From e30638e32ba923fa6ecac634bded293f93d0d21b Mon Sep 17 00:00:00 2001 From: zeramorphic <50671761+zeramorphic@users.noreply.github.com> Date: Mon, 5 Feb 2024 12:56:22 +0000 Subject: [PATCH] Lectures 13 --- iii/forcing/02_constructibility.tex | 122 +++++++++++++++++++++++++-- iii/lc/02_measurable_cardinals.tex | 126 +++++++++++++++++++++++++++- 2 files changed, 241 insertions(+), 7 deletions(-) diff --git a/iii/forcing/02_constructibility.tex b/iii/forcing/02_constructibility.tex index f116d17..c8743fa 100644 --- a/iii/forcing/02_constructibility.tex +++ b/iii/forcing/02_constructibility.tex @@ -115,8 +115,8 @@ \subsection{G\"odel functions} Hence, these formulas are absolute. \end{proposition} \begin{lemma}[G\"odel normal form] - For every \( \Delta_0 \) formula \( \varphi(x_1, \dots, x_n) \) with free variables contained in \( x_1, \dots, x_n \), there is a term \( \mathcal F_\varphi \) built from the symbols \( \mathcal F_1, \dots, \mathcal F_{10} \) such that - \[ \mathsf{ZF} \vdash \forall a_1, \dots, a_n.\, \mathcal F_\varphi(a_1, \dots, a_n) = \qty{\langle x_n, \dots, x_1 \rangle \mid a_n \times \dots \times a_1 \mid \varphi(x_1, \dots, x_n)} \] + For every \( \Delta_0 \) formula \( \varphi(x_1, \dots, x_n) \) with free variables contained in \( \qty{x_1, \dots, x_n} \), there is a term \( \mathcal F_\varphi \) built from the symbols \( \mathcal F_1, \dots, \mathcal F_{10} \) such that + \[ \mathsf{ZF} \vdash \forall a_1, \dots, a_n.\, \mathcal F_\varphi(a_1, \dots, a_n) = \qty{\langle x_n, \dots, x_1 \rangle \in a_n \times \dots \times a_1 \mid \varphi(x_1, \dots, x_n)} \] \end{lemma} \begin{remark} \begin{enumerate} @@ -127,17 +127,97 @@ \subsection{G\"odel functions} However, it cannot be formed using \( \langle x_1, x_2 \rangle \) and \( x_3 \). \end{enumerate} \end{remark} -% Proof goes here +\begin{proof} + We show this by induction on the class \( \Delta_0 \). + We call a formula \( \varphi \) a \emph{termed formula} if the conclusion of the lemma holds for \( \varphi \); we aim to show that every \( \Delta_0 \)-formula is a termed tormula. + We will only use the logical symbols \( \wedge, \vee, \neg, \exists \), and the only occurrence of existential quantification will be in formulas of the form + \[ \varphi(x_1, \dots, x_n) \equiv \exists x_{n+1} \in x_j.\, \psi(x_1, \dots, x_{n+1}) \] + where \( j \leq m \leq n \). + For example, we allow \( \exists x_3 \in x_1.\, (x_1 \in x_2 \wedge x_3 = x_1) \), but we disallow \( \exists x_1 \in x_2.\, \psi \) and \( \exists x_3 \in x_1.\, (x_3 \in x_2 \wedge \exists x_4 \in x_1.\, \psi) \). + Every \( \Delta_0 \)-formula is equivalent to one of this form. + We allow for dummy variables, so \( \varphi(x_1, x_2) \equiv x_1 \in x_2 \) and \( \varphi(x_1, x_2, x_3) \equiv x_1 \in x_2 \) are distinct. + This proof will take place in four parts: first some logical points, then we consider propositional formulas, then atomic formulas, and finally bounded existentials. + + \emph{Part (i): logical points.} + We make the following remarks. + \begin{itemize} + \item If \( \mathsf{ZF} \vdash \varphi(\vb x) \leftrightarrow \psi(\vb x) \) and \( \varphi(\vb x) \) is a termed formula, then \( \psi \) is also a termed formula. + This is immediate from the definition, since we can let \( \mathcal F_\psi = \mathcal F_\varphi \). + \item For all \( m, n \), if \( \varphi(x_1, \dots, x_n) \equiv \psi(x_1, \dots, x_m) \) and \( \psi \) is a termed formula, then so is \( \varphi \). + If \( n \geq m \), we can show this by induction on \( n \). + The base case \( n = m \) is trivial. + For the inductive step, suppose + \[ \varphi(x_1, \dots, x_{n+1}) \equiv \psi(x_1, \dots, x_m) \] + Then, we can write + \[ \varphi(x_1, \dots, x_{n+1}) \equiv \theta(x_1, \dots, x_n) \] + where \( \theta \) is a termed formula. + Then + \[ \mathcal F_\varphi(a_1, \dots, a_n, a_{n+1}) = a_{n+1} \times \mathcal F_\theta(a_1, \dots, a_n) = \mathcal F_4(a_{n+1}, \mathcal F_\theta(a_1, \dots, a_n)) \] + giving the result by the inductive hypothesis. + This is the reason for reversing the order: because the ordered triple \( \langle x, y, z \rangle \) is \( \langle x, \langle y, z \rangle \rangle \), the map + \[ \qty{\langle x_1, x_2 \rangle \in a_1 \times a_2 \mid \theta(x_1, x_2)} \mapsto \qty{\langle x_1, x_2, x_3 \rangle \in a_1 \times a_2 \times a_3 \mid \theta(x_1, x_2)} \] + is much more complicated to implement in G\"odel functions. + We prove the case \( n \leq m \) by induction; if + \[ \varphi(x_1, \dots, x_{n-1}) \equiv \psi(x_1, \dots, x_m) \] + then + \[ \varphi(x_1, \dots, x_{n-1}) \equiv \theta(x_1, \dots, x_n) \] + and + \[ \qty{0} = \qty{\mathcal F_3(a_1, a_1)} = \mathcal F_1(\mathcal F_3(a_1, a_1), \mathcal F_3(a_1, a_1)) \] + Then + \begin{align*} + \mathcal F_\varphi(a_1, \dots, a_{n-1}) &= \qty{\langle x_{n-1}, \dots, x_1 \rangle \in a_{n-1} \times \dots \times a_1 \mid \varphi(x_1, \dots, x_{n-1})} \\ + &= \operatorname{ran}(\qty{\langle 0, x_{n-1}, \dots, x_1 \rangle \in \qty{0} \times a_{n-1} \times \dots \times a_1 \mid \theta(x_1, \dots, x_{n-1}, 0)}) \\ + &= \mathcal F_6(\mathcal F_\theta(a_1, \dots, a_{n-1}), \mathcal F_1(\mathcal F_3(a_1, a_1), \mathcal F_3(a_1, a_1)), a_1) + \end{align*} + \item If \( \psi(x_1, \dots, x_n) \) is a termed formula and + \[ \varphi(x_1, \dots, x_{n+1}) = \psi(x_1, \dots, x_{n-1}, x_{n+1}/x_n) \] + then \( \varphi \) is a termed formula. + First, if \( n = 1 \), we have a termed formula \( \psi(x_1) \) and consider \( \psi(x_2/x_1) \). + Then + \begin{align*} + \mathcal F_\varphi(a_1, a_2) &= \qty{\langle x_2, x_1 \rangle \in a_2 \times a_1 \mid \psi(x_2)} \\ + &= \qty{\langle x_2, x_1 \rangle \mid x_1 \in a_1 \wedge x_2 \in \mathcal F_\psi(a_2)} \\ + &= \mathcal F_\psi(a_2) \times a_1 \\ + &= \mathcal F_4(\mathcal F_\psi(a_2), a_1) + \end{align*} + If \( n > 1 \), we have + \begin{align*} + \mathcal F_\varphi(a_1, \dots, a_{n+1}) &= \qty{\langle x_{n+1}, \dots, x_1 \rangle \mid x_n \in a_n \wedge \langle x_{n+1}, x_{n-1}, \dots, x_1 \rangle \in \mathcal F_\psi(a_1, \dots, a_{n-1}, a_{n+1})} \\ + &= \mathcal F_8(\mathcal F_\psi(a_1, \dots, a_{n-1}, a_{n+1}), a_n) + \end{align*} + \item If \( \psi(x_1, x_2) \) is a termed formula, and + \[ \varphi(x_1, \dots, x_n) \equiv \psi(x_{n-1}/x_1, x_n/x_2) \] + then \( \varphi \) is a termed formula. + This is trivial if \( n = 2 \), so we assume \( n > 2 \). + Then + \begin{align*} + \mathcal F_\varphi(a_1, \dots, a_n) &= \qty{\langle x_n, \dots, x_1 \rangle \in a_n \times \dots \times a_1 \mid \langle x_n, x_{n-1} \rangle \in \mathcal F_\psi(a_{n-1}, a_n)} \\ + &= \mathcal F_7(\mathcal F_\psi(a_{n-1}, a_n), a_{n-2} \times \dots \times a_1) + \end{align*} + \end{itemize} + + \emph{Part (ii): propositional connectives} + \begin{itemize} + \item If \( \varphi \) is a termed formula, then so is \( \neg\varphi \). + \[ \mathcal F_{\neg\varphi}(a_1, \dots, a_n) &= (a_n \times \dots \times a_1) \setminus \mathcal F_\varphi(a_1, \dots, a_n) \] + \item If \( \varphi, \psi \) are termed formulas, then so is \( \varphi \vee \psi \). + \[ \mathcal F_{\varphi \vee \psi}(a_1, \dots, a_n) = \mathcal F_\varphi(a_1, \dots, a_n) \cup \mathcal F_\psi(a_1, \dots, a_n) \] + It is easy to see that unions can be formed using G\"odel functions. + \item Conjunctions are similar to disjunctions. + \[ \mathcal F_{\varphi \wedge \psi}(a_1, \dots, a_n) = \mathcal F_\varphi(a_1, \dots, a_n) \cap \mathcal F_\psi(a_1, \dots, a_n) \] + \end{itemize} +\end{proof} \begin{definition} A class \( C \) is \emph{closed under G\"odel functions} if whenever \( x, y \in C \), we have \( \mathcal F_i(x, y) \in C \) for \( i \in \qty{1, \dots, 10} \). - Given a set \( b \), we let \( \mathrm{cl}(b) \) be the smallest set \( C \) that is closed under G\"odel functions such that \( b \cup \qty{b} \subseteq C \). + Given a set \( b \), we let \( \mathrm{cl}(b) \) be the smallest set \( C \) containing \( b \) as a subset that is closed under G\"odel functions. \end{definition} +For example, \( \mathrm{cl}(\varnothing) = \varnothing \), \( \qty{a, b} \in \mathrm{cl}(\qty{a, b}) \), and \( \mathrm{cl}(b) = \mathrm{cl}(\mathrm{cl}(b)) \). \begin{definition} Let \( b \) be a set. Define \( \mathcal D^n(b) \) inductively by - \[ \mathcal D^0(b) = b\cup\qty{b};\quad \mathcal D^{n+1}(b) = \qty{\mathcal F_i(x, y) \mid x, y \in \mathcal D^n(b), i \in \qty{1, \dots, 10}} \] + \[ \mathcal D^0(b) = b;\quad \mathcal D^{n+1}(b) = \qty{\mathcal F_i(x, y) \mid x, y \in \mathcal D^n(b), i \in \qty{1, \dots, 10}} \] \end{definition} -One can observe that \( \mathrm{cl}(b) = \bigcup_{n \in \omega} \mathcal D^n(b) \). +One can easily check that \( \mathrm{cl}(b) = \bigcup_{n \in \omega} \mathcal D^n(b) \). \begin{lemma} If \( M \) is a transitive class that is closed under G\"odel functions, then \( M \) satisfies \( \Delta_0 \)-separation. \end{lemma} @@ -157,3 +237,33 @@ \subsection{G\"odel functions} For every transitive set \( M \), the collection of definable subsets is \[ \operatorname{Def}(M) = \mathrm{cl}(M \cup \qty{M}) \cap \mathcal P(M) \] \end{theorem} +\begin{proof} + We first prove the forward direction. + Let \( \varphi \) be a formula. + Then \( \varphi^M \) is \( \Delta_0 \), so there is a term \( \mathcal G \) built from the G\"odel functions \( \mathcal F_1, \dots, \mathcal F_{10} \) such that for \( a_1, \dots, a_n \in M \), we have + \[ \qty{x \in M \mid (M, \in) \vDash \varphi(x, a_1, \dots, a_n)} = \qty{x \in M \mid \varphi^M(x, a_1, \dots, a_n)} = \mathcal G(M, a_1, \dots, a_n) \in \mathrm{cl}(M \cup \qty{M}) \] + We now show the converse. + We first claim that if \( \mathcal G \) is built from the G\"odel functions, then for any \( x, a_1, \dots, a_n \), the formulas + \[ x = \mathcal G(a_1, \dots, a_n);\quad x \in \mathcal G(a_1, \dots, a_n) \] + are \( \Delta_0 \). + This can be proven inductively using the iterative construction of \( \mathrm{cl}(M \cup \qty{M}) \). + For example, if \( X, Y \in \mathcal D^k(a_1, \dots, a_n) \), then \( x = \mathcal F_1(X, Y) \) is equivalent to the statement + \[ (\forall z \in x.\, z = X \vee z = Y) \wedge (\exists w \in x.\, w = X) \wedge (\exists w \in x.\, w = Y) \] + so the result holds for \( \mathcal F_1 \); very similar proofs show the result for both equality and membership for all other G\"odel functions. + + Let \( Z \in \mathrm{cl}(M \cup \qty{M}) \cap \mathcal P(M) \). + Since \( Z \in \mathrm{cl}(M \cup \qty{M}) \), we can fix a term \( \mathcal G \) built from the \( \mathcal F_1, \dots, \mathcal F_{10} \) such that \( Z = \mathcal G(M, a_1, \dots, a_n) \). + Let \( \varphi \) be a \( \Delta_0 \) formula such that \( x \in \mathcal G(M, a_1, \dots, a_n) \) if and only if \( \varphi(x, M, a_1, \dots, a_n) \). + Then \( \mathcal G(M, a_1, \dots, a_n) = \qty{x \in M \mid \varphi(x, M, a_1, \dots, a_n)} \) as \( Z \subseteq M \). + It therefore remains to prove that there is a formula \( \psi \) such that + \[ \psi^M(x, a_1, \dots, a_n) \leftrightarrow \varphi(x, M, a_1, \dots, a_n) \] + For example, we can define \( \psi \) from \( \varphi \) by the following replacements. + \begin{enumerate} + \item \( \exists v_i \in M \mapsto \exists v_i \); + \item \( v_i \in M \mapsto v_i = v_i \); + \item \( M = M \mapsto v_0 = v_0 \); + \item \( M \in M, M \in v_i, M = v_i \mapsto v_0 \neq v_0 \). + \end{enumerate} + Finally, we obtain + \[ Z = \mathcal G(M, a_1, \dots, a_n) = \qty{x \in M \mid \psi^M(x, a_1, \dots, a_n)} \in \operatorname{Def}(M) \] +\end{proof} diff --git a/iii/lc/02_measurable_cardinals.tex b/iii/lc/02_measurable_cardinals.tex index 4a16a4e..f7dfc54 100644 --- a/iii/lc/02_measurable_cardinals.tex +++ b/iii/lc/02_measurable_cardinals.tex @@ -6,8 +6,132 @@ \subsection{The measure problem} \item (translation invariance) if \( X \subseteq \mathbb I \), \( r \in \mathbb R \), and \( X + r = \qty{x + r \mid x \in X} \subseteq \mathbb I \), then \( \mu(X) = \mu(X + r) \); and \item (countable additivity) if \( (A_n)_{n \in \mathbb N} \) is a family of pairwise disjoint subsets of \( \mathbb I \), then \( \mu\qty(\bigcup_{n \in \mathbb N} A_n) = \sum_{n \in \mathbb N}\mu(A_n) \). \end{enumerate} -The \emph{measure problem} was the question as to whether such a measure function exists. +The \emph{Lebesgue measure problem} was the question of whether such a measure function exists. Vitali proved that a measure cannot be defined on all of \( \mathcal P(\mathbb I) \). This proof requires the axiom of choice nontrivially. In 1970, Solovay proved that if \( \mathsf{ZFC} + \mathsf{IC} \) is consistent, then, there is a model of \( \mathsf{ZF} \) in which all sets are Lebesgue measurable. In 1984, Shelah showed that the inaccessible cardinal was necessary to construct this model. + +Now, replace translation invariance with the requirement that for all \( x \in \mathbb I \), we have \( \mu(\qty{x}) = 0 \), and call such measures \emph{Banach measures}. +\emph{Banach's measure problem} was the question of whether a Banach measure exists. +Note that this property is true for any Lebesgue measure. +If \( \mu(\qty{x}) > 0 \) for some \( x \), then by translation invariance, every singleton has the same measure \( \mu(\qty{x}) > 0 \). +There is some natural number \( n \) such that \( n \mu(\qty{x}) > 1 \), but this contradicts countable additivity using a set of \( n \) reals. +Observe that for any \( \varepsilon > 0 \), there can be only finitely many pairwise disjoint sets with measure at least \( \varepsilon \). + +Banach and Kuratowski proved in 1929 that the continuum hypothesis implies that there are no Banach measures on \( \mathbb I \). +We can define Banach measures on any set \( S \) by also replacing property (i) with the requirement that \( \mu(S) = 1 \) and \( \mu(\varnothing) = 0 \). +Note that if \( \abs{S} = \abs{S'} \), then there is a Banach measure on \( S \) if and only if there is one on \( S' \). +Thus, having a Banach measure is a property of cardinals. + +For larger cardinals, it may not be natural to just consider countable additivity. +\begin{definition} + A Banach measure \( \mu \) is called \emph{\( \lambda \)-additive} if for all \( \gamma < \lambda \) and pairwise disjoint families \( \qty{A_\alpha \mid \alpha < \gamma} \), then + \[ \mu\qty(\bigcup A_\alpha) = \sup\qty{\sum_{\alpha \in F} \mu(A_\alpha) \mid F \subseteq \gamma \text{ finite}} \] +\end{definition} +\begin{theorem} + If \( \kappa \) is the smallest cardinal that has a Banach measure, then that measure is \( \kappa \)-additive. +\end{theorem} + +\subsection{Real-valued measurable cardinals} +\begin{definition} + A cardinal \( \kappa \) is \emph{real-valued measurable}, written \( \mathsf{RVM}(\kappa) \), if there is a \( \kappa \)-additive Banach measure on \( \kappa \). +\end{definition} +\begin{proposition} + Every real-valued measurable cardinal is regular. +\end{proposition} +\begin{proof} + Suppose that \( \kappa \) is a real-valued measurable cardinal, and that \( C \subseteq \kappa \) is cofinal with \( \abs{C} = \lambda < \kappa \). + We can write + \[ C = \qty{\gamma_\alpha \mid \alpha < \gamma} \] + where \( \gamma_\alpha \) is increasing in \( \alpha \). + Consider + \[ C_\alpha = \qty{\xi \mid \gamma_\alpha \leq \xi < \gamma_{\alpha + 1}} \] + Then \( \bigcup_{\alpha < \gamma} C_\alpha = \kappa \) as \( C \) is cofinal, and the \( C_\alpha \) are disjoint. + Note that \( \abs{C_\alpha} \leq \abs{\gamma_{\alpha + 1}} < \kappa \). + Writing \( C_\alpha = \bigcup_{x \in C_\alpha} \qty{x} \), we observe by \( \kappa \)-additivity that \( \mu(C_\alpha) = 0 \). + But again by \( \kappa \)-additivity, \( \mu(\kappa) = 0 \), contradicting property (i). +\end{proof} +\begin{proposition}[the pigeonhole principle] + Let \( \kappa \) be regular, \( \lambda < \kappa \), and \( f : \kappa \to \lambda \). + Then there is some \( \alpha \in \lambda \) such that \( \abs{f^{-1}(\alpha)} = \kappa \). +\end{proposition} +\begin{proof} + We have + \[ \kappa = \bigcup_{\alpha \in \lambda} f^{-1}(\alpha) \] + giving the result immediately by regularity of \( \kappa \). +\end{proof} +\begin{proposition} + All successor cardinals are regular. +\end{proposition} +\begin{proposition} + If \( \mu \) is a Banach measure on \( S \), and \( C \) is a family of pairwise disjoint sets of positive \( \mu \)-measure, then \( C \) is countable. +\end{proposition} +\begin{proof} + Consider the collection + \[ C_n = \qty{A \in C \mid \mu(A) > \frac{1}{n}} \] + Observe that each \( C_n \) is finite, so \( C = \bigcup_{n \in \mathbb N} C_n \) must be countable. +\end{proof} +\begin{lemma}[Ulam] + For any cardinal \( \lambda \), there is an \emph{Ulam matrix} \( A_\alpha^\xi \) indexed by \( \alpha < \lambda^+ \) and \( \xi < \lambda \) such that + \begin{enumerate} + \item for a given \( \xi \), the set \( \qty{A_\alpha^\xi \mid \alpha < \lambda^+} \) is a pairwise disjoint family; and + \item for a given \( \alpha \), then + \[ \abs{\lambda^+ \setminus \bigcup_{\xi < \lambda} A_\alpha^\xi} \leq \lambda \] + \end{enumerate} +\end{lemma} +\begin{proof} + For each \( \gamma < \lambda^+ \), fix a surjection \( f_\gamma : \lambda \to \gamma + 1 \). + Define + \[ A_\alpha^\xi = \qty{\gamma \mid f_\gamma(\xi) = \alpha} \] + It is clear that property (i) holds. + For property (ii), suppose + \[ \gamma \in \lambda^+ \setminus \bigcup_{\xi < \lambda} A_\alpha^\xi \] + Then \( \gamma < \lambda^+ \) and for all \( \xi \), we have \( f_\gamma(\xi) \neq \alpha \). + Hence + \[ \lambda^+ \setminus \bigcup_{\xi < \lambda} A_\alpha^\xi \subseteq \alpha \] + so the size of this set is at most \( \lambda \) +\end{proof} +\begin{theorem} + If \( \kappa \) is real-valued measurable, then \( \kappa \) is weakly inaccessible. +\end{theorem} +\begin{remark} + If there is a Banach measure on \( [0,1] \), then in particular \( 2^{\aleph_0} \) is weakly inaccessible. +\end{remark} +\begin{proof} + We have already shown regularity. + Suppose \( \kappa \) is not a limit cardinal, so \( \kappa = \lambda^+ \). + Let \( (A_\alpha^\xi)_{\alpha < \lambda^+; \xi < \lambda} \) be an Ulam matrix for \( \lambda \). + By (ii), + \[ \abs{Z_\alpha} \leq \lambda;\quad Z_\alpha = \lambda^+ \setminus \bigcup_{\xi < \lambda} A_\alpha^\xi \] + so by \( \kappa \)-additivity, \( \mu(Z) = 0 \). + Hence + \[ \mu\qty(\bigcup_{\xi < \lambda} A_\alpha^\xi) = 1 \] + This is a small union of sets of measure 1, so again by \( \kappa \)-additivity there is some \( \xi_\alpha \) such that \( \mu(A_\alpha^{\xi_\alpha}) > 0 \). + Let \( f : \lambda^+ \to \lambda \) be the map \( \alpha \mapsto \xi_\alpha \). + By the pigeonhole principle, there is some \( \xi \) and a set \( A \subseteq \lambda^+ \) with \( \abs{A} = \lambda^+ \) such that for all \( \alpha \in A \), we have \( \xi_\alpha = \xi \). + By property (i), the collection \( \qty{A_\alpha^\xi \mid \alpha \in A} \) is a collection of uncountable size \( \lambda^+ \) of pairwise disjoint sets, all of which have positive measure, but we have already shown that such a collection must be countable. +\end{proof} + +\subsection{Measurable cardinals} +\begin{definition} + A Banach measure \( \mu \) is called \emph{two-valued} if \( \mu \) takes values in \( \qty{0,1} \). +\end{definition} +This removes any mention of the real numbers from the definition of a Banach measure. +\begin{remark} + Two-valued measures correspond directly to ultrafilters. + Recall that \( F \) is a \emph{filter} on \( S \) if + \begin{enumerate} + \item \( \varnothing \notin F, S \in F \); + \item if \( A \subseteq B \) then \( A \in F \to B \in F \); + \item if \( A, B \in F \) then \( A \cap B \in F \). + \end{enumerate} + We say that \( F \) is an \emph{ultrafilter} if \( A \in F \) or \( S \setminus A \in F \) for all \( A \subseteq S \). + \( F \) is \emph{nonprincipal} if for all \( x \in S \), the singleton \( \qty{x} \) is not in \( F \). + An ultrafilter is \( \lambda \)-complete if for all \( \gamma < \lambda \) and all families \( \qty{A_\alpha \mid \alpha < \gamma} \subseteq F \), we have \( \bigcap_{\alpha < \gamma} A_\alpha \in F \). + In this way, the collection of sets of a two-valued Banach measure \( \mu \) that are assigned measure \( 1 \) form a nonprincipal ultrafilter. + This filter is \( \lambda \)-complete if and only if \( \mu \) is \( \lambda \)-additive. +\end{remark} +\begin{definition} + A cardinal \( \kappa \) is \emph{measurable} if there is a \( \kappa \)-complete nonprincipal ultrafilter on \( \kappa \). +\end{definition}