*This is a spinoff of [this earlier question of mine](https://mathoverflow.net/questions/256998/potentially-club-filters-on-omega-2).*


**Short version:** 



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



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



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> What are the potentially club measures?
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(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 $\delta^1\_2$](https://mathoverflow.net/questions/257614/cardinal-arithmetic-in-l-mathbbr) - 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:



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> What are the potentially club measures on $\omega\_2$?
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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.*



Getting measures (especially on $\omega_2$) from potential clubs
The claim you are trying to prove is false: $L(\mathbb R)$ has no measurable cardinals greater than $\Theta$. Work in $L(\mathbb R).$ We will use Woodin's theorem that $\text{HOD} = L[\mathbb P]$ for $\mathbb P\subseteq \Theta$ a partial order encoded as a subset of $\Theta$ and that $L(\mathbb R)$ is an inner model of $\text{HOD}[G]$ for a $\text{HOD}$-generic filter $G\subseteq \mathbb P$. (We will use the second half of this at the end of our answer but for now it suffices to accept that $\text{HOD} =L[A]$ for $A\subseteq \Theta$.) If $\kappa>\Theta$ is measurable, and $U$ is a $\kappa$-complete ultrafilter on $\kappa$ then we have $\text{Ult}(\text{HOD},U) = L[j(\mathbb P)] = L[\mathbb P] = \text{HOD}$ where $j$ is the ultrapower map. Now $j:\text{HOD}\to \text{HOD}$ witnesses $\mathbb P^\#$ exists. But then $\mathbb P^\#$ is in $\text{HOD}$, contradicting $\text{HOD} = L[\mathbb P]$.


The issue with your proof is that in general $L(\mathbb R)[G]$ is not equal to $L(\mathbb R)^{V[G]}$ even when $G$ is a $V$-generic real. (There is no reason that names for reals should be reals themselves. Furthermore there is no reason that $\mathbb R^V$ should be in $L(\mathbb R)^{V[G]}$.) It fails in particular when $G\subseteq \text{Col}(\omega,\kappa)$ (assuming a proper class of Woodin cardinals in $V$). At least, if $\kappa \geq |\mathbb R|$, then in $V[G]$ there is a real $r$ coding an enumeration of $\mathbb R^V$ in order type $\omega$, but if there is such a real in $L(\mathbb R)[G]$, then $L(\mathbb R)[G]$ satisfies AC (since then $L(\mathbb R)[G] = L[r,G]$). Thus assuming $L(\mathbb R)[G] = L(\mathbb R)^{V[G]}$ we obtain $L(\mathbb R)^{V[G]}\vDash \text{AC}$, a contradiction in the context of our large cardinal hypothesis. (If $V = L$ we of course have no contradiction at all.)


Incidentally, if $L(\mathbb R)\vDash \text{AD}$ then $\Theta$ is not measurable either. Suppose it were. Consider $j:\text{HOD}\to \text{Ult}(\text{HOD},U)$ where $U\in L(\mathbb R)$ is a $\Theta$-complete ultrafilter on $\Theta$. Since $\Theta$ is inaccessible in $\text{HOD}$, $j(\Theta)$ is inaccessible in $\text{Ult}(\text{HOD},U)$ and hence in $\text{HOD}$ since $\text{HOD} = L[\mathbb P] \subseteq L[j(\mathbb P)] = \text{Ult}(\text{HOD},U)$. On the other hand $j(\Theta)$ is not a cardinal in $L(\mathbb R)$ since $|j(\Theta)| = | P(\Theta)\cap \text{HOD}| = |(2^\Theta)^{\text{HOD}}|$, while $(2^{\Theta})^{\text{HOD}} < j(\Theta)$ (as ordinals) again since $j(\Theta)$ is inaccessible in $\text{HOD}$. Now $L(\mathbb R)$ is an inner model of $\text{HOD}[G]$ for a $\text{HOD}$-generic filter $G\subseteq \mathbb P$. In $\text{HOD}[G]$, $j(\Theta)$ remains inaccessible by the usual Levy-Solovay analysis. This contradicts the fact that $j(\Theta)$ is not a cardinal in the inner model $L(\mathbb R)\subseteq \text{HOD}[G]$.


