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 $2^\mathbb{R}$)



How additive is Lebesgue measure in ZF+AD ?
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](http://ams.org/bookstore-getitem/item=CONM-533)**, 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](http://andrescaicedo.files.wordpress.com/2008/04/conm10510.pdf), 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](https://math.stackexchange.com/a/372218/462) 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 $\tau$.




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$(\*)$ 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](https://math.stackexchange.com/a/88811/462); 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](http://www.essex.ac.uk/maths/people/fremlin/mt.htm) 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 $\mathsf{AD}$.


