I seem to remember reading somewhere that ZF+AD proves that $\omega\_1$ and $\omega\_2$ are measurable cardinals.


Is that right?


If so, can someone [point me to or give here] a [sketch or proof] of these results?



Measurable cardinals under Axiom of Determinacy
An alternative proof for $\omega\_1$ goes via the set $D$ of Turing degrees (also known as degrees of unsolvability). $D$ is upward directed by the ordering $\leq\_T$ of Turing reducibility, so the cones $C\_d=\{p\in D:d\leq\_Tp\}$ generate a filter on $D$. The axiom of determinacy implies that this filter is an ultrafilter, i.e., that every subset $X$ of $D$ either includes a cone or is disjoint from a cone. (Proof: Define a game where the two players alternately choose natural numbers, producing an infinite sequence $x$; player 1 wins iff the Turing degree of $x$ is in $X$. Any strategy for either player, being a function from finite sequences of natural numbers to natural numbers, can be coded as a sequence $s$ of natural numbers. Any Turing degree $p$ in the cone $C\_s$ [pedantically, the subscript should be the degree of $s$] is the degree of some play in which the player who has strategy $s$ uses it, because the other player can play any sequence of degree $p$ and then the overall play will also have degree $p$. So, if $s$ is a winning strategy for player 1 (resp. player 2), then all (resp. none) of the degrees in $C\_s$ must be in $X$.) So AD gives an ultrafilter on $D$, called the cone ultrafilter. It's non-principal (because there is no highest Turing degree) and countably complete (because AD implies that all ultrafilters are countably complete --- this follows from the fact that there are, under AD, no non-principal ultrafilters on $\omega$). 


To get from $D$ to $\omega\_1$, use the function $f$ assigning to each Turing degree $d$ the ordinal usually called $\omega^{CK}(d)$, the first ordinal that is not the order-type of an ordering of $\omega$ of degree $\leq\_Td$. The image of the cone ultrafilter under this map $f$ is again a countably complete ultrafilter, now on $\omega\_1$. It is non-principal because $f$ takes arbitrarily large countable ordinals as values on any cone.


The ultrafilter obtained in this way is actually the same as the club filter mentioned in other answers, but the proof via Turing cones needs somewhat less recursion theory than Solovay's proof (which used Kleene's boundedness theorem for $\Sigma^1\_1$ sets).


One can also prove the measurability of $\omega\_2$ by projecting the cone filter from $D$ to $\omega\_2$ via a suitable map. (I don't remember right now whether the suitable map sends a degree $d$ to the successor cardinal in $L[d]$ of (genuine) $\omega\_1$ or to the next Silver-indiscernible over $d$ after (genuine) $\omega\_1$; maybe both of them work.)


