*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:** *EDIT: Most of this is completely wrong, see the answer below.*

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 there a description of which measures on $\omega_2^{L(\mathbb{R})}$ are potentially club? A reasonable guess is that the above argument with $Col(\omega,\omega_1)$ constructs such a measure; and under appropriate hypotheses (the nonstationary ideal on $\omega_1$ is $\omega_2$-saturated and $\mathcal{P}(\omega_1)^\sharp$ exists) this is verified by calculating the projective ordinal $\delta^1_2$ - since $L(\mathbb{R})$ computes $\omega_2$ correctly, $\omega_1^{L(\mathbb{R})[G]}=\omega_2^V=\omega_2^{L(\mathbb{R})}$. However, I don't know a snappy description for this measure. So we may ask:

What are the potentially club measures on $\omega_2$?

In particular, is the measure in the usual proof of the measurability of $\omega_2$ in $L(\mathbb{R})$ the same as the potentially club filter gotten from $Col(\omega,\omega_2)$?

*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.*