**Definition:**$\mathcal{B}$ is a

*measure algebra*if it is a complete Boolean algebra equipped with some function $\mu:\mathcal{B}\rightarrow [0,1]$ with $\mu(0)=0, \mu(1)=1, \mu(b)>0$ for $b\neq 0$, and $\mu$ countably additive (i.e., if $\langle b_i:i<\omega\rangle$ is an antichain, then $\mu(\sum b_i)=\sum \mu(b_i)$.

*Exercise:*All measure algebras are c.c.c.

*Example:*Let $\kappa$ be a cardinal. We will describe a topology on ${}^\kappa 2$. Fix some $x\in [\kappa]^{<\omega}$ and $s:x\rightarrow 2$ (i.e., $s$ is a finite domain partial function from $\kappa$ to $2$). Then basic open sets are of the form $\mathcal{O}_s=\{r\in {}^\kappa 2: \forall \alpha\in x(r(\alpha)=s(\alpha))\}$. Let $\mathcal{B}_\kappa$ be the $\sigma$-algebra generated by these basic open sets, and define $\mu$ on $\mathcal{B}_\kappa$ by setting $\mu(\mathcal{O}_s=\frac{1}{2^{|s|}}$ (standard theorems from real analysis give that $\mu$ extends uniquely to a countably additive probability measure on $\mathcal{B}_\kappa$.

Define $\mathrm{Null}=\{A\in\mathcal{B}_\kappa: \mu(A)=0\}$. Then $\mathcal{R}_\kappa:=\mathcal{B}_\kappa/\mathrm{Null}$ is a measure algebra.

*Exercise:*Prove that $\mathcal{R}_\kappa$ forces $2^\omega\ge \kappa$.

*Exercise:*If $\mathcal{A}$ is a measure algebra and $\Vdash_A \dot{\mathcal{B}}$ is a measure algebra, then $\mathrm{r.o.}(\mathcal{A}\ast \dot{\mathcal{B}})$ is a measure algebra. (Note: this is not as easy as it may seem at first since for example $\mathcal{A}$ might even add new reals which can be measures of elements of $\mathcal{B}$! We use r.o. for the Boolean completion here since the letter $\mathcal{B}$ is overloaded).

*Exercise:*If $\mathcal{A}$ is a complete sublagebra of a measure algebra $\mathcal{B}$, then $\mathcal{A}$ is a measure algebra.

Continuing along this line,

**Theorem:**If $\mathcal{B}$ is a measure algebra, $\mathcal{A}$ a complete subalgebra of $\mathcal{B}$, and $G\subseteq \mathcal{A}$ is generic over $V$, then in $V[G]$ we have that $\mathcal{B}/G$ is a measure algebra.

Note that in $V[G]$, $G$ is a filter on $\mathcal{B}$, so $\mathcal{B}/G=\{[b]_G:b\in \mathcal{B}\}$. We use $G^*$ for the dual ideal. It's important to distinguish between the orderings of the two Boolean algebras here, and will be good to see how to translate between them using the forcing relation.

**Lemma:**If $\mathcal{B}$ is complete and $\mathcal{A}$ is a complete subalgebra and $G\subseteq \mathcal{A}$ is generic, then $\mathcal{B}/G$ is complete in $V[G]$.

*Proof of Lemma:*Suppose $\langle [b_\alpha]_G:\alpha<\kappa\rangle \in P(\mathcal{B}/G)\cap V[G]$. For each $\alpha<\kappa$, let $X_\alpha:=\{b:1\Vdash_{\mathcal{A}} [b]_{\dot{G}}\le [\dot{b}_\alpha]_{\dot{G}}\}$. Let $c_\alpha=\sum X_\alpha\in V$ (meet taken in $\mathcal{B}$).

We claim that $1\Vdash_{\mathcal{A}} [c_\alpha]_{\dot{G}}=[\dot{b}_\alpha]_{\dot{G}}$ for each $\alpha$--this suffices to prove the lemma. The proof of the claim is straightforward but a little tedious. First we show $1\Vdash_{\mathcal{A}} [c_\alpha]_{\dot{G}}\le[\dot{b}_\alpha]_{\dot{G}}$. If this doesn't hold, then there are $d,p$ so that $p\in \mathcal{A}$, $d\wedge p\neq 0$, and

$$p\Vdash [\check{d}]\le [\check{c}_\alpha] \textrm{ and }[\check{d}]\wedge [\dot{b}_\alpha]=0.$$

The first conjunct implies that $p\wedge d\le c_\alpha$ in $\mathcal{B}$, and since $c_\alpha$ is a lub for $X_\alpha$, there is some $b\in X_\alpha$ so that $p\wedge d\wedge b\neq 0$. So $1\Vdash [p\wedge d\wedge b]\le [b_\alpha]$ by the definition of $b\in X_\alpha$. But the second conjunct gives $p\wedge d\wedge b_\alpha=0$, contradiction.

Now to show $1\Vdash_{\mathcal{A}} [c_\alpha]_{\dot{G}}\ge[\dot{b}_\alpha]_{\dot{G}}$, assume for a contradiction that there are $p,a\in \mathcal{A}$ so that $p\Vdash [\check{a}]\le [\dot{b}_\alpha]$ and $p\Vdash [a\wedge \neg c_\alpha]\neq 0$. Now $p$ forces $[p\wedge a \wedge \neg c_\alpha]\le [\dot{b}_\alpha]$. Trivially, $\neg p$ forces $[p\wedge a \wedge \neg c_\alpha]=0$. So it's just outright forced that $[p\wedge a \wedge \neg c_\alpha]\le [b_\alpha]$ and thus $p\wedge a\wedge \neg c_\alpha \in X_\alpha$, which contradicts $c_\alpha$ is an upper bound for $X_\alpha$, completing the proof of the claim and the lemma. $\Box$

*Proof of Theorem:*Let $\mu$ be a measure on $\mathcal{B}$, $\mathcal{A}$ a complete subalgebra of $\mathcal{B}$. Define

$$\mu(b\mid a)=\frac{\mu(a \wedge b)}{\mu(a)}.$$

(We say the measure of $b$ conditioned on $a$).

**Definition**: For $a\in A,b\in B, \epsilon>0$, say $a$ is $\epsilon$-

*stable for $b$*if for all $x\le a$ in $\mathcal{A}$, $|\mu(b\mid x)-\mu(b\mid a)|<\epsilon$.

**Lemma:**For all $b\in B$ and for all $\epsilon>0$ the set $\{a\in A:a \textrm{ is }\epsilon-\textrm{stable for} b\}$ is dense in $A$.

*Proof:*Exercise. An interesting one.

In $V[G]$, let $\nu:\mathcal{B}/G\rightarrow [0,1]$ be given by $\nu([b])=r$ if for every $\epsilon>0$ there is some $a\in G$ so that $a$ is $\epsilon$-stable for $b$ and $|\mu(b\mid a)-r|<\epsilon$. The idea is that $G$ could add new reals, so we can only have approximations to the measure of $[b]$ using ground model reals attached to the members of $\mathcal{A}$.

We can check that this is well-defined: if $[b]_G=[c]_G$, then some $a\in G$ forces $b\Delta c\in G^*$. This means that $a\perp (b\Delta c)$, so $\mu(a\wedge (b\Delta c)=0$. Therefore $\mu(b\mid x)=\mu(c\mid x)$ for all $x\le a$. Suppose $r_0\neq r_1$ both satisfy $\nu(b)=r_i$. Take $\epsilon<|r_1-r_0|$. Let $a_0,a_1\in G$ be $\epsilon/4$-stable for $b$ with $|\mu(b\mid a_i)-r_i|<\epsilon/4$. Now take $a\le a_0, a_1$ in $\mathcal{A}$. By a triangle inequality argument, we have $|r_1-r_0|<\epsilon$, a contradiction.

*Exercise*: Check that $\nu(b)>0$ for all $b\neq_G 0$.

*Exercise*: Check that $\nu$ is countably additive. First prove that it is finitely additive.

$\Box$.

Now suppose $P(Z)/I$ is a measure algebra. The duality theorem (ccc case) says that for any $\theta$, $\mathcal{R}_\theta\ast P(Z)/\bar{I}\cong P(Z)/I\ast j(\mathcal{R}_\theta)$,

where $j:V\rightarrow M$ is the generic embedding in $V[G]$, $G$ generic for $P(Z)/I$.

The right hand side of this isomorphism is a measure algebra, since $P(Z)/I$ is a measure algebra by assumption, and $j(\mathcal{R}_\theta)$ is a measure algebra of $M$, a model which is closed under countable sequences (and so has all the countable sequences to witness countable additivity and completeness of the Boolean algebra).

We have a map $e:\mathcal{R}_\theta\rightarrow \mathrm{r.o.}(P(Z)/I\ast j(\mathcal{R}_\theta)$, so $\mathcal{R}_\theta$ is isomorphic to a complete subalgebra of the right hand side. Now if $H\subseteq \mathcal{R}_\theta$ is generic, then $B/e''H$ is a measure algebra. Therefore $P(Z)/\bar{I}$ is also a measure algebra.

A real-valued measurable cardinal is a cardinal $\kappa$ which carries a $\kappa$-additive probability measure on all subsets of $\kappa$ which gives measure 0 to singletons. It is atomless if every set of positive measure has a subset of strictly smaller positive measure.

**Corollary:**If $\langle \kappa_i:i<\theta\rangle$ is a sequence of measurable cardinals, then if $\gamma=\sup \kappa_i$, $\mathcal{R}_\gamma$ forces all $\kappa_i$ to be atomless real-valued measurable cardinals (RVMs).

We note the fact that if $\kappa$ is atomlessly RVM, then $2^\omega\ge \kappa$, so we can't get class many RVMs.

However, if $\kappa$ is strongly compact, then $\mathcal{R}_\kappa$ forces that for all regular $\lambda\ge \kappa$, there is a

*countably additive*real-valued probably measure $\mu_\lambda$ on $\lambda$ giving measure 0 to all subsets of size $<\lambda$.
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