数理解析研究所講究録 1202 巻 2001 年 38-60

38Models of real-valued measurability
北見工業大学 情報システム工学科 渕野 昌 (Saka\’e Fuchino)This is a preliminary version of the joint paper:

Models of real-valued measurability I
bySaka\’e Fuchino and Saharon Shelah

Any comments are appreciated.

March 16, 2001
2000 Mathematics Subject Classification: 03E35,03E55,03E65,28E15
keywords: real-valued measurability, random reals, club principle

Abstract

Starting from amodel of ZFC with ameasurable cardinal , we constructageneric extension in which is real-valued measurable,
and clubprinciple for holds. This gives apositive answer to aquestion of.
Fremlin asking the existence of models of real-valued measurability withsome combinatorial behavior different ffom that of Solovay’s model of realvalued measurablility. Some other models related to this question will begiven in the forthcomming [4].
$\kappa$$2^{\aleph_{0}}>\kappa$$\kappa$$\kappa$$\mathrm{D}.\mathrm{H}$1Introduction
In his celebrated paper [9], Solovay proved that, if is ameasurable cardinal,
then, by forcing with measure algebraof Maharam typefor appropriate$\kappa$$B_{\lambda}$

1 The$\lambda\geq\kappa$

first author is supported by Grant-in-Aid for Scientific Research (C) No. 10640099 andNo. 12640098 of the Ministry of Education, Science, Sports and Culture, Japan.
The second author is supported by “The Israel Science Foundation”

39
is real-valued measurable. Recall that is said
to be real-valued measurable if there is a additive atomless measure :$[0,1]$ . Existence of such acardinal is equivalent to the extendability of Lebesque–which, of course, cannot bemeasure to a -additive measure on the wholetranslation invariant under the axiom of choice due to Vitali’s theorem.
Some properties of Solovay’s model follow simply from the fact that the model isobtained by adding random reals; for example we have the equations $cov(null)=\lambda$
in the model while and remain as in the ground model.
and non(null)On the other hand, it is also known that the existence of areal-valued measurable cardinal alone implies alot of combinatorial consequences. For example:$\lambda$, we obtain amodel in which$\kappa$$\kappa$$\mu$$\kappa$$P(\kappa)arrow$$\mathrm{V}(\mathrm{R})$$\mathrm{c}\mathrm{r}$$\mathrm{d}$$\mathrm{b}$$=\aleph_{1}$is real-valued measurable. Then:
Theorem 1.1 Suppose that.
,, $cov(null)\geq\kappa$ ,(1) (see [2]) non(null)(2) has the tree property.
holds for a lot of stationholds (actuallythen(3) (Kunen) If(Ketonen) .
ary$\kappa$$\leq 2^{\aleph_{0}}$$=\aleph_{1}$$\mathrm{b}\neq\kappa$$\mathrm{d}\neq\kappa$$\kappa$$\kappa=2^{\aleph_{0}}$$S\subseteq\kappa$$\phi_{\kappa}(S)$$\phi_{\kappa}$$\square$$)$

For more complete list of such implications see e.g. [2].
. Fremlin asked if there is amodel of real-valued
Against this back-ground,measurability which is intrinsically different from Solovay’s models. As one of thepossible answers to this question, we present here anew model in which we have areal-valued measurable strictly less than the continuum while club principle for$\mathrm{D}.\mathrm{H}$$\kappa$$\kappa$

holds.

2Preliminaries
For aset let $B_{(u)}=Borel(^{u}2)$ , the set of all Borel sets in the generalized Cantor
is also seen as the Boolean algebra with usual set operations.
discontinuum U2.
the complement of , i.e. $-a=2u\backslash a$ .
In particular, for $a\in B_{(u)}$ we denote
Strictly speaking $-a$ depends on and hence this notation is rather ambiguous.
This ie because we often identify with the corresponding element of $Borel(^{u’}2)$
should denote the complement of the(see below): in thisfor some
Borel subset of $u’2$ which corresponds to . Nevertheless it should be always clear$u$$B_{(u)}$$\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}-a$

$a$

$u$

$a$

$u’\supseteq u$$\mathrm{c}\mathrm{a}\mathrm{s}\mathrm{e}-a$$a$from the context what is meant with this notation.
, $r>0$ , amapping $\mu:B_{(u)}arrow[0, r]$ is said to be a $[0, r]$ measure ifFor$r\in \mathbb{R}$(a)$\mu(\emptyset)=0;\mu(^{u}2)=r$;40(b) for any pairwise disjoint$s_{i}\in \mathrm{B}(\mathrm{u})$, $i<\omega$ ,$\mu(\bigcup_{\dot{\iota}<\omega}s:)=\sum_{\dot{l}<\omega}\mu(s_{i})$.

Kolmogoroff’s extension theorem can be formulated as follows:
Theorem 2.1 (Kolmogoroff) Suppose thatis a measurable space for $i\in I$, $i\in E$ (seen as
andis $a[0,1]$ -measure over the product spapce ofby the canonical embedding) for
the subspace of the product spaceare pairwise compatible (as mappings).
eachsuch that ,. such that
Suppose that for each $i\in I$ , there is$\langle S_{\dot{l},\vee}4.\rangle$$\langle S_{i}, A.\rangle$$\mu_{E}$$\langle\Pi_{:\epsilon I}S_{\dot{1}}, \otimes_{i\in I\vee}\{.\rangle$$E\in[I]^{<\aleph_{0}}$$E\in[I]^{<\aleph_{0}}$$\mu_{E}$$\mathrm{C}_{i}\subseteq \mathrm{A}$

(i)

if

$c_{n}\in c_{:}$$\bigcap_{n\in n0}c_{n}=\emptyset$

(ii)

for

and

$n\in\omega$$\bigcap_{n\in\omega}c_{n}=\emptyset$$n_{0}\in\omega$such that, and$\mu\{:\}(a)=\sup\{\mu\{:\}(c) :c\in \mathrm{C}_{i}, c\subseteq a\}$Then there is the unique $[0, 1]$ -measure over$E\in[I]^{<\aleph_{0}}$, then there isfor all $a\in A.$ .
extending all$\langle\prod_{:\in I}S_{\dot{1}}, \otimes:\in IA.\rangle$$\mu_{E}$,.

In our context we may apply the theorem in the following form:
Corollary 2.2 Suppose thatis a family of sets closed under union of finitely
many members and. If ,
areis $a[0,1]$ -measure onfor, there is the unique $[0, 1]$ -measurecompatible to each other, then, letting.
: $B(u^{*})arrow[0,1]$ extending all ,$\mathcal{U}$$u\in \mathcal{U}$$\mathrm{B}(\mathrm{u})$$\mu_{u}$$u\in \mathcal{U}$$\mu_{u}$$u^{*}=\cup \mathcal{U}$$\mu^{*}$$u\in \mathcal{U}$$\mu_{u}$, letbe such thatand $\mu_{E}=$Proof Let $I=u^{*}$ . For, we obtain a $[0, 1]-$[
. Applying Kolmogoroff’s theorem to ,becausemeasure on
which is an extension of each ,(Corollary 22).
of the uniqueness of [$E\in[I]^{<\aleph_{0}}$$\mu_{u}$$E\subseteq u$$u\in \mathcal{U}$$B_{(E)}$$E\in[I]^{<\aleph_{0}}$$\mu_{E}$$\mu^{*}$$B(u^{*})\cong\otimes:\in IB(\{i\})$$\mu^{*}$$\mu_{u}$$\square$$\mathrm{B}(\mathrm{u})$, let null(ij) $=\{a\in B(u) :For a $[0, 1]$ -measure on,
defined byis the $[0, 1]$ -measure on$\mu$$a\in B_{(u)}\backslash null(\mu)$$u\in \mathcal{U}$$\mathrm{B}(\mathrm{u})$$\mu||a$\mu(a)=0\}$ .

For$B_{(u)}$$\mu||a(b)=\frac{\mu(b\cap a)}{\mu(a)}$for $b\in B_{(u)}$ .
The following is easily seen:
Lemma 2.3 For $a[0,1]$ -measure
if and an if $a\triangle a’\in null(\mu)$ .
$y$

$\mu$

on

$B_{(u)}$

and ,
$a$

$a’\in B_{(u)}\backslash null(\mu)$, $\mu||a=\mu||a’$41) and [0, 1] measure[0,1] and j12
withdenote the product measure of and[0,1], let ) )/12. /}) @/12 is characterized by:For$\ovalbox{\tt\small REJECT}$$\ovalbox{\tt\small REJECT}/$$u_{l}C^{\ovalbox{\tt\small REJECT}^{I}}u_{2}\ovalbox{\tt\small REJECT}$$/\mathrm{J}\mathrm{t}\ovalbox{\tt\small REJECT}$$B(\mathrm{n}\mathrm{i})\ovalbox{\tt\small REJECT}$$u_{\mathrm{I}\mathrm{t}}\ovalbox{\tt\small REJECT} \mathrm{u}_{2}$$\ovalbox{\tt\small REJECT} s$$B_{(\cdot)}2\ovalbox{\tt\small REJECT}$$7\mathrm{J}^{\ovalbox{\tt\small REJECT}}$$\# 2$$\ovalbox{\tt\small REJECT}_{(u_{1}\mathrm{U}u_{2})}\ovalbox{\tt\small REJECT}$$[0_{\mathrm{t}}1]$$1^{\mathrm{L}}\mathrm{r}$$\mu_{1}\otimes\mu_{2}(a_{1}\cap a_{2})=\mu_{1}(a_{1})\cdot\mu_{2}(a_{2})$for all$a_{1}\in \mathrm{B}(\mathrm{n}\mathrm{i})$Lemma 2.4 (1),and$a_{2}\in B(u_{2})$.

For $a[0,1]$ measure$\mu$

on $B(u)$ and , $b\in B_{(u)}\backslash null(ii)$ with$a$$b\subseteq a$$\mu||b=(\mu||a)||b$(2) Suppose that$[0, 1]$ -measures,$u_{1}\cap u_{2}=\emptyset$

,

:
and

$\mu_{1}$$a_{1}\in \mathrm{B}(\mathrm{n}\mathrm{i})\backslash null(\mu_{1})$.

$\mathrm{B}(\mathrm{n}\mathrm{i})arrow[0,1]$

and$\mu_{2}$$a_{2}\in \mathrm{B}(\mathrm{n}\mathrm{i})\backslash null(\mu_{2})$$(\mu_{1}\otimes\mu_{2})||(a_{1}\cap a_{2})=(\mu_{1}||a_{1})\otimes(\mu_{2}||a_{2})$Proof (1): Let$c\in \mathrm{B}(\mathrm{u})$:$\mathrm{B}(\mathrm{n}\mathrm{i})arrow[0,1]$

are. Then.

. Then$(( \mu||a)||b)(c)=\frac{(\mu||a)(c\cap b)}{(\mu||a)(b)}$$= \frac{\mu(c\cap b\cap a)}{\mu(a)}\cdot\frac{\mu(a)}{\mu(a\cap b)}$$= \frac{\mu(c\cap b)}{\mu(a)}\cdot\frac{\mu(a)}{\mu(b)}$$= \frac{\mu(c\cap b)}{\mu(b)}=(\mu||b)(c)$(2): It is enough to show that the both sides of the equation have the samefor some $c_{1}\in B(u_{1})$ and $c_{2}\in B(u_{2})$ .
of the form
value for elements of$c_{1}\cap c_{2}$$B(u_{1}\cup u_{2})$This can be shown as follows:$(( \mu_{1}\otimes\mu_{2})||(a_{1}\cap a_{2}))(c_{1}\cap c_{2})=\frac{(\mu_{1}\otimes\mu_{2})(c_{1}\cap c_{2}\cap a_{1}\cap a_{2})}{(\mu_{1}\otimes\mu_{2})(a_{1}\cap a_{2})}$$= \frac{\mu_{1}(c_{1}\cap a_{1})\cdot\mu_{2}(c_{2}\cap a_{2})}{\mu_{1}(a_{1})\cdot\mu_{2}(a_{2})}$$=(\mu_{1}||a_{1})(c_{1})\cdot(\mu_{2}||a_{2})(c_{2})$$=((\mu_{1}||a_{1})\otimes(\mu_{2}||a_{2}))(c_{1}\cap c_{2})$$\coprod_{(\mathrm{L}\mathrm{e}\mathrm{m}\mathrm{m}\mathrm{a}2.4)}$

For u

a

$\in B_{(u)}$by identifying eachcan be embedded canonically intowith the same “definition”. In the followinwith the element of$\subseteq v$,$B_{(v)}$$B_{(u)}$$B_{(v)}$$\mathrm{g}$42
we always regard
particular, e.g. if

as the subalgebra ofby this canonical embedding. In
and $b\in B(v)$ , then we mean with $a\cap b$ the intersectionof with the element of $B(v)$ which corresponds to by this embedding.
denotes the Borel measure on $B(u)$ , i.e. the $[0, 1]$ -measure which makes$B_{(v)}$$\mathrm{B}(\mathrm{u})$$a\in \mathrm{B}(\mathrm{u})$

$b$

$a$

$m_{u}$, $x\in u$ independent events with

, or
for
denotes the basic clopen set: $\{f\in u2 : t\subseteq f\}$ . Let $B_{u}=$$B(u)/null(m_{u})$ . null(m,) is also denoted simply by null.
Concerning forcing, we use “the reverse Jerusalem notation”. I.e., in ap.o.-set$P$ ,acondition $p\in P$ is stronger than another condition $q\in P$ if $p\leq_{P}q$ . P-nameare denoted by, etc. We assume that $P$-names are constructed’
as in [8].
For ground model set
denotes its standard $P$-name. $V$ denotes the groundmodel andstandard name of generic set over the p.o.-set.
$[\{\langle x, 0\rangle\}]$$m_{u}([ \{\langle x, 0\rangle\}])=\frac{1}{2}$

where

$[t]$

$[t]_{B_{(u)}}$$t\in \mathrm{F}\mathrm{n}(u, 2)$$\mathrm{s}$$X,\mathrm{Y},.,f,g\sim\sim..\sim\sim$$\ldots$$X,\check{X}$$V$$G\mathrm{t}\mathrm{h}\mathrm{e}\sim$

3Free amalgamation of measures
The following theorems are used in later sections.

Theorem 3.1 (D. Fremlin; an instance ofin [3]) Suppose that,
: $B(u_{1})arrow[0,1]$ ,and: $B(u_{2})arrow[0,1]$ are $[0, 1]$ -measures such that[
. Then there is $a[0,1]$ measure : $B(u_{1}\cup u_{2})arrow[0,1]$ extending both andsuch thatandwhich
for any
are independent events over $B(Uo)$ with respect toandrespectively.
$456\mathrm{N}$$\mu_{1}$$\mu_{2}$$\mu_{2}$$\mu_{1}$$B(u_{0})$$B_{(u_{0})}=$$\mu_{1}$$\mu_{2}(a_{2})$$a_{1}\in \mathrm{B}(\mathrm{u})$$\mu_{1}$Proof Let 72 be the subslgebra ofof the form:
$(*)$

$[$

$\tilde{\mu}$$\tilde{\mu}(a_{1}\cap a_{2})=\mu_{1}(a_{1})\cdot$$\mu_{2}$$u_{0}=u_{1}\cap u_{2}$$a_{0}\cap a_{1}\cap a_{2}$where$B_{(u_{1}\cup u_{2})}$$a_{0}\in \mathrm{B}(\mathrm{M})$,$a_{2}\in \mathrm{B}(\mathrm{u})$$\mu_{2}$consisting of finite union of elements$a_{1}\in B_{(u_{1}\backslash u\mathrm{o})}$and$a_{2}\in B_{(u_{2}\backslash u\mathrm{o})}$.

For $c\in R$ , let
be the set of the partitions of consisting of elements of the form$(*)$ . We consider
as apartial ordering with the ordering:$\Delta_{c}$$c$$\Delta_{c}$Pfor P,$P’\in\Delta_{c}$$\leq P’$

P is arefinement of $P’$.

Now, for c $\in R$ and P(f)$\Leftrightarrow$$\in\Delta_{c}$, let$\mu_{P}^{*}(c)=\sup\{\sum_{a_{0}\cdot a_{1}\cdot a_{2}\in P’,\mu 1(a\mathrm{o})\neq 0}\frac{\mu_{1}(a_{0}\cap a_{1})\cdot\mu_{2}(a_{0}\cap a_{2})}{\mu_{1}(a_{0})}$$P’\in\Delta_{c}$, $P’\leq P\}$ .

:43Finally, for$c\in \mathcal{R}$

, let

(I)

$\mu^{*}(c)=\mathrm{t}\mathrm{h}\mathrm{e}$inverse limit of$\mu_{P}^{*}(c)$,$P\in\Delta_{c}$.

is a-additive measure on 72. Hence, by Hopf’s Extension
We can show that
$B(u_{1}\cup u_{2})arrow[0,1]$ with the
can be extended to a $[0, 1]$ -measure
Theorem,(Theorem 3.1)
desired property.
$\mu^{*}$$\mu\sim:$$\mu^{*}$$\square$We call as constructed in the proof of Theorem 3.1, the free amalgamation of, then. Note that, if
over and denote it with
and
and .
is just the usual product measure of
Theorem 3.1 can be extended to the following amalgamation theorem for in$\mu_{1}$$\tilde{\mu}$$u_{0}=\emptyset$$\mu_{1}\otimes_{u0}\mu_{2}$$\mu_{1}\otimes_{u\mathrm{o}}\mu_{2}$$u_{0}$$\mu_{2}$$\mu_{2}$$\mu_{1}$finitely may measures.

and, for each $u\in S$ ,Theorem 3.2 Suppose that is a -system with the root
[ $B(u^{*})=\mu_{u’}$ [for any: $B_{(u)}arrow[0,1]$ be $a[0,1]$ -measure such thatlet,
$B_{(\cup S)}arrow[0,1]$ extending each of, $u’\in S$ . then there is $a[0,1]$ -measure, $a_{n-1}\in B(u_{n-1})$ , if,
, $u_{n-1}\in S$ and$u\in S$ such that for any $u0$ ,,,, $B_{(n-1)}$ with respect to,are independent over,,.
respectively, then$S$$u^{*}$$\Delta$$B(u^{*})$$\mu_{u}$$\mu_{u}$$a_{0}\in B(u\mathrm{o})$$\ldots$$a_{0}$$\mu_{u}$$\mu\sim:$$u$$\ldots$$B_{(0)}$$a_{n-1}$$\ldots$$\mu_{u_{0}}$$\ldots$$\tilde{\mu}(a_{0}\cap\cdots\cap a_{n-1})=\mu_{u\mathrm{o}}(a\mathrm{o})\cdot\cdots\cdot$$\mu_{u_{n-1}}$$\ldots$$\mu_{u_{n-1}}(a_{n-1})$

Proof The construction of measures in the proof of Theorem 3.1 can be extended
below). By thisandto amalgamation of measures for all $n\geq 2$ (seeconstruction, we obtain asystem $\{\mu_{U} : U\in[S]^{<\aleph_{0}}\}$ of $[0, 1]$ measures where: $B_{\cup U}arrow[0,1]$ such that$(\dagger n)$$n$$(\ddagger n)$$\mu_{U}$are pairwise compatible;,(1)
for all $u\in U$ ;extends(2) each,, $a_{n-1}\in B_{(u_{n-1})}$ , if ,,, $u_{n-1}\in S$ and(3) for any ,respectively,,,
are independent over $B(0)$ , , $B_{(n-1)}$ with respect towith , ,forthen$\mu_{U}$$U\in[S]^{<\aleph_{0}}$$\mu_{u}$$\mu_{U}$$u_{0}$$a_{0}\in B_{(u\mathrm{o})}$$\ldots$$\mu_{u0}$$\ldots$$\tilde{\mu}_{U}(a_{0}\cap\cdots\cap a_{n-1})=\mu_{u_{0}}(a\mathrm{o})\cdot\cdots\cdot$$u_{n-1}\in U$$a_{0}$$\ldots$$\mu_{u_{n-1}}(a_{n-1})$$\ldots$$\ldots$$a_{n-1}$$\mu_{u_{n-1}}$$U\in[S]^{<\aleph_{0}}$$u_{0}$$\ldots$.

Applying Corollary 2.2 to these$\mu_{U}’ \mathrm{s}$, we obtain a $[0, 1]$ -measure as desired.
$\square$(Theorem 3.2)

We shall call the $[0, 1]$ -measure constructecd as in the proof of Theorem 3.2 thefree amalgamation of , $u\in S$ over . The free amalgamation is characterized
by the following equations which correspond (\dagger ) and (\ddagger ):consisting of finite unions of elements of the form
Let 7?’ be the subset of$\mu\sim$$u^{*}$$\mu_{u}$$B_{(\cup S)}$$\tilde{\mu}$44$(**)$$a_{*}\cap a_{u_{0}}\cap\cdots\cap a_{u_{n-1}}$$a_{u_{0}}\in B_{(u_{0}\backslash u^{\mathrm{s}})}$

For,$\ldots$,for$n\in\omega$where$a_{u_{n-1}}\in B_{(u_{n-1}\backslash u^{\mathrm{s}})}$$u_{0}$,$\ldots$,$u_{n-1}\in \mathcal{U}$and$a_{*}\in B_{(u^{*})}$

,

.

, let
be the set of partitions of consisting of elements of the form$(**)$ . $P\leq P’$ for,, let $P’\leq P$ be defined just as before.
For $c\in R’$ of the form $(**)$ and, let$c\in \mathcal{R}’$$\Delta_{c}$

$P$

$c$

$P’\in\Delta_{c}$$P\in\Delta_{c}$$( \uparrow_{n})\mu_{P}^{*}(c)=\sup\{$$\sum$$\frac{\mu_{u0}(a_{*}’\cap a_{0})\cdot\cdots\cdot\mu_{u_{n-1}}(a_{*}’\cap a_{n-1}’)}{\mu_{1}(a_{*}’)}.\cdot$’$a_{\acute{*}} \cap a_{u_{0}}’\bigcap_{1}\cdots\bigcap_{a},a_{\acute{u}_{\hslash}}\in P’\mu(.)\neq 0^{-1}$$P’\in\Delta_{P}$, $P’\leq P\}$ .

Then$(\ddagger n)$$\mu^{*}(c)=\mathrm{t}\mathrm{h}\mathrm{e}$4The model
For cardinals(A)$\aleph_{1}<\kappa$$\mu_{P}^{*}(c)$, P$\in\Delta_{c}$.

$V^{Q}$$\leq\lambda$, let$p\in Q\Leftrightarrow p=\langle u^{p}, \mu^{p}\rangle$$Q=Q_{\kappa,\lambda}$

be the$\mathrm{p}.0$.-set defined as follows:

where:.

(a)$u^{p}\in[\lambda]^{<\kappa}$(b)$|u\cap\theta^{+}|<\theta$(c)$\mu^{p}$for all strongly inaccessible .
$\theta$

is a $[0, 1]$ -measure on(B) For , $q\in Q$ ,inverse limit of$B_{(u^{p})}$.

there is an $r\in Q$ such that
where
and
are defined as follows where”and”stand for “pure” and“anti-pure” respectively:$p$$q\leq_{Q}p\Leftrightarrow$$q\leq_{a\mathrm{p}t}r\leq_{pr}p$$\leq_{apt}$$\leq_{\mathrm{P}^{f}}$$” \mathrm{p}\mathrm{r}$(C) For , $q\in Q$ ,(a) $q\leq_{pr}p$is the free product of$” \mathrm{a}\mathrm{p}\mathrm{r}$$p$$\Leftrightarrow$(b)$u^{p}\subseteq u^{q}$$\mu^{p}$

and

and

$\mu^{q}$$q\leq_{apr}p\Leftrightarrow u^{p}=u^{q}$$[$$\mu^{q}$extends$B_{((u^{q}\backslash u^{p})\cap\gamma)}$and$\mu^{p}$. For$\gamma$$= \sup$up,$\mu^{q}$ $[$$B_{((u^{q}\cap\gamma)\cup u^{p})}$.

$\mu^{q}=\mu^{p}||a$for some$a\in B_{(u^{p})}\backslash null(/j1)$.

We shall call as in (C)(b) a witness of. Note that the witness ofis unique upto elements of $null(\mu^{p})$ by Lemma 2.3. We shall also say thatwitnesses $q\leq_{apr}p$ and denote $q=p||a$ .
For $p\in Q$ , let $B(p)=B(u^{p})/null(\mu^{p})$ . For,(or $a/null(\mu^{p})$ ) denotes
the equivalence class of modulo null $(/jP)$ .
The following lemma shows that the relation
is apartial ordering on :$a$$q\leq_{ap\mathrm{r}}p$$q\leq_{ap\mathrm{r}}p$$a$$a\in B_{(u^{p})}$$[a]_{p}$$a$$\leq_{Q}$

$Q$

45
Lemma 4.1 (1)(2) For p, q, r $\in Q$ , if$\leq_{pr}$$q’\leq_{pr}p$(3)$and\leq_{ap\mathrm{r}}$are transitive relations.

$r\leq_{pr}q\leq_{ap\mathrm{r}}p$, then there is $q’\in Q$ such that$r\leq_{ap\mathrm{r}}$.
$\leq_{Q}$is a partial ordering on Q.

. For (2), suppose that $r\leq_{pr}q\leq_{apt}$
andProof (1) is clear by definition of
witness $q\leq_{apr}p$ . Without loss of generality, we may. Let$(-a)arrow$:assume that $-a\in B(up)\backslash null(\mu^{p})$ as well. Put $u^{q’}=u^{r}$ . Let[ $(B_{(u^{p})} [ (-a))$ freely and let$[0, \mu^{p}(-a)]$ be any $[0, \mu^{p}(-a)]$ -measure extending$\leq_{apr}$$\leq_{pr}$$p$$a\in B_{(u^{p})}\backslash null(\mu^{p})$$\mu^{*}$$B(u^{r})$ $[$$\mu^{p}$$\mu^{q’}$: $B_{(u^{q’})}arrow[0,1]$ be defined by$\mu^{q’}(b)=\mu^{r}(b\cap a)\cdot\mu^{p}(a)+\mu^{*}(b\backslash a)$by, we have $q’\leq_{pr}p$ andfor $b\in B_{(u^{r})}$ . Then, letting$r=q’||a$ .
. Suppose that , , $r\in P$ are(3): It is enough to show the transitivity of, there are , $s’\in Q$ such thatsuch that $p\leq_{Q}q\leq_{Q}r$ . By definition of$r\leq_{ap\mathrm{r}}q’$$q’=\langle u^{q’}, \mu^{q’}\rangle$$p$$\leq Q$$q$$s$$\leq_{Q}$$p\leq_{apr}s\leq_{pr}q\leq_{apr}s’\leq_{pr}r$.

By (2) we can find an $s’\in Q$ such that$p\leq_{apr}s\leq_{apr}s’\leq_{pr}s’\leq_{pr}r$.

It follows by (1) that $p\leq_{apr}s’\leq_{pr}r$ . Thus $p\leq Qr$ as desired.

$\square$(Lemma 4.1)

For $q\in Q$ , let$AP_{q}^{Q}=\{p\in Q : p\leq_{apr}q\}$.

Lemma 4.2 Suppose that , $q\in Q$ with $p\leq_{pr}q$ and $a\in B(up)\backslash null(\mu^{q})$ . Then$p||a\leq_{pr}q||a$ .
$p$. It is enough to show that
Proof Let $\gamma=\sup u^{q}$ and $u=$.
$(q||a)$ [is the free product of $p||a$ and $(q||a)($. Then by Lemma 2.4,(2) we have:andSuppose that$(u^{p} \backslash u^{q})$$\mathrm{B}(\mathrm{u}\mathrm{p})$$B_{((u^{q}\cap\gamma)\cup u^{p})}$$b_{1}\in B_{(u^{p})}$$\cap\gamma$$b_{2}\in B_{(u\cap\gamma)}$$(q||a)(b_{1}\cap b_{2})=(q[ B_{((u^{q}\cap\gamma)\cup u^{\mathrm{p}})})||a(b_{1}\cap b_{2})$$=(p\otimes(q( B_{(u\cap\gamma)}))||a(b_{1}\cap b_{2})$$=((p||a)\otimes((q( B_{(u\cap\gamma)})||a))(b_{1}\cap b_{2})$$=p||a(b_{1})\cdot(q[ B_{(u\cap\gamma)})||a(b_{2})$.
$\coprod_{(\mathrm{L}\mathrm{e}\mathrm{m}\mathrm{m}\mathrm{a}4.2)}$

46
Lemma 4.3 (1) For qE Q,
sure algebra of Maharam type(2)$/’ AP^{Q}$$\ovalbox{\tt\small REJECT}$,$\ovalbox{\tt\small REJECT}_{a},$.

$|y\ovalbox{\tt\small REJECT}^{Q}|$$I\langle’ AP\ovalbox{\tt\small REJECT}\rangle\ovalbox{\tt\small REJECT}_{apv^{=}}\rangle\ovalbox{\tt\small REJECT}\ovalbox{\tt\small REJECT} Z^{Q\ovalbox{\tt\small REJECT}}’\rangle q\ovalbox{\tt\small REJECT}|tq$) is equivalent to the forcing by the mea-.

Proof (1): The mapping$\Phi$:$AP_{q}^{Q}arrow B_{(q);}q||a\vdasharrow[a]_{q}$is well-defined and adense embedding of
into.
(2): First we show that, , $p’\in AP_{q}^{Q}$ are compatible inif and only ifthey are compatible in.
If , $p’\in AP_{q}^{Q}$ are compatible inthen clearly they are also compatible in.
Conversely, suppose that , $p’\in AP_{q}^{Q}$ are compatible in (and let $r^{*}\in Q$
be such that
, . Let and be the witnesses of $p\leq_{apr}q$ and $p’\leq_{apr}q$respectively. Let , $r’\in Q$ be such that $r^{*}\leq_{apt}r\leq_{pr}p$ and,and let ,be witnesses ofandrespectively. Then$\mu^{r^{*}}(a\cap b)=\mu^{r^{*}}(a’\cap b’)=1$ . It follows
that $a\cap a’\in B(u^{q})\backslash null(\mu^{q})$ . So letting, we have, . Thus and are compatible
andin.
$AP_{q}^{Q}$$B_{(q)}$$p$$\langle Q, \leq_{Q}\rangle$$\langle AP_{q}^{Q}, \leq_{apr}\rangle$$p$$\langle AP_{q}^{Q}, \leq_{apr}\rangle$$\langle Q, \leq_{Q}\rangle$$p$$r^{*}\leq_{Q}p$$p’$$\mathrm{Q},$

$a$

$a’$

$r$

$b$

$r^{*}\leq_{ap\mathrm{r}}r’\leq_{pr}p’$$b’\in B_{u^{r^{*}}}$$r^{\mathrm{t}}=q||(a\cap a’)$$\leq_{Q}\rangle$$r^{*}\leq_{apr}r$$r^{\uparrow}\leq_{apt}p$$r^{\uparrow}\in AP_{q}^{Q}$$r^{*}\leq_{a\mathrm{p}_{t}}r’$

$p’$

$p$

$p’$

$\langle AP_{q}^{Q}, \leq_{apr}\rangle$

that I is predense below

$q$

in

$\langle Q, \leq_{Q}\rangle$

is predense below , we have$q$$\sum^{B_{(q)}}\{[a]_{q} :.

$\backslash$q||a\in I\}_{\ovalbox{\tt\small REJECT}}^{\backslash }1_{B_{(q)}}$.

Hence there issuch that $q||a\in I$ and $c=a\cap b\in B_{(u\cdot)}\backslash null(fis)$ . Now$s||a\leq_{pr}q||a$ by Lemma 4.2. Hence, by Lemma 2.4,(1), $s||c=(s||a)||c\leq_{Q}q||a$.
Similarly, $s||c=(s||a)||c=r||c\leq_{ap\mathrm{r}}r$ .
$a\in B_{(u^{q})}$$\coprod_{(\mathrm{L}\mathrm{e}\mathrm{m}\mathrm{m}\mathrm{a}4.3)}$

Lemma 4.4 Suppose that p, q, r $\in Q$ , $q\leq_{pr}p$ and $r\leq_{apr}p$ . Then q and r arecompatible in Q. Furthermore, there is an s $\in Q$ such that $s\leq_{apr}q$ and $s\leq_{pr}r$ .
Proof Let abe awitness of $r\leq_{apt}p$. Then $q||a\leq_{pr}p||a=r$ by Lemma
4.2 and $q||a\leq_{apr}q$ . Thus s $=q||a$ is as desired.
$\in B_{(u^{p})}$$\coprod_{(\mathrm{L}\mathrm{e}\mathrm{m}\mathrm{m}\mathrm{a}44)}$

Lemma 4.5 Suppose that , $q\in Q$ and $u_{0}=u^{p}\cap u^{q}$ . If
and are compatible
$p$

$p$

$q$

$\mu^{p}$[ $B(u_{0})=\mu^{q}$ [$B_{u_{0}}$

then

47
Proof Let $u^{r}=u^{p}\cup u^{q}$ and$\mu^{r}=\mu^{p}\otimes_{u_{0}}\mu^{q}$. Then $r\leq_{pt}p$ , q.

[
for $u_{0}=u^{p}\cap u^{q}$ ,For , $q\in Q$ with [4.5 is denoted by (&q.
let [For $p\in Q$ and
, let
. For
[ $X\leq_{pr}p$ for any $p\in Q$ and$B_{(u_{0})}=\mu^{q}$$\mu^{p}$

$p$

$r$

$B_{u0}$$\coprod_{(\mathrm{L}\mathrm{e}\mathrm{m}\mathrm{m}\mathrm{a}4.5)}$

as in the proof of Lemma$p$$X\subseteq\lambda$$p$$X=\langle u^{p}\cap X, \mu^{p} [ B(u^{\mathrm{p}}\cap X)\rangle$$p$[ $X\in Q$ and$X\subseteq\lambda$$X\subseteq\lambda$$p$.

$Q\downarrow X=\{p\in Q : u^{p}\subseteq X\}$.

Clearly $Q\downarrow X=\{p[ X : p\in Q\}$ .

.
Lemma 4.6 Suppose that
, if $q\leq_{Q}p$ [ , then and are compatible.
(1) For $p\in Q$ and.
(2) If is a strong limit or a successor of a strong limit then$\alpha<\lambda$$q\in Q\downarrow\alpha$$p$$\alpha$$q$$Q\downarrow\alpha\leq 0Q$$\alpha$be suchbe such that $q\leq_{apr}q’\leq_{pr}p$ [ . Let
Proof (1): Letoverandbe the free amalgamation ofthat $q=q’||a$ , $u^{r}=u^{q}\cup u^{p}$ and let
and $r’=r||a$ . Then $r’\leq_{pr}q$ by Lemma(see Theorem 3.1). Let
4.2 and $r’\leq_{apr}r\leq_{pr}p$ .
$q’\in Q\downarrow\alpha$$a\in B_{(u^{q’})}$$\alpha$$\mu^{q’}$$\mu^{r}$$u^{p}\cap\alpha$$\mu^{p}$$r=\langle u^{r}, \mu^{r}\rangle$

are compatible in then they are(2): We first show that if ,. So suppose that $r\in Q$ is such that $r\leq_{Q}q$ , . Let ,compatible in$s’\in Q$ be such that $r\leq_{apr}s\leq_{pr}q$ and $r\leq_{apr}s’\leq_{pr}q’$ withand $a’\in B_{u^{s’}}\backslash null(\mu^{s’})$ witnessing $r\leq_{apr}s$ and $r\leq_{apr}s’$ respectively. Let
and , $a’\in B(u)$ .
be such that $|u|=|u^{q}|+|u^{q’}|+\aleph_{0}$ ,Then we have [ $u\leq_{apr}s$ [ $u\leq_{pr}q$ and [ $u\leq_{apr}s’[u\leq_{pr}q’$ . By (A)(b) we have,$|u|=|u^{q}|+|u^{q0}|+\aleph_{0}<\alpha$ . Hence we can findsuch that$q$$Q$$q’\in Q\downarrow\alpha$$q’$$Q\downarrow\alpha$$s$$a\in B_{u^{s}}\backslash null(\mu^{s})$$u^{q}\cup u^{q’}\subseteq u$$u\subseteq u^{r}$

$a$

$r$

$r$

$u^{q}\cup u^{q’}\subseteq\overline{u}$$\overline{u}\subseteq\alpha$and so thatis an end-extension of$(u^{q}\cup u^{q’})=$
be abijection withsatisfy the requirement (A)(b). Let :such that. induces an isomorphism of [ , [ and [ to some
. This shows that andand.
are compatible inis predense in [ . We show that I is also
Now suppose that
predense in . Let $p\in Q$ be arbitrary. By the assumption there is $q\in I$ such thatbe such that $r\leq_{Q}p$ [ , . By (1), and[ and are compatible. Let(Lemma 4.6)
are compatible. Hence and are compatible.
$u^{q}\cup u^{q’}$$|u\backslash (u^{q}\cup u^{q’})|=|\overline{u}\backslash (u^{q}\cup u^{q’})|,\overline{u}$$f$$id_{u^{q}\cup u^{q’}}$$f$$f$ $[$$uarrow\overline{u}$

$u$

$r$

$u^{\overline{r}}=u^{\overline{s}}=u^{\overline{s}’}=\overline{u},\overline{r}\leq_{apr}\overline{s}\leq_{pr}q$

$s$

$u$

$\overline{u}$

$u$

$r$

$\overline{r},\overline{s},\overline{s}’$$\overline{r}\leq_{apr}\overline{s}’\leq_{pr}q’$

$q’$

$q$

$Q\downarrow\alpha$$Q$$I\subseteq Q\downarrow\alpha$$\alpha$$Q$$p$$\alpha$$r\in Q\downarrow\alpha$$q$$\alpha$$r$$q$$\square$

$p$

$q$

$p$

Lemma 4.7 Suppose that
there is a q $\in Q$ , $q\leq_{pr}p$ such that I$p|\vdash_{Q}$$”\sim\tau\in V$

$=$

”

{r

and $Q- name\tau\sim$ . Then
: r decides } is predense below

for some p$\in AP_{q}^{Q}$$\in Q$$\sim\tau$48
Proof Let$q_{i}$, $r_{i+1}\in Q$ ,(0) $q_{0}=p$ ;(1)(2) For limit$i<\omega_{1}$be defined inductively such thatis adecreasing sequence with respect to
;
, let
where
and
is themeasure on
generated from;
(3) For all $i<\omega_{1,:+1}r\leq_{apt}q:+1$ ,decides and $r:+1$ is incompatible with, $j<i$ provided that there are suchall
, $r:+1$ ;otherwise we let $q_{i+1}=q_{i}$and.
$\langle q_{i} :i<\omega_{1}\rangle$$\leq_{pr}$$\gamma<\omega_{1}$$q_{\gamma}=\langle u^{q_{\gamma}}, \mu^{q_{\gamma}}\rangle$$B_{(u^{q_{\gamma)}}}$$u^{q_{\gamma}}= \bigcup_{:<\gamma}u^{q:}$$\mu^{q_{\gamma}}$$\bigcup_{i<\gamma}\mu^{q:}$$r_{i+1}$$\sim\tau$$r_{j+1}$$q_{\dot{\iota}+1}$$r_{\dot{|}+1}=\langle\emptyset, \emptyset\rangle$

Note that (2) is possible by Corollary 2.2.
Claim 4.7.1 There is asuch that$\delta<\omega_{1}$$r_{\delta+1}=\langle\emptyset, \emptyset\rangle$.

Otherwise $r:+1$ ,are pairwise incompatible. For eachletbethe witness of
. Let
and
. Thenisa $[0, 1]$ measure on $B(u*)$ and $a:+1/null(\mu^{*})$ ,are pairwise disjoint non-zeroelements of. This is acontradiction to the c.c.c. of.
$\vdash$$i<\omega_{1}$$i<\omega_{1}$$r:+1\leq_{ap\mathrm{r}}q:+1$$u^{*}= \bigcup_{:<\omega_{1}}u^{q:}$$a_{i+1}$$\mu^{*}=\cup:<\omega_{1}\mu^{q:}$$\mu^{*}$$i<\omega_{1}$$B_{(u^{*})}/null(\mu^{*})$$B_{(u^{*})}/null(\mu^{*})$$\dashv$(Claim 47$1\rangle$

Now, letbe minimal withand let. We show that
this is as required. By Lemma 4.4, for each
, we can find
such
that
and
. Let $I’=\{r_{i+1}^{+} : i<\delta^{*}\}$ . By (3), every elements
of $I’$ decides .
Hence it is enough to show that is predense below .
Suppose not. Then by Lemma 4.3,(2), there issuch that is incompatible with every. It follows that is also incompatible with every. Let$r\leq_{Q}r’$ be such thatand $q’\in Q$ be such that
. Then we could
have choosen and at’st stage of construction asand. This isacontradiction to.
$\delta^{*}<\omega_{1}$$r_{\delta^{\mathrm{r}}+1}=\langle\emptyset, \emptyset\rangle$$q=q_{\delta^{*}}$$i<\delta^{*}$$q$$r_{i+1}^{+}\leq_{ap\mathrm{r}}q$$r_{i+1}^{+}\in Q$$r_{\dot{l}+1}^{+}\leq_{pr}r_{i+1}$$\sim\tau$

$I’$

$q$

$r’\leq_{a\mathrm{p}t}q$$r’$$r_{\dot{l}+1}^{+}$$r_{i+1}$$r\leq_{ap\mathrm{r}}q’\leq_{pr}q$.

$\sim\tau$

$q’$

$r$

$\delta^{*}+1$$r$$q_{\delta+1}$$r_{\delta^{\mathrm{r}}+1}$$r_{\delta^{\mathrm{r}}+1}=\langle\emptyset, \emptyset\rangle$

Proposition 4.8$Q$$\coprod_{(\mathrm{L}\mathrm{e}\mathrm{m}\mathrm{m}\mathrm{a}47)}$

preserves all cardinals

Proof Suppose not. Then there are $q\in Q$ ,$q|\vdash Q$

where$\delta^{+}$$q_{\dot{1}}$:$\check{\delta}^{+}arrow\check{\delta}$.

$\delta<\kappa$

and aQ-namef\sim such thatis aone to one mapping”

denotes the successor cardinal of in the ground model. Let$\delta$$\nu=\sup$Let$”\sim f$$\leq\kappa$, $s:\in Q$ ,($\delta\cap\{\theta$$a:\in B(u^{q}\cdot)$,for$\theta^{+}$:$\theta$$i<\delta^{+}$is strongly inaccessible}).
be such that

49
(0)
(1)
(2)

$q_{0}=q$$\langle q_{i}[(\lambda \backslash \nu) :$s_{i}\leq_{apr}q_{i}$(3)$a_{i}$

Now since(4)$q_{i}$(5)is adecreasing sequence with respect to;decidesi<\delta^{+}\rangle$

and$s_{i}$$\leq_{pr}$;$\sim f(i)$witnesses $s:\leq_{apr}q:$ .
$|Q\downarrow\nu|\leq\delta$, there are$X\in[\delta^{+}]^{\delta^{+}}$,$p\in Q\downarrow\nu$

and$n\in\omega$$\backslash \{0\}$such that[ $\nu=p$ for all $i\in X$ ;$\mu^{q_{i}}(a_{i})\geq\frac{1}{n}$for all $i\in X$ .

: $Xarrow\delta$ be defined by $f(i)=j$ for $i\in X$ and $j\in\delta$ such that
to 1. But this is impossibleis$”\sim f(i)=j$ ”. Then by (4) and (5),Let$s_{i}|\vdash_{A}$

;

$f$

$f$

$\leq n$since $|X|>\delta$ .

$\square _{(\mathrm{P}\mathrm{r}\circ \mathrm{p}\circ\epsilon \mathrm{i}\mathrm{t}\mathrm{i}\circ \mathrm{n}4.8)}$

Lemma 4.9 Suppose that$\kappa$

is strongly incaccessible. Then Q has the strong$\kappa^{+}-$c.c.
By -system lemma, there is an, for
Proof Suppose that
, $i\in S$ form a -system, say with root . Since $|^{B_{(u^{*})}}[0,1]|<\kappa$ ,such that[ $B(u)*=\mu^{*}$ for all $i\in S’$ .
such that
andthere isobtained as:and be the $[0, 1]$ -measure on
For , $i’\in S’$ , let, .
, we haveand ’over . Then, forthe free product of$i<\kappa^{+}$$q_{i}\in Q$$S\in[\kappa^{+}]^{\kappa^{+}}$$\Delta$$u^{*}$$\Delta$$u^{q}\cdot.$$S’\in[S]^{\kappa^{+}}$$\mu^{q:}$$\mu^{*}$$\tilde{u}=u^{q:}\cup u^{q}$$i$$u^{*}$$\mu^{q}:$$\mu^{q:}$$B_{(\overline{u})}$$\tilde{\mu}$$\tilde{q}\leq_{pr}q_{i}$$\tilde{q}=\langle\tilde{u},\tilde{\mu}\rangle$$q_{i’}$$\square _{(\mathrm{L}\mathrm{e}\mathrm{m}\mathrm{m}\mathrm{a}4.9)}$

5Real-valued measurability in$V^{Q}$.-set introduced in the last section forces real-valued meaWe show that thehas enough large cardinal property.
surability of provided that in
Let us begin with introducing the following notation: Suppose that $p\in Q$ andis aformula in the forcing language over . Let$\mathrm{p}.0$$Q$$\kappa$$\kappa$$V$$Q$$\varphi$$I_{p,\varphi}=\{[a]_{p} :Let$a^{*}\in B_{(u^{\mathrm{p}})}$

a\in B_{(u^{\mathrm{p}})}\backslash null(\mu^{p}), p||a|\vdash_{Q} " \varphi"\}$be such that$[a^{*}]_{p}= \sum^{B_{(p)}}I_{p,\varphi}$

$(*)$

and
$[\varphi \mathrm{I}_{p}=\mu^{p}(a^{*})$

Note that [$\varphi \mathrm{I}_{p}$.

does not depend on the choice of$a^{*}$

.

.

50
For a

$Q- \mathrm{n}\mathrm{a}\mathrm{m}\mathrm{e}\tau\sim$of an element of V and p $\in Q$ , let$[\tau \mathrm{I}_{p}^{*}=\mu^{p}(a^{\uparrow})\sim$with:, $p||a$ desides }.
{We identify each formula in the forcing language forwith the Q-namewhich gives the truth value of the formula (i.e. either 0or 1depending on whether
is forced or not.) and consider [under this identification. Thus

for$a^{\uparrow}\in B(u^{p})$$[a^{\uparrow}]_{p}=\Sigma^{B_{(\mathrm{p})}}$$[a]_{p}$$a\in B_{(u^{p})}\backslash null(\mu^{p})$$\sim\tau$$Q$$\varphi$$\varphi$$\varphi \mathrm{J}_{p}^{*}$$\mathbb{I}\varphi \mathrm{J}_{p}+[\neg\varphi \mathrm{J}_{p}$$[\varphi \mathrm{I}_{p}^{*}=$.

Lemma 5.1 Suppose that $p\in Q$ and
Q. Letbe as in $(*)$ above. If$a^{*}$$\varphi$

is a formula in the forcing language over
, then$a^{*}\not\in null(\mu^{p})$$p||a^{*}|\vdash_{Q}$“ $\varphi"$.

“
Proof Otherwise there is some $q\leq_{Q}p||a^{*}$ such that
. Let besuch thatand let $b\in B_{(u^{q’})}$ be such that $q=q’||b$ . Since, there issuch thatby $(*)$ . Hence by
the same argument as in the proof of Lemma 4.3,(2), we can show thatand“are compatible. But. This is acontradiction.
$q|\vdash_{Q}$$q’$$\neg\varphi"$$q\leq_{apt}q’\leq_{\mathrm{P}^{f}}p||a^{*}$$b\cap a^{*}\not\in null(\mu^{q’})$$b\cap a_{0}\not\in null(\mu^{q’})$$a_{0}\in I_{p,\varphi}$$p||a_{0}$$p||a_{0}|\vdash_{Q}$$q$$\varphi"$$\coprod_{(\mathrm{L}\mathrm{e}\mathrm{m}\mathrm{m}\mathrm{a}51)}$

Lemma 5.2 For , $p’\in Q$ , if $p\leq_{pr}p’$ then$p$$[\varphi]_{p},$$\leq \mathrm{I}\varphi]_{p}$.

“Proof Suppose. Then. Since $p||a\leq_{P^{f}}p’||a$ by Lemma“
4.2,. As $p||a\leq_{apt}p$ , it follows that
. Thus.
$[a]_{\mu}\in I_{p’,\varphi}$$p||a|\vdash_{Q}$$p’||a|\vdash_{Q}$$\varphi"$$\varphi"$$[a]_{p}\in I_{p,\varphi}$$\mathrm{I}\varphi \mathrm{J}_{p},$$\leq[\varphi \mathrm{J}_{p}$$\coprod_{(\mathrm{L}\mathrm{e}\mathrm{m}\mathrm{m}\mathrm{a}52)}$

Theorem 5.3 Suppose thatwith crit(j)and$=\kappa$$2^{\kappa}>M\subseteq M$$|\vdash Q$

and j : V $arrow M$ is an elementary embedding. Then$\lambda=\kappa^{+}$“$\kappa$

is real-valued measurabl\"e.

Proof In M, we have Q
by (A)(b). Hence, by Lemma 4.6, Q
.
Let
: k
be an emumeration of -names of mappings from to some$=j(Q)\downarrow\lambda$$\langle\tau_{k}\sim$$\gamma<\kappa$$\leq\circ \mathrm{j}(\mathrm{Q})$$Q$$<2^{\kappa}\rangle$$\kappa$.

Let
for allthat$\eta:2^{\kappa}arrow Q\cross 2^{\kappa};i\mapsto*\langle\eta_{0}(i), \eta_{1}(i)\rangle$$2^{\kappa}$(a)$\langle q, i\rangle\in Q\cross 2^{\kappa}$. Let$\langle q:$q:\in j(Q)\downarrow(j(\lambda)\backslash \lambda)$

sequence with respect to$\leq_{\mathrm{P}^{f}}$;:be asurjection such thatbe asequence of elements of $j(Q)$ such$|\eta^{-1\prime\prime}\{\langle q, i\rangle\}|=$

i<2^{\kappa}\rangle$

for all$i<2^{\kappa}$

and$\langle q::i<2^{\kappa}\rangle$

is adecreasing

51(b) for all i

and

$q’$

C

$\ovalbox{\tt\small REJECT} 7^{\ovalbox{\tt\small REJECT}}(Q)+$$<2\mathrm{K}$, there is$\ovalbox{\tt\small REJECT}\ovalbox{\tt\small REJECT}(\mathrm{A})_{\ovalbox{\tt\small REJECT}}^{\ovalbox{\tt\small REJECT} S}$

Awith$\ovalbox{\tt\small REJECT}_{i}\ovalbox{\tt\small REJECT} \mathrm{E}$Q\rangle$p’\mathrm{S}_{1^{\mathrm{t}}}\ovalbox{\tt\small REJECT}_{i}$$\ovalbox{\tt\small REJECT}_{i}\ovalbox{\tt\small REJECT}_{pr}\mathrm{r}/\mathrm{o}(\ovalbox{\tt\small REJECT})$andsuch that, in M, for any

$p’$

E $Q$

$q’\ovalbox{\tt\small REJECT}_{pr}(i+\mathit{1}$$[j(\tau_{\eta 1(i)})(\kappa)\mathrm{I}_{\tilde{p}\dot{.}\otimes q.+1}^{*}\sim.=[j(\tau_{\eta 1(i)})(\kappa)\mathrm{I}_{p\otimes q’}^{*}\sim’\cdot$

The construction is possible by closedness of $M$ (for (b), aconstruction similar tothe one in the proof of Lemma 4.7 is to be applied).
asubset of and $q\in Q$ , let
For a -name$Q$$\kappa$$X\mathrm{o}\mathrm{f}\sim$:$q’\in Q, q’\leq_{pr}q,$ i$\mu_{q}(X)=\mathrm{r}\mathrm{n}\mathrm{a}\mathrm{x}\{[\check{\kappa}\in j(X)\mathrm{I}_{q’\otimes q:}\sim\sim$$<2^{\kappa}\}$The following is immediate from Lemma 5.1 and the definition ofClaim 5.3.1 Suppose that $p\in Q$ ,a

$X\sim is$$Q$-name of a subset of$\mu_{q}$$\kappa$.
:and$i^{*}<\omega_{1}$. If$[\kappa\vee\in j(X)\mathrm{J}_{p\otimes q_{t}*}^{*}\sim=[\check{\kappa}\in j(X)\mathrm{I}_{p\otimes q’}^{*}\sim’$

for any $p’\in Q$ and$q’\in j(Q)\downarrow(j(\lambda\backslash \lambda))$with $p’\leq_{pr}p$ and$q’\leq_{pr}q_{i}*$, then we have
(Claim 5.3.1)$\dashv$$\mu_{p}(X)\sim=[\check{\kappa}\in j(X)\mathrm{I}_{p\otimes q_{i^{*}}}\sim\cdot$

Claim 5.3.2 (1) For, if(2) For(3) For $q\in Q$ and, if $\in j(X)$ , there
for all $q\in Q$ .
, thenfor all $q\in Q$ .
“, then-namesof subsets of , if

$X\subseteq\kappa$$X\subseteq\kappa$$\mu_{q}(X)\sim\leq\mu_{q}(\mathrm{Y})\sim$(4)$\mu_{q}(\check{X})=1$$\kappa$$\mu_{q}(\check{X})=0$$\not\in j(X)$$\kappa$$Q$$\kappa$

For $q\in Q$ ,$\gamma<\kappa$

and

$Q$

-names

$X\sim’ X\ell\sim’\ell<\kappa$$\kappa$$\in j(X)$, then$\mu_{q}(X)\geq\mu^{q\otimes qi}(u^{q}\cup u^{q}\cdot)=1$, if“ $X=\cup\cdot\{X\ell\sim\sim : \ell<\gamma\}"$then for any $q’\leq_{Q}q$ there is $q’\leq_{pr}q’$ such that(1): If$X\sim\subseteq \mathrm{Y}\sim"$.

$q|\vdash_{Q}$$\vdash$$q|\vdash_{Q}$$X\sim’ \mathrm{Y}\sim$$q\otimes q_{i}|\vdash_{j(Q)}$,$\mu_{q’}(X)=\Sigma_{\ell<\gamma}\mu_{q’}(X\ell)\sim\sim$“ $\check{\kappa}\in j(\check{X})$ ”

for any.

(2): Simlilarly to (1).

(3): Clear.
(4): Suppose that $q’\leq_{Q}q$ . Let$|\vdash_{Q}$$k<2^{\kappa}$$”\sim\tau_{k}$:be such that$\kappaarrow\gamma+1$”

and$q|\vdash_{Q}$“ $\forall i<\kappa$ $(\tau_{k}(i)=\ell\simrightarrow i\in X\ell)\sim$ ”$i<2^{\kappa}$.

. Hence$1\geq$

52
for allLet$\langle q:

and:$\ell<\gamma$.
be such thatand
. By (b) in the definition of, the assumption of Claim 5.3.1 is satisfied forand each of. Hence$i^{*}<2^{\kappa}$i<2^{\kappa}\rangle$$X_{\ell}\sim$’$\eta_{0}(i^{*})=q’$$\eta_{1}(i^{*})=k$$p’=p_{\dot{l}^{*}}$$X\sim$$\ell<\gamma$.

$\mu_{p}\prime\prime(X)=\mathrm{I}\check{\kappa}\in j(X)\mathrm{I}_{p’\otimes q*}=\sum_{\ell<\gamma}\sim\sim:+1\mathbb{I}\check{\kappa}\in j(X_{\ell})\mathrm{J}_{p’\otimes q*}\sim:+1=\sum_{\ell<\gamma}\mu_{t’}(X\ell)\sim$$\dashv$(Claim 532)

From this point on, the proof is very similar to the one for real-valued measurability in arandom model by Solovay [9]. We shall follow closely the versionof the proof given in Kanamori [7] (the proof of Theorem 17.5 in [7]). The lastparagraphs of our proof in particular are almost identical with the corresponding
part of the proof in [7]. Nevertheless, we also include them for convenience of thereader.
Claim 5.3.3 Suppose that p $\in Q$ , X is a$0\leq r\leq 1$ .
If Q

$Q$

-name of a subset ofthen we have$\models\forall q\leq_{pt}p\exists s\leq Qq(\mu_{s}(X)\sim\geq r)$

Suppose that, let $p:\in Q$ ,Forsuch that$\vdash$(0) $p_{0}=p$ ;(1) ( :
(2) For alimit$j_{i}<\omega_{1}$, $a:+1$ ,and r$\in \mathbb{R}$$\mu_{p}(X)\sim\geq r$$Q\models\forall q\leq_{pr}p\exists s\leq_{Q}q(\mu,(X)\sim\geq r)$$i<\omega_{1}$$\kappa$. We show that$b_{:+1}\in B_{u^{\mathrm{p}:+1}}\Phi q_{\mathrm{j}}:+1$

with

.

.
be defined inductively$\mu_{p}(X)\sim\geq r$

is adecreasing sequence in with respect to
;
,
where
and is a $[0, 1]$ -measureonextending;(3)
is acontinuously increasing sequence of ordinals;(4) If we cannot find $p:+1$ ,, $a:+1$ ,satisfying the conditions (5) $\sim(7)$in the following, then we let $p:+1=pi$ ,and $a:+1=b_{:+1}=\emptyset$ ;otherwise:(5)
is anull set with respect tofor all $i’<i$ ;
(6)
,
and
$Pi$

$i<\omega_{1}\rangle$$B_{(u)}$$\gamma<\omega_{1}$$Q$$p_{\gamma}=\langle u,\mu\rangle$$\leq_{pr}$$u= \bigcup_{:<\gamma}u^{p:}$$\mu$$\bigcup_{:<\gamma}\mu^{p:}$$\langle j_{\dot{1}} :i<\omega_{1}\rangle$$<\omega_{1}$$j_{\dot{l}+1}$$b_{\dot{|}+1}$$j_{\dot{|}+1}=j_{\dot{1}}$$a:+1\cap a_{\dot{1}’+1}$$b_{i+1}\subseteq a:+1$$\mu^{p:+1}:+1\otimes q_{\mathrm{j}}$$b_{:+1}\in B_{(u^{p:+1^{\theta q_{\mathrm{j}}}})}\backslash :+1null(\mu^{p.\otimes q_{\mathrm{j}}}.+1:+1)$$. \cdot..\frac{\mu^{\mathrm{P}\cdot 1}\cdot+1(+\emptyset q_{j}.b_{\dot{|}+1})}{\mu^{p_{+1}\otimes q_{j}}\cdot+1(a_{\dot{|}+1})}\geq r$(7)$\langle u:p:+1\otimes q_{j}+1, \mu p:+1\otimes q_{\mathrm{j}}:+1||b:+1\rangle|\vdash_{j(Q)}$;“$\check{\kappa}\in j(X)\sim$”.

Now, as in the proof of Lemma 4.7, there is the minimal. Let $j^{*}= \sup\{j_{\dot{1}} : i<\delta^{*}\}$.

$a_{\delta+1}*=\emptyset$$\delta^{*}<\omega_{1}$$\{[a:+1]_{ps*\otimes q_{\mathrm{j}^{*}}} :i<\delta^{*}\}$

such thatis a maxima

53
antichain inthat$B_{(ps*\otimes q_{\mathrm{j}}*)}$

by the assumption of the Claim. Let$[b]_{p_{\delta}*\otimes q_{j^{*}}}= \sum^{B_{(\mathrm{p}_{\delta^{*\Phi q_{j^{*)}}}}}}\{[b:+1]_{\mathrm{P}s*\otimes q_{\mathrm{j}^{\wedge}}} :by (6) and
Then we have, it follows thatby (7) and Lemma 5.1. Sincei<\delta^{*}\}$$p_{\delta}*\leq_{pr}p$

be such.
“$\langle u^{p_{\delta}*\otimes q_{\mathrm{j}^{*}}}, \mu^{ps*\otimes q_{\mathrm{j}^{*}}}||b\rangle|\vdash_{j(Q)}$$\mu^{\mathrm{P}s*\otimes q_{j^{*}}}(b)\geq r$

Now, for $q\in Q$ and $Q$-name$b\in B_{(u^{p_{\delta^{*\Phi q_{j^{*})}}}}}$$\mu_{p}(X)\sim\geq r$.

”$\check{\kappa}\in j(X)\sim$$\dashv(\mathrm{C}\mathrm{l}*\mathrm{i}\mathrm{m}5.3.3)$

asubset of , let$\kappa$$X\mathrm{o}\mathrm{f}\sim$$\mu_{p}^{*}(X)=\inf\sim(\{\mu_{q}(X)\sim :.

q\leq_{Q}p\})$By this definition the following is clear:

Claim 5.3.4 For , $p’\in Q$ and
.

$Q$

$p$

-name

$X\sim$of a subset of$\kappa$, if $p\leq_{Q}p’$ , then$\dashv$$\mu_{p}^{*}(X)\geq\mu_{p}^{*}(X)\sim’\sim$Let$\mu \mathrm{b}\mathrm{e}\sim$

the

$Q$

-name such that$|\vdash_{Q}$for Q-name$Q$$X\sim \mathrm{o}\mathrm{f}$$|\vdash_{Q}$“$\mu:P\sim(\kappa)arrow[0,1]$“ $\mu(X)\sim\sim=\sup(\{\mu_{p}^{*}(X)\sim :subset of$\kappa$and the standard.

The well-definedness of“
moment we consider each

$Q$

and

”

p\in G\})\sim$

-name

”

$G\mathrm{o}\mathrm{f}\sim$the generic set over

yet to be established in course of the proof: At the”merely as an abbreviation of aQ-name of a$\mu \mathrm{i}\mathrm{s}\sim$$\sim r$$\mu(X)\sim\sim$

real such that$|\vdash_{Q}$$” \sim r=\sup(\{\mu_{p}^{*}(X)\sim :Claim 5.3.5 Suppose that $q\in Q$ ,Then we have$\mu_{q}^{*}(X)\geq r\sim$$X\sim$is a $Q$ -name of a subset of$q|\vdash_{Q}$$\Leftrightarrow$“$\mu(X)\sim\sim\geq\check{r}$and since:This is clear by definition of”“and let:Suppose that
. By definition of
It is enough to show that$\vdash$$\mu^{*}(X)\sim\sim$$(\Rightarrow)$$(\Leftarrow)$$q|\vdash_{Q}$$\mu(X)\sim\sim\geq\check{r}$“$\exists s\in G\sim(\mu_{s}^{*}(X)\sim\geq\check{r}_{0})$and $r\in[0,1]$ .

“$q\in G\sim"$.

be such that $0<r0<r$ .
’we have$\mu(X)\sim\sim$”.

By Claim 5.3.4, it follows that$\forall p\leq_{Q}q\exists s\underline{<}_{Q}p(\mu_{s}^{*}(X)\sim\geq r_{0})$$\kappa$”.

$q|\vdash Q$$r_{0}\in \mathrm{R}$$\mu_{q}^{*}(X)\sim\geq r_{0}$$q|\vdash_{Q}$”.

p\in G\})\sim$.

54
Hence by definition of
we havefor any $q’\leq_{Q}q$ we havefor any $q’\leq_{Q}q$ . It follows that$\mu_{s}^{*}$$\forall p\leq_{Q}q’\exists s\leq_{Q}p(\mu_{s}(X)\sim\geq r_{0})$$\mu_{q’}(X)\sim\geq r\circ$In the rest of the proof, we show that$|\vdash Q$. In particular. Hence by Claim 5.3.3,$\forall p\leq_{Q}q\exists s\leq_{Q}p(\mu_{s}(X)\sim\geq r_{0})$$\mu_{q}^{*}(X)\sim\geq r_{0}$$\mu \mathrm{i}\mathrm{s}\sim$.

well-defined anda -additive [0,$1]$ -measure on“$\kappa$$\mu \mathrm{i}\mathrm{s}\sim$(Claim 5.35)$\dashv$$P(\kappa)$”.

“”
For any, if $\in j(X)$ , then we haveby Claim 5.3.2(1);“”otherwise, i.e. if
thenby Claim 5.3.2(2). In particular,
“
”
“
and
”.
“
”
If
then
for any $q\leq_{Q}p$ by Claim 5.3.2(3).
“
”
Hence
: Otherwise we can findandsuch that$X\subseteq\kappa$$\kappa$$\kappa\not\in j(X)$$|\vdash Q$$\mu(\check{\kappa})=1\sim$$p|\vdash Q$$q^{*}|\vdash_{Q}$$|\vdash_{Q}$$|\vdash_{Q}$$\mu(\check{X})\sim=0$$\mu(\emptyset)\sim=0$$\mu_{q}(X)\sim\leq\mu_{q}(\mathrm{Y})\sim$$X\sim\subseteq\sim \mathrm{Y}$$p|\vdash_{Q}$$\mu(\check{X})\sim=1$$|\vdash_{Q}$$\mu(X)\sim\sim\leq\sim\sim\mu(X)$“$\mu(X)\sim\sim>\check{r}>\mu(\mathrm{Y})\sim\sim$

5.3.5. It follows that
In particular,$q^{*}\leq_{Q}p$”. But then we havefor some”
whenever$\mu_{q}\cdot\cdot(X)\sim>\mu_{q}**(\mathrm{Y})$$p|\vdash_{Q}$$r\in \mathbb{Q}$by Claim$\mu_{q^{*}}^{*}(X)\sim>r\geq\mu_{q^{\mathrm{c}}}^{*}(\mathrm{Y})\sim$$q^{**}\leq_{Q}q$“ $\mu(X)=\mu(\mathrm{Y})\sim\sim\sim\sim\sim$$p|\vdash_{Q}$“. This is acontradiction.
. This shows the$X\sim=\sim \mathrm{Y}"$well-definedness of .
Next, we show that
finitely additiv\"e. Suppose that“”. We show that
follows.
Let $p\in Q$ and ,be such that
. Then,by Claim 5.3.2(4), Claim 5.3.3 and by definition of , we have$\mu\sim$$|\vdash_{Q}$“$\mu \mathrm{i}\mathrm{s}\sim$$|\vdash_{Q}$“$X,\mathrm{Y}\sim\sim\subseteq$$\kappa\Lambda X\cap \mathrm{Y}=\emptyset\sim\sim$$|\vdash_{Q}$$r_{1}$$\mu(X\sim\sim\cup \mathrm{Y})=\mu(X)\sim\sim\sim+\mu(\mathrm{Y})\sim$$r_{2}\in \mathbb{Q}$$p|\vdash_{Q}$

”

“

”

$\mu(X)\sim\sim\geq\check{r}_{1}\Lambda\mu(\mathrm{Y})\sim\sim\geq\check{r}_{2}$$\mu_{p}^{*}$$\mu_{p}^{*}(X\cup \mathrm{Y})\geq r_{1}+r_{2}\sim\sim$Hence by Claim 5.3.5,$p|\vdash_{Q}$“$\mu(X\sim\sim\cup \mathrm{Y})\sim$$”\geq\check{r}_{1}+\check{r}_{2}$.

. By density argument it follows

that
(
“
Hence
.
“
For the opposite inequality, assume that
we can find $p\in Q$ and , , ,such that$|\vdash_{Q}$“$\forall r_{1},r_{2}\in \mathbb{Q}$$|\vdash_{Q}$$\mu(X\sim\sim\cup \mathrm{Y})\sim$$(r_{1}<\mu(X)\sim\sim\wedge r_{2}<\mu(\mathrm{Y}))\sim\sim$) .

$\sim\sim\sim"$$”\geq\mu(X)\sim\sim+\mu(\mathrm{Y})\sim\sim$$|\vdash_{Q}$$\mathrm{f}_{3}$$p|\vdash_{Q}$$arrow r_{1}+r_{2}\leq\mu(X\cup \mathrm{Y})$$r_{4}$$r_{5}$”$\mu(X\cup \mathrm{Y})\sim\sim\sim>\mu(X)+\mu(\mathrm{Y})\sim\sim\sim\sim$. Then$r_{6}\in \mathbb{Q}$“$\mu(X)<\check{r}_{3}<\check{r}_{4}\sim\sim\wedge\sim\sim\mu(\mathrm{Y})<\check{r}_{5}<\check{r}_{6}\wedge\mu(X\cup \mathrm{Y})\geq\check{r}_{4}+\check{r}_{6}\sim\sim\sim$”.

For all $q\leq Qp$ we haveby Claim 5.3.5. Hence there is ansuch that. By Claim 5.3.3 with ”replaced by, it follows that. Similarly, we get. We may assume that is taken so thatthe additivity in Claim 5.3.2(4) holds for “. So$\mu_{q}^{*}(X)\sim<r_{3}$$s$$\mu_{s}(X)\sim<r_{3}$$\mu_{p}(X)\sim\leq r_{3}$$”\geq$$”\leq"$$\mu_{p}(\mathrm{Y})\sim\leq r_{5}$$p$$X\sim\cup\cdot \mathrm{Y}\sim"$$\mu_{p}^{*}(X\cup \mathrm{Y})\leq\mu_{p}(X\cup \mathrm{Y})=\mu_{p}(X)+\mu_{p}(\sim\sim\sim\sim\sim\sim \mathrm{Y})\leq r_{3}+r_{5}<r_{4}+r_{6}$But by Claim 5.3.5, this is acontradiction to$p|\vdash_{Q}$.

“$\mu(X\sim\sim\cup \mathrm{Y})\sim\geq\check{r}_{4}+\check{r}_{6}"$.

$\leq_{Q}q$55
Finally, to show -additivity, suppose that$\kappa$of subests of$\kappa$such that$|\vdash_{Q}$“$X=\sim\cup\cdot\{X\ell\sim :$|\vdash_{Q}$and$\gamma<\kappa$$X\sim$$\ell\sim<\gamma$$X_{\ell},$’be Q-names”. By finite additivity, we have\ell<\gamma\}$“$\mu(X)\geq\sum_{\ell<\gamma}\mu(X\ell)\sim\sim\sim\sim$”.

To show the opposite inequality, suppose that there are $p\in Q$ and
that
“
”.

$r_{1}$,$r_{2}\in \mathbb{Q}$such$p|\vdash_{Q}$$\sum_{\ell<\gamma}\mu(X\ell)<\check{r}_{1}\sim\sim<\check{r}_{2}\leq\mu(X)\sim\sim$”“. Hencefor all
By finite additivity we have
. Without loss of generality
for any
by Claim 5.3.5,
we may assume that is taken so that the additivity in Claim 5.3.2(4) holds for. But this is. Hence. Then we have$p|\vdash_{B}$$t\in[\gamma]^{<\aleph_{0}}$$\mu(\sum_{\ell\in t}X_{\ell})\sim<\check{r}_{1}$$t\in[\gamma]^{<\aleph_{0}}$$\mu_{p}(\Sigma_{\ell\in t}X\ell)\sim\leq r_{1}$$p$($‘ X\sim=\cup\cdot\{X\ell\sim :\ell<\gamma\}"$acontradiction to$\mu_{p}^{*}(X)\sim\leq r_{1}$$\mu_{p}(X)\sim\leq r_{1}$$p|\vdash_{Q}$by Claim 5.3.5.

”“$\check{r}_{2}\leq\mu(X)\sim\sim$$\square$(Theorem 5.3)

6Stick and club principles
For cardinals$\nu<\kappa$

the stick principle$\exists X\underline{\subseteq}[\kappa]^{\nu}(|X|=\kappa$$(\uparrow\nu\kappa)$

For astationary$E\subseteq\kappa$

is defined as follows:$\wedge\forall \mathrm{Y}\in[\kappa]^{\kappa}\exists x\in X(x\subseteq \mathrm{Y}))$, the club principle$\exists\langle x_{i} :$(l_{\kappa}(E))$$\uparrow\nu\kappa$

i\in E\rangle(\forall i\in E$($x_{i}$$l_{\kappa}(E)$.

is:is acofinal subset of )$i$$\forall \mathrm{Y}\in[\kappa]^{\kappa}\exists i\in E(x_{i}\subseteq \mathrm{Y}))$.

:implieswhereandimplies
Clearly.
implies. For $\nu<\nu’<\kappa,$
Maharam type
Random forcing destroys stick principle for every cardinalof the forcing. For uncountable this is because of the following result by J.
$l_{\kappa}(E)$$\phi_{\kappa}(E)$$\uparrow\kappa\nu’$$\nu\}$$l(E_{\kappa}^{\nu})$$\uparrow\kappa\nu$$E_{\kappa}^{\nu}=\{i<\kappa$$\mathrm{c}\mathrm{f}(i)=$$\uparrow\kappa\nu$$<\mathrm{t}\mathrm{h}\mathrm{e}$$\nu$Brendle.
For$\nu\leq\kappa$, let$r( \nu, \kappa)=\min\{|\mathcal{F}| :It is clear that$\uparrow\kappa\nu$

F \subseteq[\kappa]^{\nu}, \forall A\subseteq\lambda \exists b\in \mathcal{F}(b\subseteq A\vee b\subseteq\lambda \backslash A)\}$implies that$r(\nu, \kappa)\leq\kappa$.

.

Theorem 6.1 (J. Brendle, [1]) For uncountable$\nu\leq\kappa$, we habe $cov(null)\leq$56
The random forcing
forces the value of $cov(null)$ to befor every uncountable.
For
we have the following:$B_{\lambda}$$\geq\lambda$

hence$|\vdash_{B_{\lambda}}$“,$\neg\uparrow\kappa\nu,$$<\lambda$$\nu\leq\kappa$$\nu=\aleph_{0}$Lemma 6.2 (Folklore) Suppose thatare$\aleph_{0}<\kappa<\lambda$. Then$|\vdash_{B_{\lambda}}$$x_{\dot{*}}\sim’ i<\kappa$:

does not hold ”.

$B_{\lambda}$$\kappa$“thatsuch that , $i<\kappa$, $j<\kappa$ be pairwise distinct andLet$\{x\sim:$\uparrow\kappa\aleph_{0}$-names of countable subsets of . we show
is not a-set”. By the c.c.c. of , there is an
are all -names.

Proof Suppose that$|\vdash_{B_{\lambda}}$“i<\kappa\}$$\mathrm{f}_{\kappa}^{\aleph_{0}}$$S\in[\lambda]^{\kappa}$$B_{\lambda}$$B_{S}$$x\sim:$$\xi_{j}\in\lambda\backslash S$$a_{j}=[\{\langle\xi_{j}, 1\rangle\}]$,$b_{j}=[\{\langle\xi_{j}, 0\rangle\}]$for $j<\kappa$ . Let$X=\{\check{\mathrm{U}^{\cdot}}, a_{j}\rangle\sim :

Then

“

$|\vdash_{B_{\lambda}}$$X\sim\in[\kappa]^{\kappa}$

forced not to be
Claim 6.2.1$\mathrm{a}\uparrow\kappa\aleph_{0}$$|\vdash_{B_{\lambda}}$j<\kappa\}$.

”. Hence the following claim implies that $\{x\sim: : i<\kappa\}$ is

-set.

“
$x\sim:\not\subset$$X\sim$”for all i$<\kappa$.

Let be ageneric filter over . In $V[G]$ , let $B=B_{\lambda}/G$ .
induces afinitely
additive measure $m$ on . Let , $j<\kappa$ be the elements of $B$ corresponding to ,$j<\kappa$ respectively. Note that
, $j<\kappa$ are independent events inof measure
1/2. By Borel-Cantelli theorem (which holds also for finitely additive measure) we
have
$G$

$\vdash$$B_{S}$$B$$m_{\lambda}$$b_{j}’$$b_{j}$$\langle B, m\rangle$$b_{j}’$$\sum b_{j}’=1$.

$j\in x^{G}\dot{\sim}$Hence for any $b\in B^{+}$ there is $j\in x$ :such thatHence$|\vdash_{B_{\lambda}}$$b\cdot$$b_{j}’|\vdash_{B}$“$x\sim:\not\subset$$X\sim"$“$”\tilde{X}$$x\sim:\not\subset$. This shows$|\vdash_{B}$$b\cdot$“$x_{i}\sim\not\subset$$X\sim$$b_{j}’\neq 0$. But”and, since“$b_{j}’|\vdash_{B}$$G$$j\not\in X\sim"$.

was arbitrary,.

$\dashv$(Claim 6.21)$\coprod_{(\mathrm{L}\mathrm{e}\mathrm{m}\mathrm{m}\mathrm{a}62)}$

Stick or club principle at the cardinality of the partial ordering depends rather
on the ground model:
Proposition 6.3 (Folklore) Suppose that is a regular cardinal and a stationary
subset of such that
. If is a pc-c. . .-set of cardinality, then“”.
$S$$\kappa$$\kappa$$|\vdash_{P}$$\phi_{\kappa}(S)$$\phi_{\kappa}(S)$

$P$

$c$

$\mathrm{p}.0$$\leq\kappa$

57
Proof Without loss of generality, we may assume that the underlying set of $P$. I.e.
-sequence which guesses subsets of
be ais . Let, $\{i\in S : X\cap(i\cross i)=x_{i}\}$ is stationary. Each :and for every. We show
can be naturally associated with the -name$\langle x_{i} :$\kappa$i\in S\rangle$$\phi_{\kappa}(S)$$\kappa$$\cross\kappa$$x$$X\subseteq\kappa\cross\kappa$$x_{i}\subseteq i\cross i$$P$$x\sim:=\{\langle\check{i},p\rangle :\langle i,p\rangle\in x:\}$-sequence”.
is a: $i\in$
a $P$-name of asubset of . We may assume that
. By theis incompatible for all, $p\in P\}$ and that
. Let be a -name of aclub subset
x-c.c. of this set then has cardinality“.
such thatof . Again by the x-c.c. of , there is aclub set
”
“
. Let
$x\sim:=X\cap i\sim$
It is enough to show that there is an $i\in C$ such that

that
Let

“

$|\vdash_{P}$$\langle x_{i}\sim :$\phi_{\kappa}(S)$i\in S\rangle$$X\sim\subseteq\{\langle\check{i},p\rangle$$\kappa$$X\sim \mathrm{b}\mathrm{e}$$i\in\kappa$$\{p\in P : \langle\check{i},p\rangle\in X\}\sim$$\kappa$$P$$<\kappa$$P$$\kappa$$P$$C$$C\subseteq\kappa$$|\vdash_{P}$$C\subseteq C\sim"$$|\vdash_{P}$$X=\{\langle i,p\rangle :\langle\check{i},p\rangle\in X\}\sim$$D=\{i<\kappa :. Then\forall j<i\forall\gamma<\kappa(\langle j, \gamma\rangle\in Xarrow\gamma<i)\}$.

is aclub subset of . For $i\in C\cap D\cap S$ such that $x_{i}=X\cap(i\cross i)$ , we have“.
$\kappa$$\square _{(\mathrm{P}\mathrm{r}\mathrm{o}\mathrm{p}\circ 8\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}6.3)}$$|\vdash_{P}$$x_{i}=X\cap i\sim\sim"$

In contrast to the situation in the generic extension by arandom algebra discribed in Lemma 6.2, it is possible to have stick and club principles for in the.-set .
generic extension by the$\kappa$$Q$$\mathrm{p}.0$

Theorem 6.4 Assume that is strongly Mahlo cardinal and
:
”.
(a)
“
then
is stationary and(b) if$\kappa$$\nu<\kappa$$\leq\lambda$. Then for$Q=Q_{\kappa,\lambda}$$|\vdash_{Q}$$”\uparrow\nu\kappa$$V\models\phi_{\kappa}(S_{*})$$S_{*}\subseteq E_{\kappa}^{\nu}$$|\vdash Q$$l_{\kappa}(S_{*})$

in the ground model is forced to be
Proof (a): We show that).
aset as $X$ in the definition of
Suppose that $p^{*}\in Q$ and is a -name such that”.
$\mathrm{a}\uparrow\kappa\nu$$[\kappa]^{\nu}$set (i.e.

$\uparrow\kappa\nu$$\sim \mathrm{Y}$$p^{*}|\vdash_{Q}$“ $\langle\alpha_{i}\sim : i<\kappa\rangle$$Q$$p^{*}|\vdash Q$

is an increasing sequence and$”\sim \mathrm{Y}\in[\kappa]^{\kappa}$$\mathrm{Y}=\sim\{\alpha_{i}\sim :”, sayi<\kappa\}"$strongly inaccessible cardinal. By assumption {For $i<\kappa$ , let be theis stationary subset of .
For $i<\kappa$ , let , , be chosen inductively so that$i’ \mathrm{t}\mathrm{h}$$\xi_{i}$$\kappa$$\kappa\}$$q_{i}$

(1)

(2)

(3)
(4)

$a_{i}$$\gamma_{i}$$q_{i}\leq_{pr}p^{*};$$\gamma_{i}<\kappa$;$a_{i}\in B_{(u^{q}:)}\backslash null(\mu^{q}\cdot)$$q_{i}||a_{i}|\vdash_{Q}$“ $\alpha_{i}=\check{\gamma}_{i}"$

.

;

.

$\xi_{i}$: $i<$

58
By Fodor’s lemma, there are$I\subseteq\kappa$,$\alpha^{**}<\kappa$, $p^{**}\in Q$ such that(5) I is cofinal in ;(6)
for all $i\in I$ ;(7)
for all , $j\in I$ with $i<j$ ;(9)for all $i\in I$ ;(10) the order preserving mapping:$\kappa$$q:\mathrm{r}$$\xi_{i}=p^{**}$$\sup(u^{q}\cdot)<\xi_{j}$$i$$otp(u^{q}\cdot)=\alpha^{**}$

induces an isomorphism from$u^{q}\cdotarrow u^{q_{j}}$$p_{\dot{*},j}$$\langle B(uq:), \mu^{q:}\rangle$

to$\langle B_{(u^{q_{j}})}, \mu^{q_{j}}\rangle$

sending$a_{i}$

to$a_{j}$

for all , $j\in I$ .
$i$be an increasing sequence in . Then
, $n<\nu$ arecompatible to each other. Hence by Theorem 3.2, there is the free amalgamation, $n<\nu$ over
of
Let

Let

$\mu^{q^{*}}$$\langle i_{n} :$I$n<\nu\rangle$$\mu^{q:_{n}}||a:_{n}$$u^{p^{\mathrm{e}}}$.

$q^{*}= \langle\bigcup_{n<\nu}u^{q:_{n}}, \mu^{q^{*}}\rangle$

Then we haveBy (4), it follows that$q^{*}\leq_{pr}q:_{\hslash}||a_{\dot{\iota}_{n}}$$\gamma_{i_{n}}$$\mu^{q:_{n}}||a:_{n}$.

for all $n<\nu$ . In particular, $n<\nu$ are all distinct and$q^{*}|\vdash_{Q}$“$\{\check{\gamma}_{i_{n}} :n<\nu\}$$”\subseteq \mathrm{Y}\sim$$q^{*}\leq_{Q}p^{*}$by (1).

.

This shows that$|\vdash_{Q}$for$X=[\kappa]^{\nu}$“ $\forall \mathrm{Y}\subseteq[\kappa]^{\kappa}\exists x\in\check{X}(x\subseteq \mathrm{Y})$ ”(in $V$ ).

(b): The proof is dsimilar to (1). Using, , :for $i<\kappa$ as well as,
and $p^{**}\in Q$ as in the proof of (1), we can show that thesequence in the
ground model becomes a-sequence in the generic extension.
(Theorem 6.4)
$q_{\dot{\iota}}$$a:$$I\subseteq\kappa$$\gamma$$\theta_{\kappa}(S^{*})$$l_{\kappa}(S^{*})$$\square$

7Concluding remarks
We can put together the results obtained in the previous sections to get:
Theorem 7.1 If

ZFC

$+$

“there exists ameasurable cardinal”

is consistent, then so is the theory:

ZFC

$+$

$+$

“there is areal-valued measurable cardinal“”“$cov(null)=\kappa"$ .
for all$\kappa$$l(E_{\kappa^{\nu}}^{\aleph})$$\nu<\kappa$$+$$<2^{\aleph_{0}}$”59Proof Let be ameasurable cardinal. By moving to the inner model of meathe p.o.-set. Let
surability $V[U]$ , we may assume that
introduced in section 4.
By Proposition 4.8 and Lemma 4.9, preserves all cardinals. By Theorem 5.3),
forces to be areal-valued-measurable cardinal. By(here we needfor all–actually also
is forced for all
Theorem 6.4,
holds in the ground model. By Lemma 4.3, $cov(null)\geq\kappa$ . Bysuch thatand Theorem 6.1, it follows that
. By
Lemma 4.3 and $|Q|=\kappa^{+}$ ,$\kappa$$2^{\kappa}=\kappa^{+}$$Q=Q_{\kappa,\kappa}+\mathrm{b}\mathrm{e}$$Q$$2^{\kappa}=\kappa^{+}$$Q$$\kappa$$\nu<\kappa$$\mathrm{J}_{\kappa}(E_{\kappa}^{\nu})$$E\subseteq\kappa$$l_{\kappa}(E)$$\phi_{\kappa}(E)$$2^{\aleph_{0}}=\kappa^{+}$$cov(null)=\kappa$$l(E_{\kappa}^{\omega_{1}})$.

$\square$(Theorem 7.1)

References[1] J. Brendle, Cardinal invariants of the continuum and combinatorics of uncountable cardinals, preprint.
[2] D.H. Fremlin, Real-valued measurable cardinals. In: Judah, Haim (ed.) Set
Theory of the Reals. Israel Mathematical Conference Proceedings vol. 6.
Providence, American Mathematical Society 1993, 151-304.

[3] D.H. Fremlin, Measure Theory, Chapter 45, preprint.
[4] S. Fuchino and S. Shelah, Models of real-valued measurability II, in preparation.
[5] S. Fuchino, S. Shelah and L. Soukup, Sticks and clubs, Annals of Pure andApplied Logic 90, no.l (1997), 57-77.

[6] A. Kamburelis, On cardinal numbers related to the Baire property, dissertation (1989).
[7] A. Kanamori, The Higher Infinite, Springer-Verlag (1994/1997).
[8] K. Kunen, Set Theory, An Introduction to Independence Proofs, NorthHolland (1980).
[9] R. Solovay, Real-valued measurable cardinals. In: Scott (ed.) Axiomatic Set
Theory. Proceedings of Symposia in Pure Mathematics vol. 13, part 1. Providence, American Mathematical Society (1971), 397-428.
1 The first author would like to thank Jorg Brendle and D. Fremlin for valuable comments toearlier versions of the paper

60

Authors’ addresses

Saka6 Fuchino
Kitami Institute of Technology, Kitami, Japan,uchinoQmath. cs kitami-it. .jp$\mathrm{a}\mathrm{c}$Saharon Shelah

Institute of Mathematics, The Hebrew University ofJerusalem, 91904 Jerusalem, Israel.
Department of Mathematics, Rutgers University, New
Brunswick, NJ 08854, U.S.A.
$\mathrm{s}\mathrm{h}\mathrm{e}\mathrm{l}\mathrm{a}\mathrm{h}\emptyset \mathrm{m}\mathrm{a}\mathrm{t}\mathrm{h}$. huji.ac. il