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What is known about the additivity of Lebesgue measure under the Axiom of Determinacy?
That is, for what cardinals $\kappa$ do we have

with $|I| = \kappa$, for all functions $f : I \to 2^\mathbb{R}$, $\; \lambda(\displaystyle\bigcup_{i\in I}\ f(i)) = \operatorname{sup}(\{\lambda(\displaystyle\bigcup_{i\in J}\ f(i)) : J\subseteq I \land |J|<\infty\})$.

(This is equivalent to the usual definition when $\kappa \leq \aleph_0$.)



For example:


Does it hold for $\aleph_1$?

If so, what alephs does it hold for?

Does it hold nontrivially for any infinite cardinals that are not alephs?
(nontrivially meaning sets of that cardinality can be mapped onto non-well-orderable subsets of $2^\mathbb{R}$)

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I already know ZF+AD proves "For all cardinals κ, if $\kappa \leq \aleph_0$, then Lebesgue measure is κ-additive". I don't need anyone to point that out to me. :-) –  Ricky Demer Dec 7 '10 at 2:56
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2 Answers

up vote 4 down vote accepted

Ricky:

I think I see how to answer the problem under a stronger assumption. Rather than $\mathsf{AD}$, work in $$ {\sf AD}^+ + V=L({\mathcal P}({\mathbb R})). $$ This is a bit unsatisfying, since it is very possible the question can be answered assuming only $\mathsf{AD}$. In any case, $\mathsf{AD}^+$ is potentially harmless, since it may be that $\mathsf{AD}$ implies $\mathsf{AD}^+$; but it seems strange to need here the additional machinery that $\mathsf{AD}^+$ allows us. The assumption on the form of $V$ is more immediately troublesome, since it actually makes some (non-well-ordered) cardinals "invisible". (Meaning, there may be cardinals $\tau$ that inject into $2^{|{\mathbb R}|}$ but not in $L({\mathcal P}({\mathbb R}))$, so our assumption is potentially simplifying the question.)

Using $\mathsf{AD}^+ + V=L({\mathcal P}({\mathbb R}))$, Richard Ketchersid and I proved (in A trichotomy theorem in natural models of ${\sf AD}^+$, in Set Theory and Its Applications, Contemporary Mathematics, 533, Amer. Math. Soc., Providence, RI, 2011, pp. 227-258) that every set either is well-orderable, or is at least as large as ${\mathbb R}$. (The paper is available here, and it also provides an introduction to $\mathsf{AD}^+$.)

Obviously, Lebesgue measure is not ${\mathbb R}$-additive, considering singletons, so this means it is enough to answer the question for well-ordered cardinals. But given a well-ordered union of sets of reals, we may assume they are pairwise disjoint (arguing inductively). And then it is standard that only countably many of them can be of positive measure. $(*)$

This shows (under the listed assumptions) that Lebesgue measure is $\kappa$-additive for all (well-orderable) $\kappa$ but not $\tau$-additive for any non-well-orderable cardinal $\tau$.


$(*)$ To see that the last paragraph follows, I need to argue that a well-ordered union of measure zero sets has measure zero. In fact, under $\mathsf{AD}$, a bit more holds, namely, $$ \iint_{[0,1]^2}f(x,y)dxdy=\iint_{[0,1]^2}f(x,y)dydx $$ for any bounded $f:[0,1]^2\to{\mathbb R}$ (in particular, the integrals are defined).

From this it is easy to conclude the claim about measure zero sets, e.g., by looking at minimal counterexamples and adapting the standard Sierpiński argument; note that this only uses $\mathsf{DC}({\mathbb R})$ (or even just $\mathsf{AC}_\omega(\mathbb R)$, which follows from $\mathsf{AD}$) and that all sets of reals are Lebesgue measurable.

(Unfortunately, although this Fubini theorem is not too hard, I am not sure of a reference for it. I remember vol. 5 of Fremlin's treatise has it as an exercise.) Let me add that this is a folklore result that is frequently used, as is the dual fact that a well-ordered union of meager sets is meager under $\mathsf{AD}$.

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How do you get from your second-to-last sentence to your last sentence? –  Ricky Demer Dec 7 '10 at 4:14
    
No, it follows that if we have uncountably many disjoint sets, all but countably many have measure zero. –  Ricky Demer Dec 7 '10 at 7:02
    
@Ricky : Sorry for the obvious gap. I added a few lines. –  Andres Caicedo Dec 7 '10 at 7:52
    
What is "the standard Sierpinski argument"? Also, would Fubini hold for all nonnegative functions too? –  Ricky Demer Dec 7 '10 at 7:58
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In case one wants to keep track of the assumptions: The $\kappa$-additivity for $\kappa$ a well-ordered cardinal only uses AD; the amount of choice needed (usually called $AC_\omega({\mathbb R})$) is actually provable from AD. The failure of $\tau$-additivity for $\tau$ non-well-orderable requires the additional assumptions AD${}^++V=L({\mathcal P}({\mathbb R}))$ (in particular, our proof that ${\mathbb R}$ embeds into $\tau$ uses $DC_{\mathbb R}$, and an induction through the "constructible" levels of $V$). –  Andres Caicedo Dec 7 '10 at 8:25
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If $\kappa$ is a well orderable cardinal (i.e. an $\aleph_\alpha$) I don't think there is a need to involve determinacy. If all sets of reals are Lebesgue measurable (and DC holds), then Lebesgue measure is $\aleph_\alpha$ additive for every $\alpha$. This is sufficient (at least granting DC) since AD certainly implies all sets of reals are Lebesgue measurable.

First argue that if $\kappa$ is a well orderable cardinal (i.e. an $\aleph_\alpha$) and Lebesgue measure is not $\kappa$ additive, then this can be witnessed in such a way that the sets in the union are all null sets (there is a countable subcollection whose union has measure equal to the sup in your equation). Without loss of generality, we may arrange that $$ \lambda ( \bigcup_{\xi \in \alpha} f(\xi) ) = 0 $$ for all $\alpha \in \kappa$. Define $Y = \bigcup_{\alpha \in \kappa} f(\alpha)$. Now define $$ X = \{(x,y) \in Y^2 : \alpha_x < \alpha_y\} $$ where $\alpha_x = \min\{\alpha \in \kappa : x \in f(\alpha)\}$. The set $X$ now violates Fubini's theorem: the set of all horizontal sections are the union of fewer than $\kappa$ many of the sets $f(\xi)$ and the vertical sections have their complement (in $Y$) being the union of fewer than $\kappa$ sets of the form $f(\xi)$. Thus by integrating in one direction, $X$ has measure 0, while in the other direction, its measure is $\lambda(Y)^2$.

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Hi Justin. How do you argue when $\kappa$ is a non-well-orderable cardinality? –  Andres Caicedo Apr 2 '11 at 21:26
    
@Andres: You are right -- my answer just concerns the $\aleph$'s. I understood that to be the main part of the question but upon reading it again, I see that isn't necessarily the case. –  Justin Moore Apr 2 '11 at 23:41
    
@Andres: I edited appropriately. –  Justin Moore Apr 2 '11 at 23:43
    
Justin, as a minor technical comment, all you need is ${\sf DC}_{\mathbb R}$, DC for sets of reals, rather than the full version of DC; this holds in all known models of determinacy, while DC fails is some of them. –  Andres Caicedo Apr 2 '11 at 23:49
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