counterexample to the question I asked, since the set of reals certainly has subsets of arbitrarily large submeasure (trivially, $[0,N)$). – user361424 Jan 05 '18 at 07:11
  • When you say "arbitrarily large", do you specifically mean "finite but arbitrarily large"? Just removing a singleton should still leave you with a set of infinite Lebesgue outer submeasure. – Julian Newman Jan 08 '18 at 15:05
  • (Also, is it obvious that in ZF, one can prove that the Lebesgue outer submeasure of $[0,N)$ is unbounded as a function of $N$? The only thing that seems obvious to me is that the Lebesgue outer submeasure of $[0,N)$ is $\leq N$. In fact, how do we even know that Lebesgue outer submeasure of $\mathbb{R}$ is nonzero?) – Julian Newman Jan 08 '18 at 18:05
  • You're right that I should have specified finite. I'll change that. As for the other question, yes, ZF is strong enough to show that the submeasure of $[a,b)$ is $b - a$, and that the submeasure of the real line is infinite; for the former, see Fremlin's book on measure theory, lemma 114B, and the commentary on it in proposition 565B(b). The latter follows from the former and the submeasure being a submeasure, proposition 565B(a) of the same. – user361424 Jan 09 '18 at 02:52
  • @Julian Yes, the standard (classical) definition results in measure $|b-a|$ for any of the 4 intervals with endpoints $a $ and $b $. Choice is not needed here. – Andrés E. Caicedo Jan 09 '18 at 12:43
  • @Julian I review the classical definition in comments to this answer. But note the restriction there to bounded sets. – Andrés E. Caicedo Jan 09 '18 at 12:56

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Let $[0,1]=\bigcup_{i=0}^{\infty}A_i$, where the $A_i$ are countable. Let $B=\bigcup_{i=0}^{\infty}A_i+i$. I argue below that $B$ is a counterexample to your question. Write $m(X)$ for the outer submeasure of a set $X$.

  1. The union of finitely many countable sets is countable. So, every bounded subset of $B$ is countable and hence null.
  2. Assume $m(C)$ is finite. Then $C\subseteq\bigcup_{i=0}^{\infty} C_i$, where each $C_i$ is an open segment and $\sum_{i=0}^{\infty}m(C_i)<\infty$. Now for every $\varepsilon>0$, there is a finite index $n$ such that $m(\bigcup_{i=n+1}^{\infty}C_i)<\varepsilon$. On the other hand, $\bigcup_{i=0}^{n}C_i$ is bounded. Therefore, $$m(B\cap C)\le m(B\cap\bigcup_{i=0}^{n}C_i)+m(B\cap\bigcup_{i=n+1}^{\infty}C_i)<0+\varepsilon, $$ so $m(B\cap C)=0$.
  3. Caratheodory's criterion for $B$ follows from (2).
  4. Let $B\subseteq\bigcup_{i=0}^{\infty}C_i$, where each $C_i$ is an open segment and $m(C_i)<1$. For $i\in\mathbb{N}$, let $D_{2i}=C_i-\lfloor\sup(C_i)\rfloor$ and $D_{2i+1}=D_{2i}+1$. Now $m(D_{2i})=m(D_{2i+1})=m(C_i)$. Consider $x\in[0,1]$. There is some $k$ such that $x\in A_k$, and hence $k+x\in B$. So, there is an index $i$ such that $k+x\in C_i$. Then $k<\sup(C_i)<k+2$, and consequently $\lfloor\sup(C_i)\rfloor=k$ or $\lfloor\sup(C_i)\rfloor=k+1$, which implies that $x\in D_{2i}$ or $x\in D_{2i+1}$. In either case, $x\in\bigcup_{i=0}^{\infty}D_i$. Hence, $[0,1]\subseteq\bigcup_{i=0}^{\infty}D_i$, and therefore $$\sum_{i=0}^{\infty}m(C_i)=\sum_{i=0}^{\infty}m(D_i)/2\ge 1/2. $$ Therefore, $m(B)>0$.
  5. By (2) and (4), it follows that $m(B)=\infty$.
Taneli Huuskonen
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