11
$\begingroup$

Question. Is it consistent with ZF that every (countably additive, non-negative) measure $\mu: \Sigma \to \bf R$, where $\Sigma$ is a sigma-algebra on a given set $X$, extends to a (countably additive, non-negative) measure $\tilde{\mu}: \mathcal P(X) \to \bf R$?

What I seem to know: Let us first work in the frame of ZFC. Given a set $X$ of cardinality $\ge \aleph_1$, we take $\Sigma$ to be the smallest sigma-algebra on $X$ containing all the countable subsets of $X$. It is seen that the function $$\mu: \Sigma \to {\bf R}: A \mapsto \left\{ \begin{array}{ll} \! 0 & \text{if } |A| \le \aleph_0 \\ \! 1 & \text{otherwise} \end{array} \right. $$ is a measure on $X$. Thus, if $\mu$ extends to a (countably additive, non-negative) measure $\mathcal P(X) \to \bf R$, then $|X|$ is a real-valued measurable cardinal, which implies the existence of a weakly inaccessible cardinal by an old theorem of S. Ulam, see

S. Ulam, Zur Masstheorie in der allgemeinen Mengenlehre, Fund. Math. 16 (1930), 140-150 (in German),

or Corollary 10.15 in:

T. Jech, Set Theory - The Third Millennium Edition, Revised and Expanded, Springer Monogr. Math., Springer, Berlin, 2006 (corrected 4th printing).

However, the existence of a weakly inaccessible cardinal is unprovable from the axioms of ZFC.

On the other hand, we have from:

R. M. Solovay, ``Real-valued measurable cardinals'', pp. 397-428 in: D. Scott (ed.), Axiomatic set theory, Proc. Sympos. Pure Math. (Univ. California, Los Angeles, CA, 1967), Vol. XIII, Part I, Amer. Math. Soc.: Providence, RI, 1987 (reprinted ed.)

that the existence of measurable cardinals in ZF is equiconsistent with the existence of measurable (resp., real-valued measurable) cardinals in ZFC. However, this doesn't answer the question I'm posing, as far as I can say (in particular, it occurs to me that Solovay's result doesn't cover the case of real-valued measurable cardinals in ZF, does it?).

$\endgroup$
  • $\begingroup$ Actually Solovay's original result was that it is equiconsistent with ZFC that in ZF there is a total extension of the Lebesgue measure. I'm not sure about its additivity, though, because you are losing countable choice there and the whole thing about additivity becomes quite fussy then. $\endgroup$ – Asaf Karagila Jan 25 '16 at 22:42
  • $\begingroup$ Your title is also misleading. You ask "true or false" but you actually ask "consistent or inconsistent". If something is not provable from ZFC it will certainly not be provable from ZF. $\endgroup$ – Asaf Karagila Jan 25 '16 at 22:48
  • $\begingroup$ Right. I'm editing the title according to your comment. $\endgroup$ – Salvo Tringali Jan 25 '16 at 22:51
  • $\begingroup$ (Solovay's original result is from 1965, by the way.) $\endgroup$ – Asaf Karagila Jan 25 '16 at 22:52
  • $\begingroup$ Really? I'm relying on second-hand information concerning Solovay's result. So, please, feel free to fix the OP and provide the correct reference! $\endgroup$ – Salvo Tringali Jan 25 '16 at 22:54
6
$\begingroup$

EDIT: This answer is wrong. I am not deleting it at the request of the OP. See the counterexample at the bottom.

Here is a terrible mathematician answer that you didn't know that you didn't care for. Yes, it is consistent, at least assuming the consistency of very large cardinals.

In Gitik's model, where every limit ordinal has countable cofinality we also have the following: If we start with the class of singletons and close it under countable unions, we obtain the entire universe. And I claim that in the presence of this axiom, every countably additive measure is either atomic (and thus extends naturally) or trivial (and thus extends naturally).

To see why, if the measure gives $0$ to singletons, then by $\sigma$-additivity it has to give $0$ to every countable set, and so to every set which can be generated by a countable union of countable sets. Which is every set. If there are singletons which have positive measure, then they are necessarily the atoms and we're done.

Some relevant papers:

  1. Arnold W. Miller, Long Borel hierarchies, MLQ Math. Log. Q. 54 (2008), no. 3, 307--322.

  2. M. Gitik, All uncountable cardinals can be singular, Israel J. Math. 35 (1980), no. 1-2, 61--88.


Edit. So here is a counterexample, and some consequence of it.

In Gitik's model, $\Bbb R$ is a countable union of sets $A_n$, such that each $A_n$ is the countable union of countable sets. In particular, there is a countable sequence of countable sets $X_n$ whose union is uncountable.

Now, if $X_n$ is countable $R_n=\{r\in\Bbb R\mid r\text{ codes an enumeration of }X_n\}$ gives us a family without a choice function. If we could choose $r_n\in R_n$ then we could have enumerated all the $X_n$'s uniformly and their union would have been countable.

Now consider $\Sigma$ as the $\sigma$-algebra generated by the $R_n$'s, and since they are obvious pairwise disjoint (a real cannot code more than one countable set), $\Sigma$ is isomorphic to $\mathcal P(\Bbb N)$ with $R_n$ as atoms.

If we consider the atomic measure on $\mathcal P(\Bbb N)$ such that $\mu(\{n\})=1/2^{n+1}$, it induced a measure on $\Sigma$. Now I claim that $\mu$ cannot be extended to a measure on $\mathcal P(\Bbb R)$.

Suppose that it could have been extended, then it is impossible that every singleton has measure $0$, since $\sigma$-additivity would have implied that every countable set has measure $0$, thus every $A_n$ (from the partition of $\Bbb R$) has measure $0$ as a countable union of null sets; and finally $\Bbb R$ has measure $0$. But this is impossible since $\mu(\Bbb R)=1$.

So each $R_n$ has some countable subset whose elements have all positive measure summing to $1/2^{n+1}$. But if $\sum_n r_n<\infty$ it has to be the case that for every $k$, at most finitely many $r_n$'s satisfy $\frac1k < r_n$. So there is a finite set of reals with maximal weight inside each $R_n$.

So we have $R'_n\subseteq R_n$, finite sets of reals defined as above. But finite sets of reals have canonical choice functions in them: take the minimum. And this is a contradiction to the assumption that there is no choice function from $\{R_n\mid n\in\Bbb N\}$.

As a consequence we see that the "total extension principle" cannot be true if there exists an uncountable set $X$ generated by iterated countable unions from its singletons; and there is a choice function from finite subsets of $X^\omega$.

$\endgroup$
  • $\begingroup$ I will need a couple of days or so to go through the references you have provided. In the meanwhile, thank you for your answer. $\endgroup$ – Salvo Tringali Jan 25 '16 at 23:35
  • $\begingroup$ Sure. It's pretty late, so I might have had some serious oversight. Let me know if you notice anything fishy. $\endgroup$ – Asaf Karagila Jan 25 '16 at 23:38
  • $\begingroup$ Asaf, perhaps I misunderstand, but you seem to imply in your remarks that every set in Gitik's model is a countable union of countable sets. But that is stronger than what I think is true there. Without that, you would seem to iterate the countable unions, but since the cardinals in Gitik's model are not well-founded (or are they?), I don't see how the iteration argument goes through. Can you explain a bit more? $\endgroup$ – Joel David Hamkins Jan 26 '16 at 0:06
  • $\begingroup$ @Joel: Define a rank function, singletons have rank $0$, and now step up with countable unions of sets of lower rank. Then in Gitik's model you get the entire universe like that. The point being that generally, sigma algebras are already more or less power sets to begin with in this model. $\endgroup$ – Asaf Karagila Jan 26 '16 at 0:24
  • 1
    $\begingroup$ @Joel: You're probably right. But it will have to wait for the morning. :-) in any case, this is true in Gitik's model, and Arnie Miller gives some survey of this axiom (due to Specker if my memory serves me right) in his paper also. $\endgroup$ – Asaf Karagila Jan 26 '16 at 0:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.