[proofplan] We begin with a nonprincipal $\kappa$-complete ultrafilter $U$ on $\kappa$ and form the ultrapower preorder on functions $\kappa \to \kappa$. Countable completeness, which follows from $\kappa$-completeness because $\kappa$ is uncountable, makes the strict ultrapower order well-founded, so we may choose a minimal class lying above every constant and below the identity class. Pushing $U$ forward along a representative of this minimal class gives a new ultrafilter $W$. The minimality of the chosen class forces every regressive function on a $W$-large set to be constant on a $W$-large subset, which is precisely normality. [/proofplan]

[step:Record the consequences of nonprincipality and $\kappa$-completeness] Let $U$ be a nonprincipal $\kappa$-complete ultrafilter on $\kappa$. We view $\kappa$ as the initial ordinal of cardinality $\kappa$, so every ordinal $\alpha < \kappa$ has cardinality less than $\kappa$. First, every subset of $\kappa$ of cardinality less than $\kappa$ is $U$-null. Indeed, if $A \subset \kappa$ and $|A| < \kappa$, then for each $\xi \in A$ the singleton $\{\xi\}$ is not in $U$, because $U$ is nonprincipal. Since $U$ is an ultrafilter, $\kappa \setminus \{\xi\} \in U$ for each $\xi \in A$. By $\kappa$-completeness, \begin{align*} \kappa \setminus A = \bigcap_{\xi \in A}(\kappa \setminus \{\xi\}) \in U. \end{align*} Thus $A \notin U$. In particular, for each $\alpha < \kappa$, the cofinal tail \begin{align*} T_\alpha := \{\xi < \kappa : \alpha < \xi\} \end{align*} belongs to $U$, because its complement $\alpha + 1$ has cardinality less than $\kappa$. [/step]

[step:Make the ultrapower order well-founded]For functions $a,b:\kappa \to \kappa$, write $a =_U b$ when \begin{align*} \{\xi < \kappa : a(\xi)=b(\xi)\} \in U. \end{align*} Let $[a]_U$ denote the equivalence class of $a$ modulo $=_U$. Define a strict order on equivalence classes by \begin{align*} [a]_U < [b]_U \quad \text{if and only if} \quad \{\xi < \kappa : a(\xi)<b(\xi)\} \in U. \end{align*} This is well-defined because $U$ is an ultrafilter. We also write $[a]_U \le [b]_U$ to mean that either $[a]_U < [b]_U$ or $[a]_U = [b]_U$; equivalently, $\{\xi < \kappa : a(\xi) \le b(\xi)\} \in U$. The collection of equivalence classes is the quotient of the set $\kappa^\kappa$ by $=_U$, so it is a set. For a strict relation on a set, ill-foundedness is equivalent to the existence of an $\omega$-indexed descending chain, where $\omega$ is the first infinite ordinal. We prove that no such chain exists. Suppose, toward a contradiction, that there are functions \begin{align*} g_n:\kappa \to \kappa \quad \text{for each } n < \omega \end{align*} such that \begin{align*} [g_0]_U > [g_1]_U > [g_2]_U > \cdots. \end{align*} For each $n < \omega$, define \begin{align*} A_n := \{\xi < \kappa : g_{n+1}(\xi)<g_n(\xi)\}. \end{align*} Then $A_n \in U$ for every $n < \omega$. Since $\omega$ has cardinality less than $\kappa$ and $U$ is $\kappa$-complete, \begin{align*} A := \bigcap_{n < \omega} A_n \in U. \end{align*} In particular, $A$ is nonempty. Choose $\xi \in A$. Then \begin{align*} g_0(\xi) > g_1(\xi) > g_2(\xi) > \cdots \end{align*} is an infinite strictly descending sequence of ordinals, impossible. Therefore the ultrapower order is well-founded.[/step]

[guided]The point of this step is to justify the later phrase “choose a minimal equivalence class.” We need well-foundedness of the ultrapower order, and the only set-theoretic input needed is countable completeness. For functions $a,b:\kappa \to \kappa$, define $a =_U b$ to mean that $a$ and $b$ agree on a $U$-large set: \begin{align*} \{\xi < \kappa : a(\xi)=b(\xi)\} \in U. \end{align*} The equivalence class of $a$ is denoted $[a]_U$. The strict ultrapower order is defined by \begin{align*} [a]_U < [b]_U \quad \text{if and only if} \quad \{\xi < \kappa : a(\xi)<b(\xi)\} \in U. \end{align*} This definition does not depend on the chosen representatives: changing $a$ or $b$ on a $U$-null set does not change whether the comparison holds on a $U$-large set. We also write $[a]_U \le [b]_U$ to mean that either $[a]_U < [b]_U$ or $[a]_U = [b]_U$; equivalently, $\{\xi < \kappa : a(\xi) \le b(\xi)\} \in U$. We are working with a set-sized relation: the ultrapower classes form the quotient of the set $\kappa^\kappa$ by the [equivalence relation](/page/Equivalence%20Relation) $=_U$. For such a strict relation, non-well-foundedness is equivalent to the existence of a descending chain indexed by $\omega$, where $\omega$ denotes the first infinite ordinal. Thus it is enough to rule out an $\omega$-sequence of strict decreases. Assume for contradiction that the order is not well-founded. Then there is an infinite descending sequence represented by functions \begin{align*} g_n:\kappa \to \kappa \quad \text{for } n < \omega \end{align*} with \begin{align*} [g_0]_U > [g_1]_U > [g_2]_U > \cdots. \end{align*} For each $n < \omega$, the comparison $[g_{n+1}]_U < [g_n]_U$ means that the set \begin{align*} A_n := \{\xi < \kappa : g_{n+1}(\xi)<g_n(\xi)\} \end{align*} belongs to $U$. Since $\kappa$ is uncountable, $\omega$ has cardinality less than $\kappa$. Therefore $\kappa$-completeness applies to the countable family $(A_n)_{n < \omega}$ and gives \begin{align*} A := \bigcap_{n < \omega} A_n \in U. \end{align*} Because an ultrafilter never contains the empty set, $A$ is nonempty. Choose $\xi \in A$. By the definition of $A$, the ordinal values satisfy \begin{align*} g_0(\xi) > g_1(\xi) > g_2(\xi) > \cdots. \end{align*} This is an infinite strictly descending sequence of ordinals, contradicting the well-ordering of the ordinals. Hence the ultrapower order is well-founded.[/guided]

[step:Choose a minimal class above every constant and below the identity] For each $\alpha < \kappa$, define the constant map \begin{align*} c_\alpha:\kappa \to \kappa,\quad \xi \mapsto \alpha. \end{align*} Define the identity map \begin{align*} \operatorname{id}_\kappa:\kappa \to \kappa,\quad \xi \mapsto \xi. \end{align*} Let $\mathcal{C}$ be the collection of ultrapower classes \begin{align*} \mathcal{C}:=\{[f]_U : [c_\alpha]_U < [f]_U \text{ for every } \alpha<\kappa \text{ and } [f]_U \le [\operatorname{id}_\kappa]_U\}. \end{align*} The collection $\mathcal{C}$ is nonempty, because $[\operatorname{id}_\kappa]_U \in \mathcal{C}$. Indeed, for each $\alpha < \kappa$, \begin{align*} \{\xi < \kappa : c_\alpha(\xi)<\operatorname{id}_\kappa(\xi)\} = \{\xi < \kappa : \alpha < \xi\}=T_\alpha \in U. \end{align*} Since the ultrapower order is well-founded, choose a function \begin{align*} f:\kappa \to \kappa \end{align*} such that $[f]_U$ is a minimal element of $\mathcal{C}$. [/step]

[step:Push the ultrafilter forward along the minimal function] Define $W$ on subsets $X \subset \kappa$ by \begin{align*} X \in W \quad \text{if and only if} \quad f^{-1}(X) \in U. \end{align*} This is the pushforward of $U$ along $f$. Since preimages preserve complements, finite intersections, and arbitrary intersections, and since $U$ is an ultrafilter, $W$ is an ultrafilter on $\kappa$. It is $\kappa$-complete: if $\lambda < \kappa$ and $(X_i)_{i<\lambda}$ is a family of sets in $W$, then $f^{-1}(X_i) \in U$ for all $i<\lambda$, so \begin{align*} f^{-1}\left(\bigcap_{i<\lambda} X_i\right)=\bigcap_{i<\lambda} f^{-1}(X_i) \in U. \end{align*} Hence $\bigcap_{i<\lambda} X_i \in W$. It remains to check nonprincipality. Fix $\alpha < \kappa$. Since $[c_\alpha]_U < [f]_U$, \begin{align*} \{\xi < \kappa : \alpha < f(\xi)\} \in U. \end{align*} This set is contained in $\kappa \setminus f^{-1}(\{\alpha\})$, so $f^{-1}(\{\alpha\}) \notin U$. Thus $\{\alpha\} \notin W$. Since no singleton belongs to $W$, the ultrafilter $W$ is nonprincipal. [/step]

[step:Convert a regressive function on a $W$-large set into a smaller ultrapower class] Let $S \in W$, and let \begin{align*} h:S \to \kappa \end{align*} be regressive, meaning $h(\eta)<\eta$ for every nonzero $\eta \in S$. Since $W$ is nonprincipal, $\{0\} \notin W$, so $S \setminus \{0\} \in W$. Replacing $S$ by $S \setminus \{0\}$ if necessary, we may assume $0 \notin S$. Define the total map $\widehat{h}:\kappa \to \kappa$ by setting $\widehat{h}(\eta)=h(\eta)$ for $\eta \in S$ and $\widehat{h}(\eta)=0$ for $\eta \notin S$. Define \begin{align*} g:\kappa \to \kappa,\quad \xi \mapsto \widehat{h}(f(\xi)). \end{align*} Because $S \in W$, we have $f^{-1}(S) \in U$. For every $\xi \in f^{-1}(S)$, the ordinal $f(\xi)$ lies in $S$ and is nonzero, so regressiveness gives \begin{align*} g(\xi)=h(f(\xi))<f(\xi). \end{align*} Therefore \begin{align*} \{\xi < \kappa : g(\xi)<f(\xi)\} \in U, \end{align*} and hence $[g]_U < [f]_U$. [/step]

[step:Use minimality to force a large fibre]We prove that $h$ has a $W$-large fibre. Suppose not. Then for every $\beta < \kappa$, \begin{align*} h^{-1}(\{\beta\}) \notin W. \end{align*} Since $W$ is an ultrafilter, $S \setminus h^{-1}(\{\beta\}) \in W$ for every $\beta < \kappa$. Fix $\alpha < \kappa$. Define \begin{align*} B_\alpha := S \cap \{\eta < \kappa : \alpha < \eta\} \cap \bigcap_{\beta \le \alpha}\left(S \setminus h^{-1}(\{\beta\})\right). \end{align*} The set $S$ belongs to $W$. The tail $\{\eta < \kappa : \alpha < \eta\}$ belongs to $W$, because its complement has cardinality less than $\kappa$ and $W$ is nonprincipal and $\kappa$-complete. For each $\beta \le \alpha$, the set $S \setminus h^{-1}(\{\beta\})$ belongs to $W$. Since $\alpha+1$ has cardinality less than $\kappa$ and $W$ is $\kappa$-complete, $B_\alpha \in W$. For every $\eta \in B_\alpha$, we have $\eta \in S$, $\alpha<\eta$, and $h(\eta) \ne \beta$ for all $\beta \le \alpha$. Since $h(\eta)$ is an ordinal, this implies $\alpha < h(\eta)$. Therefore \begin{align*} B_\alpha \subset \{\eta < \kappa : \alpha < \widehat{h}(\eta)\}. \end{align*} Thus $\{\eta < \kappa : \alpha < \widehat{h}(\eta)\} \in W$. By the definition of $W$, \begin{align*} \{\xi < \kappa : \alpha < g(\xi)\} = f^{-1}(\{\eta < \kappa : \alpha < \widehat{h}(\eta)\}) \in U. \end{align*} Hence $[c_\alpha]_U < [g]_U$ for every $\alpha < \kappa$. Also $[g]_U \le [\operatorname{id}_\kappa]_U$. Indeed, the set $\{\xi < \kappa : g(\xi)<f(\xi)\}$ belongs to $U$, and the set $\{\xi < \kappa : f(\xi)\le \xi\}$ belongs to $U$ because $[f]_U \le [\operatorname{id}_\kappa]_U$. Their intersection belongs to $U$, and on that intersection $g(\xi)<f(\xi)\le \xi$, so $\{\xi < \kappa : g(\xi)\le \xi\}\in U$. Therefore $[g]_U \in \mathcal{C}$. This contradicts the minimality of $[f]_U$, since $[g]_U < [f]_U$. Consequently, there exists $\beta < \kappa$ such that $h^{-1}(\{\beta\}) \in W$.[/step]

[guided]We now use the minimality of $[f]_U$. The previous step produced a function $g:\kappa \to \kappa$ with $[g]_U < [f]_U$. If we can also show that $[g]_U$ still lies above every constant class $[c_\alpha]_U$, then $[g]_U$ belongs to the same collection $\mathcal{C}$ as $[f]_U$, contradicting the choice of $[f]_U$ as minimal. Assume, toward a contradiction, that no fibre of $h$ is $W$-large. This means that for every $\beta < \kappa$, \begin{align*} h^{-1}(\{\beta\}) \notin W. \end{align*} Since $W$ is an ultrafilter and $h^{-1}(\{\beta\}) \subset S$, it follows that \begin{align*} S \setminus h^{-1}(\{\beta\}) \in W. \end{align*} Fix an ordinal $\alpha < \kappa$. We want to prove $[c_\alpha]_U < [g]_U$, which means that $g(\xi)>\alpha$ on a $U$-large set of $\xi$. Because $g=\widehat{h}\circ f$, it is enough to find a $W$-large set of $\eta$ on which $\widehat{h}(\eta)>\alpha$. Define \begin{align*} B_\alpha := S \cap \{\eta < \kappa : \alpha < \eta\} \cap \bigcap_{\beta \le \alpha}\left(S \setminus h^{-1}(\{\beta\})\right). \end{align*} Each factor in this intersection is $W$-large. The set $S$ is $W$-large by hypothesis. The tail $\{\eta < \kappa : \alpha < \eta\}$ is $W$-large because its complement has cardinality less than $\kappa$, and nonprincipal $\kappa$-complete ultrafilters contain complements of all sets of cardinality less than $\kappa$. Finally, for each $\beta \le \alpha$, the set $S \setminus h^{-1}(\{\beta\})$ is $W$-large by the assumption that the $\beta$-fibre is not $W$-large. There are only $|\alpha+1|<\kappa$ many such $\beta$, so $\kappa$-completeness gives $B_\alpha \in W$. Now take $\eta \in B_\alpha$. Since $\eta \in S$, the total extension satisfies $\widehat{h}(\eta)=h(\eta)$. Since $\eta$ avoids every fibre $h^{-1}(\{\beta\})$ for $\beta \le \alpha$, the value $h(\eta)$ is not any ordinal $\beta \le \alpha$. Because ordinal order is linear, this implies \begin{align*} \alpha < h(\eta)=\widehat{h}(\eta). \end{align*} Thus \begin{align*} B_\alpha \subset \{\eta < \kappa : \alpha < \widehat{h}(\eta)\}. \end{align*} Since $B_\alpha \in W$, the larger set $\{\eta < \kappa : \alpha < \widehat{h}(\eta)\}$ is also in $W$. By the definition of the pushforward ultrafilter $W$, \begin{align*} f^{-1}(\{\eta < \kappa : \alpha < \widehat{h}(\eta)\}) \in U. \end{align*} But $g(\xi)=\widehat{h}(f(\xi))$, so this preimage is exactly \begin{align*} \{\xi < \kappa : \alpha < g(\xi)\}. \end{align*} Therefore $[c_\alpha]_U < [g]_U$. Since $\alpha<\kappa$ was arbitrary, $[g]_U$ lies above every constant class. We already know $[g]_U < [f]_U$ and $[f]_U \le [\operatorname{id}_\kappa]_U$. To justify the resulting comparison with the identity, let $E:=\{\xi < \kappa : g(\xi)<f(\xi)\}$ and $F:=\{\xi < \kappa : f(\xi)\le \xi\}$. Both $E$ and $F$ belong to $U$, so $E\cap F\in U$. For every $\xi\in E\cap F$, we have $g(\xi)<f(\xi)\le \xi$, hence $g(\xi)\le \xi$. Thus $\{\xi < \kappa : g(\xi)\le \xi\}\in U$, which means $[g]_U \le [\operatorname{id}_\kappa]_U$. Hence $[g]_U \in \mathcal{C}$. This contradicts the minimality of $[f]_U$ in $\mathcal{C}$, because $[g]_U$ is a strictly smaller member of $\mathcal{C}$. The contradiction proves that the assumption was false. Hence there is some $\beta < \kappa$ such that $h^{-1}(\{\beta\}) \in W$.[/guided]

[step:Conclude that the pushforward ultrafilter is a normal measure] We have shown that $W$ is a nonprincipal $\kappa$-complete ultrafilter on $\kappa$. We have also shown that whenever $S \in W$ and $h:S \to \kappa$ is regressive, there exists $\beta < \kappa$ such that $h^{-1}(\{\beta\}) \in W$. This is the regressive-function formulation of normality for a measure on $\kappa$. Therefore $W$ is a normal measure on $\kappa$, so $\kappa$ carries a normal measure. [/step]