I found this intriguing remark at the end of Woodin's *[Supercompact cardinals, sets of reals, and weakly homogeneous trees](https://www.jstor.org/stable/32425)* (1988):



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> 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''.
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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?



Weakly homogenously Souslin sets and the measurability of $\omega_1$
This result is implied by a result in Kechris's paper ["Subsets of $\aleph\_1$ constructible from a real"](https://authors.library.caltech.edu/38904/). 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.


