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.


1 Answer 1


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 $L(\mathbb R)\subseteq \text{HOD}[G]$.

  • $\begingroup$ Ouch. That's definitely the kind of mistake I shouldn't be making anymore - both the assumption that $L(\mathbb{R})$ has all the names for reals, and missing the fact that that forcing (for large enough $\kappa$) makes $L(\mathbb{R})[G]$ satisfy choice! Thanks a ton. Incidentally, do you know if it's possible to get a measure on $\omega_2$ in the manner above? $\endgroup$ Dec 20, 2016 at 17:35
  • $\begingroup$ Did you mean to write $\text{Col}(\omega,\omega_2)$ above, or did you mean $\text{Col}(\omega,\omega_1)$ or $\text{Col}(\omega,{<}\omega_2)$? For the latter forcings, I think Asaf argued that every club in the extension contains a ground model club so you'll just get the club filter of $L(\mathbb R)$ on $\omega_2$ back. (This is not an ultrafilter since $\text{Cof}(\omega)$ and $\text{Cof}(\omega_1)$ are stationary.) You might try shooting a club through $\text{Cof}(\omega)$. (Note that the club filter on $\omega_2$ along with $\text{Cof}(\omega)$ generates an ultrafilter.) $\endgroup$ Dec 20, 2016 at 18:58
  • $\begingroup$ I meant $Col(\omega,\omega_1)$, thanks. $\endgroup$ Dec 20, 2016 at 19:00

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