$\operatorname{AD}$ and the measurability of $\omega_1$ Are there proofs of the measurability of $\omega_1$ (under $\operatorname{AD}$) that do not use Turing degrees nor the $\Sigma_1^1$ boundedness lemma? 
I've been struggling to find an "elementary" proof of this fact. Note that I consider the proof of "Assume $\operatorname{AD}$. Then every ultrafilter is $\sigma$-complete" to be elementary. 
The club filter on $\omega_1$ is thus readily seen to be a $\sigma$-complete filter, but the "ultra" part seems not to be so easy to prove (it's for an introductory course on $\operatorname{AD}$, without too much knowledge of recursivity etc.)
 A: I'm not sure this will work for you, but there's a way to recast the Turing argument so that it avoids recursion theory; if this is the reason you want to avoid the Turing argument, this might be the way to go.
Namely, instead of working with Turing degrees, work with a coarser reducibility, which is easier to explain. For anything coarser than Turing reducibility, the cone property holds by the same argument, and we can define an analogous $f$ mapping reals to ordinals.
Specifically, here are the details for one particularly nice notion, relative projectiveness (or projective reducibility). Say a real $r$ is projective relative to a real $s$, and write $r\le_{p}s$, if there is a second-order sentence $\varphi(x, Y)$ in two variables with no parameters - where $x$ is a natural number variable and $Y$ is a set variable - such that $$r=\{n: \varphi(n, s)\}$$ holds. (There are many equivalent ways to phrase this.)
It's easy to check that this is, in fact, a reducibility (in particular, that it's transitive - this isn't hard, but it's worth doing explicitly), and so yields a degree structure $\mathcal{D}_p$. And by the same proof as in the Turing case, we have that any "$\equiv_p$-invariant" set of reals either contains or is disjoint from a cone in $\mathcal{D}_p$. So the "projective cone" filter on $\mathbb{R}$ is an ultrafilter. Moreover, by taking infinite joins, it's clear that this ultrafilter is countably closed.
Now we want to port it over to $\omega_1$. We'll use the same trick as in the Turing case: for a real $r$, let $f(r)$ be the least ordinal $\alpha$ such that $\alpha$ is not projectively definable in terms of $r$ (formally, there is no well-ordering of $\mathbb{N}$ of ordertype $\alpha$ which is projective relative to $r$); such a real exists, since there are only countably many projectively definable ordinals relative to a given real.
Note that this uses a small bit of choice - namely, that $\omega_1$ is regular. But this is provable in ZF+AD, so that's fine. Note that DC is not known to be provable in ZF+AD (although it does follow from ZF+AD+V=L$(\mathbb{R})$) so we can't use it here.
By the same argument as in the Turing case, the filter gotten by "porting over" the projective cone filter via $f$ is a measure on $\omega_1$.

Another natural reducibility to consider is relative constructibility. In many ways this is actually more natural than relative projectiveness; however, it makes things a bit trickier, since you have to show that $\omega_1$ is inaccessible from reals assuming AD.
However, this isn't hard - if $\omega_1^{L[r]}=\omega_1$ for some real $r$, then we have $\vert\mathbb{R}\cap L[r]\vert=\omega_1^{L[r]}=\omega_1$ (the first equaltiy since $L[r]$ satisfies condensation appropriately); but then $\mathbb{R}\cap L[r]$ is a counterexample to the perfect set property.
