# Given a model of ZF where $\mathbb{R}$ is the countable union of countable sets, does every subset of $\mathbb{R}$ have measure zero?

The question basically says it all. It is a well-known result that there exists a model $\mathcal{M}$ of ZF with the property that $\mathbb{R}^{\mathcal{M}}$ (here, $\mathbb{R}^{\mathcal{M}}$ is the element of $\mathcal{M}$ that $\mathcal{M}$ thinks is its real-number field) is the countable union of countable sets. How can this be compatible with the assertion that the unit interval $[0,1]^{\mathcal{M}}$ (the object that $\mathcal{M}$ thinks is the unit interval) has measure $1$? Herein lies the problem. Every countable set has measure $0$, hence a countable union of countable sets is a countable union of measure-$0$ sets, which, in turn, implies that $\mathbb{R}$ has measure $0$. It then follows that $[0,1]$ has measure $0$.

Unless some form of the Axiom of Choice (AC) is needed to make sense of measure theory (i.e., some variant of AC is required to ensure that the standard Borel measure $\mu$ exists in the first place), I do not see any way to resolve this issue.

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What precisely do you mean by "measure"? ZF doesn't prove the existence of any function with all the properties we want Lebesgue measure to have. – Chris Eagle Nov 2 '12 at 18:49
I was thinking that ZF itself does not allow us to construct a measure on the Borel subsets of $\mathbb{R}$, but I couldn't be sure. Do you know if ZF + DC is sufficient to give us a measure? – Haskell Curry Nov 2 '12 at 18:54
Yes, the usual construction (and proof of sigma-additivity) of Lebesgue measure works in ZF+DC. – Chris Eagle Nov 2 '12 at 18:55
I see. Hence, can I take it that this is a proof that the model $\mathcal{M}$ mentioned above does not satisfy DC? Otherwise, the Borel measure would exist and we thus end up with a contradiction. Thanks for your help! – Haskell Curry Nov 2 '12 at 19:22
Yes, and it doesn't satisfy CC either, for the same reason. $\:$ – Ricky Demer Nov 2 '12 at 19:38

If we insist that the Lebesgue (or Borel) measure is $\sigma$-additive, then yes. It is immediate that countable sets have measure zero; and therefore the countable union of disjoint countable sets has to be of measure zero as well.

One of the ways to solve this is to give up the $\sigma$-additivity of the measure. Another way is to use Borel codes. Borel codes are effective recipes to make sets of real numbers, we take not just any unions and intersections and complements (like we would normally), but we also "keep track" of what we did so far.

Fortunately there is always a surjective map from the real numbers onto Borel codes, so even without choice there are not too many of them (even if all sets are Borel, not all of them have codes). There is quite a rich theory about these sets and how we can develop a measure on them, you can read it on Fremlin, Measure Theory, vol. 5.

I should also note that I recently discovered that it is consistent with ZF that:

1. Every countable union of countable sets of real numbers is countable;
2. Every set of real numbers is Borel (in the sense that the smallest $\sigma$-algebra containing the open sets is $\mathcal P(\mathbb R)$)
3. The cofinality of $\omega_1$ is countable. Namely, $\omega_1$ is the countable union of countable ordinals.

Therefore measure theory can get screwed even if the real numbers are not a countable union of countable sets.

In terms of choice principles, the axiom of countable choice is usually enough to conclude that measure theory behaves nicely, and dependent choice (which is slightly stronger) is certainly enough to develop almost all classical analysis and basic measure theory. Both would imply that $\omega_1$ is regular and that not all sets of real numbers are Borel.

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