Can it be shown in ZF that a set of infinite Lebesgue submeasure must have a subset of arbitrarily large finite submeasure?

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?

• Where do you think choice gets involved? – Asaf Karagila Jan 4 '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 4 '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 4 '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 4 '18 at 10:49
• 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 5 '18 at 7:11

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 $m(B)=\infty$.
• How do you argue that $B$ is not null? – Andrés E. Caicedo Jan 11 '18 at 18:24
• Integer translations of a cover of $B$ yield a cover of the unit interval. I will edit my answer to explain. – Taneli Huuskonen Jan 11 '18 at 19:35
• We do not have countable additivity, so I am not sure how that helps. Each $A_i$ has measure zero after all. I look forward to your edit. – Andrés E. Caicedo Jan 11 '18 at 20:07