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

Question

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?

Answer 1

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 $m(B)=\infty$.