This is a follow-up to this question. We say that a set $A \subseteq \mathbb{R}$ is bounded if there exists a finite interval $(a,b)$ such that $A \subseteq (a,b)$.

Working in $\mathsf{ZFC}$, the existence of a (Lebesgue) non-measurable set (of $\mathbb{R}$) easily implies the existence of a bounded non-measurable set. A proof is as follows - Let $X \subseteq \mathbb{R}$ be non-measurable. Then $X = \bigsqcup_{n \in \mathbb{Z}} X \cap (n,n + 1]$. If all $X \cap (n,n+1]$ are measurable, then $X$ must also be measurable, at it is a countable union of measurable sets ($\mathsf{AC}$ is used here). Thus, there exists an $n \in \mathbb{Z}$ in which $X \cap (n,n+1] \subseteq (n,n+2)$ is not measurable.

However, it appears to not be so clear if we only work in $\mathsf{ZF}$, since we cannot guarantee that the countable union above is measurable. I also can't seem to see an easy way around choice here. Thus, if we write:

  1. $\mathsf{NM}$ as "there exists a non-measurable set".
  2. $\mathsf{BNM}$ as "there exists a bounded non-measurable set".
  3. $\mathsf{M}_\omega$ as "countable union of measurable sets is measurable".

My questions are (assuming $\mathrm{Con}(\mathsf{ZFC})$):

  1. Is $\mathsf{ZF} + \neg\mathsf{M}_\omega$ consistent?
  2. Is $\mathsf{ZF} + \mathsf{NM} + \neg\mathsf{BNM}$ consistent?
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    $\begingroup$ @Wojowu That’s in fact a comment I originally posted on the linked answer, but on further reflection, I don’t think it works without AC. If you cover $f(X)$ by an open set $U$ whose measure is close to the measure of $f(X)$, the preimage of $U$ may still be much larger than $X$ (unless $f$ is bi-Lipschitz, in which case it can’t have bounded image). Basically, you still need to split the set to countably many pieces and approximate each piece by open sets separately, and it requires countable choice to collect the pieces together. $\endgroup$ Commented May 18, 2021 at 13:47
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    $\begingroup$ As for question 1, it is consistent with ZF that $\mathbb R$ (and therefore every subset thereof) is a countable union of countable (and therefore null) sets. This certainly implies that Lebesgue measure is not $\sigma$-additive; I’m not sure though whether it also implies the existence of a non-measurable set (i.e., $\neg M_\omega$). $\endgroup$ Commented May 18, 2021 at 13:51
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    $\begingroup$ There is a theorem (or, in some developments, a definition) ... A set $A$ is measurable if and only if $A \cap [a,b]$ is measurable for every $a<b$. Are you claiming this is cannot be proved in ZF? Perhaps improve the question by including the definition of "measurable". $\endgroup$ Commented May 18, 2021 at 14:04
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    $\begingroup$ @GeraldEdgar I believe the standard definition of Lebesgue measure in this context is as follows. (1) Define the measure of open sets: any such is a countable disjoint union of open intervals (possibly infinite), take the sum of the lengths of the intervals. (2) Define the outer measure $\lambda^*(X)=\inf\{\lambda(U):U\supseteq X\text{ open}\}$. (3) Define $X$ to be measurable if $\lambda^*(Y)=\lambda^*(Y\cap X)+\lambda^*(Y\smallsetminus X)$ for all bounded intervals $Y$ (or possibly a larger collection of $Y$, I’m not sure what is most common in this point). $\endgroup$ Commented May 18, 2021 at 14:12
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    $\begingroup$ Hmm. Though if I take only bounded intervals as test sets as I just wrote, it should actually work. I was originally thinking about arbitrary test sets $Y$. So perhaps this part of the definition is a crucial point that needs to be clarified. $\endgroup$ Commented May 18, 2021 at 14:20

2 Answers 2


The answer to your first question is yes, and the answer to your second question is no, under any of the multiple definitions of "measurable" in choiceless contexts.

We will prove a theorem relating various measure-theoretic consequences of countable choice.

(ZF) The following are equivalent. Note that (1)-(6) are about the algebra of subsets of $[0,1]$ which satisfy $\lambda^*(X)+\lambda^*([0,1] \setminus X)=1$ while (7) refers to the algebra of subsets of $\mathbb{R}$ which satisfy $\lambda^*([-n, n] \cap X) + \lambda^*([-n, n] \setminus X) = 2n$ for all $n.$

  1. Lebesgue measure is $\sigma$-additive.
  2. A countable union of measurable sets is measurable.
  3. A countable union of null sets is measurable.
  4. A countable union of null sets is null.
  5. Every null set is contained in a null $G_{\delta}$ set.
  6. For every measurable set $X,$ there is an $F_{\sigma}$ set $A$ and a $G_{\delta}$ set $B$ such that $A \subset X \subset B$ and $B \setminus A$ is null.
  7. Any of the above but for measurable subsets of $\mathbb{R}.$


(1) $\rightarrow$ (2) Clear.

(2) $\rightarrow$ (3) Clear.

(3) $\rightarrow$ (4) Suppose towards contradiction $X_i$ are null sets with $\lambda(\bigcup_{i<\omega} X_i)>0.$ Let $Y_i = \{x+\frac{m}{2^i}: m < 2^i, \exists j < i (x \in X_j)\}$ (note that addition is mod 1) and $Z_i = Y_i \setminus Y_{i-1}.$ In particular, the $Z_i$ are disjoint null sets whose union has positive measure and is closed under translation by dyadic rationals.

For $A \subset \omega,$ let $H_A=\bigcup_{n \in A} Z_n.$ By (3), each $H_A$ is measurable. We will show that for every $A \subset \omega,$ either $H_A$ or $H_{\omega \setminus A}$ has measure 1.

Suppose $H_A$ has positive measure (otherwise $H_{\omega \setminus A}$ has positive measure). Fix $\epsilon>0.$ By Lebesgue density theorem, there is some interval $I$ of length $\frac{1}{2^n}$ such that $\lambda(H_A \cap I) > \frac{1-\epsilon}{2^n}.$ Clearly $H_{A \setminus (n+1)}$ also satisfies this inequality, and is furthermore closed under translation by $\frac{1}{2^n}.$ Thus, $\lambda(H_A) = \lambda(H_{A \setminus (n+1)}) > 1-\epsilon,$ so $\lambda(H_A)=1.$

Since $H_{\omega}$ has measure 1, we see that $[0,1]$ is a countable union of null sets. By (3), every subset of $[0,1]$ is measurable. However, $\{A \subset \omega: H_A \text{ is measure 1} \}$ is a non-principal ultrafilter, so there is a nonmeasurable subset of $[0,1],$ contradiction.

(4) $\rightarrow$ (1) Let $X_i$ be measurable sets. Let $U_i$ enumerate the basic open sets. Define $S_n \subset \omega$ recursively by having $i \in S_n$ iff there is $j$ such that $\lambda(U_i \cap X_j \setminus \bigcup_{k<i, k \in S_n} U_k) > \frac{n}{n+1} \lambda(U_i).$ Let $V_n = \bigcup_{i \in S_n} U_i.$ Then $V:=\bigcap_{n < \omega} V_n$ is a $G_{\delta}$ set such that $\lambda(V)=\sup_{n<\omega} \lambda(\bigcup_{i < n} X_i)$ and $V \triangle \bigcup_{i<\omega} X_i$ is null.

(4) $\rightarrow$ (5) Let $X$ be null. By (4) we can assume $X$ is closed under translation by rational numbers. Let $U$ be an open cover of $X$ of measure less than $\frac{1}{2}.$ We can canonically cover $X \cap [0, \frac{1}{n}]$ with an open set of measure $\frac{1}{2n}$ by considering the least $m$ such that $\lambda(U \cap [\frac{m}{n}, \frac{m+1}{n}])<\frac{1}{2n}.$ We can thus recursively construct open covers of $X$ of measure less than $\frac{1}{2^n}.$

(5) $\rightarrow$ (6) Similarly to in (4) $\rightarrow$ (1), there is a $G_{\delta}$ set $B_1$ with null symmetric difference from $X.$ Let $B_2$ be a null $G_{\delta}$ set containing $X \setminus B_1.$ Then $B:=B_1 \cup B_2$ is a $G_{\delta}$ set with $X \subset B$ and $B \setminus X$ null. We can similarly construct such a $B'$ for $[0, 1] \setminus X$ and set $A = [0, 1] \setminus B'.$

(6) $\rightarrow$ (4) Let $X_i$ be null sets. Let $X=\bigcup_{i<\omega} X_i.$ Consider $Y = \{2^{-i-1}(1+x): x \in X_i\}.$ It is easy to see $Y$ is null, and thus contained in some null $G_{\delta}$ set $Y'.$ Let $U_n$ be a sequence of open sets covering $Y,$ each satisfying $\lambda(U_n)<\frac{1}{n}.$

Fix $\epsilon>0$ and $i<\omega.$ Let $n$ be least such that $\frac{1}{2^n}<\frac{\epsilon}{4^{i+1}}.$ Then $X_i$ is covered by a translation of $U_n$ scaled up by $2^{i+1},$ which has measure less than $\frac{\epsilon}{2^{i+1}}.$ Applying this construction to all $i,$ we get a cover of $X$ of measure less than $\epsilon.$ Thus, $X$ is null.

(7) Finally, it is routine to verify that assertions (1)-(6) collectively prove their generalizations to $\mathbb{R}.$ E.g., one could verify $\frac{1}{x}$ on $(0, \infty)$ sends null sets to null sets and measurable sets to measurable sets using the fact that it's Lipschitz on each $[2^{-n}, 2^n]$ and that $\frac{1}{x}$ sends $G_{\delta}$ null sets to $G_{\delta}$ null sets. $\square$

Thus, $\text{M}_{\omega}$ fails in any model of ZF + "$\mathbb{R}$ is a countable union of countable sets" since this theory negates (1), providing an affirmative answer to question 1. As for question 2, if $\neg \text{BNM}$ holds, then we immediately have (2), so every subset of $\mathbb{R}$ is measurable. In particular, all interpretations of "all sets are measurable" are equivalent.

  • $\begingroup$ Aha! This is great! However, I think the last paragraph uses DC... (Raisonnier's Theorem) $\endgroup$ Commented May 19, 2021 at 6:31
  • $\begingroup$ @FrançoisG.Dorais A non-principal ultrafilter is a nonmeasurable set because it violates Lebesgue density. I think Raisonnier's Theorem is that an $\omega_1$-sequence of reals implies there is a nonmeasurable set, which I didn't use. $\endgroup$ Commented May 19, 2021 at 6:34
  • $\begingroup$ You're right! So Lebesgue Density must use CC or DC? $\endgroup$ Commented May 19, 2021 at 6:37
  • $\begingroup$ @FrançoisG.Dorais No, Lebesgue Density is a choiceless theorem. The standard proof doesn't use any choice. $\endgroup$ Commented May 19, 2021 at 6:38
  • $\begingroup$ Ah, I think I got it! Lebesgue Density requires you to fix a notion of measurable? $\endgroup$ Commented May 19, 2021 at 6:40

This is more of a long comment than an answer.

The "right" notion of an unbounded set being measurable in ZF is less than clear. Suppose $\mathbb{R}$ is a countable union of countable sets. Let $\langle X_n: n<\omega \rangle$ be a partition of $[0, 1]$ into countable sets. Consider $X=\{x+m: n\le m<\omega, x \in X_n\}.$ It's not hard to show that $X$ has infinite outer measure and zero inner measure. Also, responding to a discussion in the comments, $\{\frac{1}{x}: x \in X\}$ is null since it can be written as a union of a countable set and a set of arbitrarily small outer measure. While this set satisfies the definition of measurability Emil provided in the comments, it violates almost everything we expect measurable sets to satisfy.

I think a better definition would be that $X \subset \mathbb{R}$ is measurable if for all $\epsilon>0,$ there is an open cover $C$ of $X$ such that for all bounded intervals $I,$ the restriction of $C$ to $I$ has measure less than $\lambda_*(X \cap I)+\epsilon,$ though I don't know if there's any literature studying this. Anyway, under this definition, your first question has an affirmative answer as demonstrated by the previous example, and the second question seems nontrivial.

Edit: Actually we need both inner and outer uniformity for this to be a finitely additive measure. Say $X$ is measurable if for all $\epsilon>0,$ there are open sets $U_1, U_2$ such that $X \subset U_1,$ $\mathbb{R} \setminus X \subset U_2,$ and $\lambda(U_1 \cap U_2)<\epsilon.$

  • $\begingroup$ While I agree now that the right definition of measurability should be the one in your edit (which I would prefer to state as: for all $\epsilon>0$ there exist open $U$ and closed $F$ such that $F\subseteq X\subseteq U$ and $\lambda(U\smallsetminus F)<\epsilon$), this is not required for finite additivity: $\lambda$ (that is, $\lambda^*$) is finitely additive on the algebra of Carathéodory measurable sets as defined in my comments above. $\endgroup$ Commented May 19, 2021 at 15:23
  • $\begingroup$ My comment was referring to my pre-edit definition, which isn’t even closed under complements. $\endgroup$ Commented May 19, 2021 at 15:41

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