TITLE: Weakly homogenously Souslin sets and the measurability of $\omega_1$
QUESTION [7 upvotes]: I found this intriguing remark at the end of Woodin's Supercompact cardinals, sets of reals, and weakly homogeneous trees (1988):

The assertion that every set of reals, in $L(\mathbb{R})$, is the projection of a weakly homogeneous tree has consequences beyond the usual regularity properties such as Lebesgue measurability. For example by results of Kechris it follows that $L(\mathbb{R})\vDash$ ''$\omega_1$ is measurable''.

Today, we would prove the last sentence by noting that if every set of reals in $L(\mathbb{R})$ is the projection of a weakly homogeneous tree (i.e., is weakly homogeneously Souslin), then $L(\mathbb{R})\vDash$ AD (and therefore $L(\mathbb{R})\vDash$ ''$\omega_1$ is measurable''). But in 1988 this was not known, so there should be a different proof, that uses just the structure of weakly homogeneously Souslin sets. Is there a source somewhere for such a proof? Or maybe it is an easy consequence of a previous result by Kechris?

REPLY [8 votes]: This result is implied by a result in Kechris's paper "Subsets of $\aleph_1$ constructible from a real". Kechris proves the following: If there is a measurable cardinal, then every subset of $\omega_1$ is constructible from a real if and only if every subset of $\omega_1$ is weakly homogeneously Suslin in the codes. This implies that the closed unbounded filter is an ultrafilter, since if $x^\#$ exists, then the closed unbounded filter restricts to an $L[x]$-ultrafilter. The measurable cardinal is unnecessary for the direction you're interested in, because if there is a weakly homogeneously Suslin set that is not $\mathbf{\Sigma}^1_1$, then there is a measurable cardinal.