This is a spinoff of this earlier question of mine.

Short version:

What measures in $L(\mathbb{R})$ can be gotten from "potentially club" filters, under appropriate hypotheses?

Long version:

Here's a very silly proof that large cardinals in $V$ imply that $L(\mathbb{R})$ has a proper class of measurables.

Of course, from some mild large cardinals we know that $L(\mathbb{R})$ has one measurable, namely $\omega_1$. We'll show how we can use this (together with more large cardinals) to get, for any cardinal $\kappa$, a measurable $\mu>\kappa$. Specifically, our large cardinal assumption is:

$(*)$ There is a proper class of Woodins.

Aaaaand that's why this is silly. But bear with me.

Fix $\kappa$; we want to show that, in $L(\mathbb{R})$, there is a measurable above $\kappa$.

Suppose there are a proper class of Woodins. Then the theory of $L(\mathbb{R})$ is invariant under set forcing. In particular, for any set-generic real $G$ we have that $L(\mathbb{R})[G]=L(\mathbb{R})^{V[G]}$ satisfies "The club filter on $\omega_1$ is an ultrafilter."

Now consider the forcing $Col(\omega,\kappa)$. A generic for this is a bijection $G: \omega\rightarrow\kappa$. The induced ordering on $\omega$ - $a<_Gb\iff G(a)<G(b)$ - is coded by a real, and determines $G$ ($G$ is the unique order-preserving bijection between $<_G$ and $\kappa$), so $L(\mathbb{R})[G]=(L(\mathbb{R}))^{V[G]}$. In particular, since $L(\mathbb{R})$ thinks the club filter on $\omega_1$ is measurable, so does $L(\mathbb{R})[G]$ by $(*)$. But $\omega_1^{L(\mathbb{R})[G]}>\kappa$. Let $\mu=\omega_1^{L(\mathbb{R})[G]}$; we'll show $\mu$ is measurable in $L(\mathbb{R})$.

Let $\mathcal{F}_G=\{x\subseteq\mu: L(\mathbb{R})[G]\models\mbox{"$x$ contains a club"}\}.$ Then it's easy to check that (in $L(\mathbb{R})$) $\mathcal{F}_G$ is a $\mu$-complete filter on $\mu$: $L(\mathbb{R})$ satisfies "the intersection of countably many clubs on $\omega_1$ is a club," so so does $L(\mathbb{R})[G]$. By the above paragraph $\mathcal{F}_g$ is an ultrafilter. So - in $L(\mathbb{R})$ - $\mu$ is measurable. $\Box$

Now, this is a very silly proof: our large cardinal assumption far outstrips what we're trying to prove! In particular, every Woodin in $V$ is measurable in $L(\mathbb{R})$.

That said, the construction itself seems neat to me: by using forcing absoluteness, we pull a specific definable measure on $\omega_1$ in the generic extension back to a measure on some large $\mu$ in the ground. This is neat!

My question is,

What are the measures which can be so recovered?

Precisely, say that a measure $U$ on a cardinal $\mu$ (in $L(\mathbb{R})$) is a potentially club measure if for some real $G$ which is set-generic over $L(\mathbb{R})$, we have $\mu=\omega_1^{L(\mathbb{R})[G]}$ and $U=\{x\subseteq\mu: L(\mathbb{R})[G]\models\mbox{"$x$ contains a club"}\}$. Then:

What are the potentially club measures?

(This is in $L(\mathbb{R})$ of course.)

In particular, is any measure on $\omega_2$ potentially club? A reasonable guess is that the above argument with $Col(\omega,\omega_2)$ demonstrates this; however, I do not see immediately that $\omega_1^{Col(\omega,\omega_2)}=\omega_2$ in $L(\mathbb{R})$.

Note: any definable measure on $\omega_1$ gives rise to an analogous question. As a computability theorist, for instance, the "potentially cone" measures actually seem more interesting. However, I suspect they're harder to analyze, so I'm beginning with the club version.