TITLE: How additive is Lebesgue measure in ZF+AD ?
QUESTION [7 upvotes]: 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 $2^\mathbb{R}$)

REPLY [3 votes]: 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 $\lambda(Y)^2$.