[proofplan] We prove both implications directly from the diagonal-intersection definition of normality. In the forward direction, if a regressive function has no measure-one fibre, then the complements of all fibres are measure-one; normality diagonalizes these complements and produces a point at which the value of the function is different from every smaller ordinal, contradicting regressiveness. In the reverse direction, if a diagonal intersection failed to be measure-one, the least index witnessing failure defines a regressive function on a measure-one set; the assumed fibre property then gives a measure-one subset contained in the complement of one of the original measure-one sets, contradicting the ultrafilter property. [/proofplan] [step:Record the elementary ultrafilter consequences used in both directions] Since $U$ is an ultrafilter on $\kappa$, for every set $B\subseteq\kappa$ exactly one of $B$ and $\kappa\setminus B$ belongs to $U$. Since $U$ is closed under finite intersections, two disjoint subsets of $\kappa$ cannot both belong to $U$. Since $U$ is nonprincipal, $\{0\}\notin U$, and therefore $\kappa\setminus\{0\}\in U$ by the ultrafilter property. [guided] We will use only three basic consequences of being a nonprincipal ultrafilter. First, because $U$ is an ultrafilter on the underlying set $\kappa$, every subset $B\subseteq\kappa$ is decided by $U$: exactly one of $B$ and its complement $\kappa\setminus B$ lies in $U$. This is the step that lets us convert “a fibre is not measure-one” into “the complement of that fibre is measure-one.” Second, if $B,C\in U$, then $B\cap C\in U$. Hence two disjoint sets cannot both lie in $U$, because otherwise their intersection would be $\varnothing\in U$, which is impossible for a proper filter. Third, nonprincipality gives $\{0\}\notin U$. Applying the ultrafilter dichotomy to the subset $\{0\}\subseteq\kappa$, we get \begin{align*} \kappa\setminus\{0\}\in U. \end{align*} This lets us ignore the unique ordinal at which the strict inequality $f(\xi)<\xi$ cannot hold. [/guided] [/step] [step:Use normality to force a measure-one fibre for every regressive function] Assume that $U$ is normal. Let $S\in U$, and let $f:S\to\kappa$ be a function such that $f(\xi)<\xi$ for every nonzero $\xi\in S$. Suppose, toward a contradiction, that no fibre of $f$ belongs to $U$. Thus for every $\alpha<\kappa$, \begin{align*} F_\alpha := f^{-1}(\{\alpha\}) \end{align*} satisfies $F_\alpha\notin U$. Since $F_\alpha\subseteq\kappa$, the ultrafilter property gives \begin{align*} B_\alpha := \kappa\setminus F_\alpha \in U \end{align*} for every $\alpha<\kappa$. By normality applied to the sequence $(B_\alpha)_{\alpha<\kappa}$, the diagonal intersection \begin{align*} D := \Delta_{\alpha<\kappa} B_\alpha = \{\xi<\kappa : \text{for every } \alpha<\xi,\ \xi\in B_\alpha\} \end{align*} belongs to $U$. Therefore \begin{align*} E := D\cap S\cap(\kappa\setminus\{0\}) \end{align*} belongs to $U$, because each of $D$, $S$, and $\kappa\setminus\{0\}$ belongs to $U$. In particular, $E$ is nonempty. Choose $\xi\in E$. Since $\xi\in S$ and $\xi\ne 0$, regressiveness gives $f(\xi)<\xi$. Put $\alpha:=f(\xi)$. Then $\alpha<\xi$. Since $\xi\in D$, the definition of $D$ gives $\xi\in B_\alpha$. But $B_\alpha=\kappa\setminus f^{-1}(\{\alpha\})$, while $f(\xi)=\alpha$ says $\xi\in f^{-1}(\{\alpha\})$. This contradiction proves that some fibre $f^{-1}(\{\alpha\})$ belongs to $U$. [/step] [step:Use the regressive-function property to prove closure under diagonal intersections] Assume now that for every $S\in U$ and every regressive function $f:S\to\kappa$, some fibre of $f$ belongs to $U$. Let $(A_\alpha)_{\alpha<\kappa}$ be a sequence with $A_\alpha\in U$ for every $\alpha<\kappa$. Define \begin{align*} D := \Delta_{\alpha<\kappa} A_\alpha = \{\xi<\kappa : \text{for every } \alpha<\xi,\ \xi\in A_\alpha\}. \end{align*} We prove that $D\in U$. Suppose, toward a contradiction, that $D\notin U$. Then \begin{align*} C := \kappa\setminus D \end{align*} belongs to $U$. Define \begin{align*} S := C\cap(\kappa\setminus\{0\}). \end{align*} By the preceding ultrafilter observations, $S\in U$. For each $\xi\in S$, we have $\xi\notin D$ and $\xi\ne 0$. Therefore, by the definition of $D$, there exists some ordinal $\alpha<\xi$ such that $\xi\notin A_\alpha$. Since ordinals are well-ordered, the least such ordinal exists. Define the function \begin{align*} f:S &\to \kappa \end{align*} \begin{align*} \xi &\mapsto \min\{\alpha<\xi : \xi\notin A_\alpha\}. \end{align*} This function is regressive, because its value is always an ordinal $f(\xi)<\xi$. By the assumed regressive-function property, there is an ordinal $\beta<\kappa$ such that \begin{align*} f^{-1}(\{\beta\})\in U. \end{align*} For every $\xi\in f^{-1}(\{\beta\})$, the definition of $f$ gives $\xi\notin A_\beta$. Hence \begin{align*} f^{-1}(\{\beta\})\subseteq \kappa\setminus A_\beta. \end{align*} Since $f^{-1}(\{\beta\})\in U$ and $U$ is upward closed, $\kappa\setminus A_\beta\in U$. But $A_\beta\in U$ by hypothesis, so the two disjoint sets $A_\beta$ and $\kappa\setminus A_\beta$ both belong to $U$, contradicting the properness of $U$. Therefore $D\in U$ for every sequence $(A_\alpha)_{\alpha<\kappa}$ of members of $U$, so $U$ is normal. [/step] [step:Conclude the equivalence] The first direction shows that normality implies the measure-one fibre property for every regressive function on every $S\in U$. The second direction shows that the measure-one fibre property implies closure under diagonal intersections of all sequences $(A_\alpha)_{\alpha<\kappa}$ in $U$. By the stated definition of normality, these two conditions are equivalent. [/step]