[proofplan] Fix a normal measure $U$ on the measurable cardinal $\kappa$. We first record the measure-theoretic consequences needed in the argument: bounded subsets of $\kappa$ are $U$-null, $\kappa$ is regular and strong limit, and every $U$-large set is stationary. We then show that the regular ordinals below $\kappa$ form a $U$-large set by applying normality first to the cofinality function and then to the coordinates of chosen cofinal sequences. Finally we show that, among these regular cardinals, the non-strong-limit cardinals cannot be $U$-large; intersecting the resulting $U$-large sets gives a stationary set of inaccessible cardinals below $\kappa$. [/proofplan]

[step:Fix the normal measure and record its basic consequences] Let $U$ be a normal $\kappa$-complete nonprincipal ultrafilter on $\kappa$. We use normality in the following Fodor form: if $X \in U$ and $f:X \to \kappa$ satisfies $f(\alpha) < \alpha$ for every nonzero $\alpha \in X$, then there are $\xi < \kappa$ and $Y \in U$ with $Y \subset X$ such that $f(\alpha)=\xi$ for every $\alpha \in Y$. For every ordinal $\delta$, let $\operatorname{cf}(\delta)$ denote the least order type of a cofinal subset of $\delta$. Let $\omega$ denote the first infinite ordinal. A subset $C\subseteq\kappa$ is called club in $\kappa$ if it is closed and unbounded in the order topology on $\kappa$, and a subset $X\subseteq\kappa$ is stationary in $\kappa$ if $X\cap C\neq\varnothing$ for every club subset $C\subseteq\kappa$. A subset $A\subseteq\kappa$ is called $U$-null if $A\notin U$. Every bounded subset of $\kappa$ is $U$-null. Indeed, for each $\xi<\kappa$, nonprincipality gives $\{\xi\}\notin U$. If $B\subset\kappa$ is bounded, choose $\beta<\kappa$ with $B\subseteq \beta+1$. Since $\beta+1<\kappa$ and the dual ideal of $U$ is closed under unions of fewer than $\kappa$ sets, the set $\beta+1=\bigcup_{\xi\leq\beta}\{\xi\}$ is $U$-null, hence so is $B$. The cardinal $\kappa$ is regular. If $\operatorname{cf}(\kappa)=\lambda<\kappa$, choose an increasing cofinal map \begin{align*} h:\lambda \to \kappa. \end{align*} For each $i<\lambda$, the initial segment $h(i)+1$ is bounded in $\kappa$, hence $U$-null. Since $h$ is cofinal, we have \begin{align*} \kappa=\bigcup_{i<\lambda}(h(i)+1). \end{align*} This is a union of fewer than $\kappa$ many $U$-null sets, contradicting $\kappa\in U$. Thus $\operatorname{cf}(\kappa)=\kappa$. The cardinal $\kappa$ is strong limit. Suppose, toward a contradiction, that there is a cardinal $\lambda<\kappa$ with $2^\lambda\geq\kappa$. Choose an injection \begin{align*} A:\kappa \to \mathcal{P}(\lambda). \end{align*} For each $i<\lambda$, define the function $X_i:\kappa\to\{0,1\}$ by $X_i(\alpha)=1$ if $i\in A(\alpha)$ and $X_i(\alpha)=0$ otherwise. Since $U$ is an ultrafilter, for each $i<\lambda$ there is a unique $\varepsilon_i\in\{0,1\}$ such that $\{\alpha<\kappa:X_i(\alpha)=\varepsilon_i\}\in U$. By $\kappa$-completeness, the intersection \begin{align*} Y=\bigcap_{i<\lambda}\{\alpha<\kappa:X_i(\alpha)=\varepsilon_i\} \end{align*} belongs to $U$. But any two elements of $Y$ have the same image under $A$, so $Y$ has at most one element, contradicting nonprincipality. Therefore $2^\lambda<\kappa$ for every cardinal $\lambda<\kappa$. Finally, every $U$-large set is stationary. Let $X\in U$, and let $C\subseteq\kappa$ be club. If $X\cap C=\varnothing$, then $\kappa\setminus C\in U$. Define \begin{align*} r:\kappa\setminus C &\to \kappa \end{align*} \begin{align*} \alpha &\mapsto \sup(C\cap\alpha). \end{align*} After removing the bounded set below $\min C+1$, this is regressive on a $U$-large subset of $\kappa\setminus C$. By normality, $r$ is constant with value $\gamma<\kappa$ on some $Z\in U$. If $\delta=\min(C\setminus(\gamma+1))$, then $Z\subseteq \delta+1$, so $Z$ is bounded, contradicting $Z\in U$. Hence $X$ meets every club, so $X$ is stationary. [/step]

[step:Show that regular cardinals below $\kappa$ form a measure-one set]Let \begin{align*} N=\{\alpha<\kappa:\alpha\neq 0 \text{ and } \operatorname{cf}(\alpha)<\alpha\}. \end{align*} We prove that $N\notin U$. Suppose instead that $N\in U$. Define the cofinality map \begin{align*} f:N &\to \kappa \end{align*} \begin{align*} \alpha &\mapsto \operatorname{cf}(\alpha). \end{align*} This map is regressive, because every $\alpha\in N$ satisfies $\operatorname{cf}(\alpha)<\alpha$. By normality, there are $\lambda<\kappa$ and $S_\lambda\in U$ with $S_\lambda\subseteq N$ such that $\operatorname{cf}(\alpha)=\lambda$ for every $\alpha\in S_\lambda$. For each $\alpha\in S_\lambda$, choose an increasing cofinal map \begin{align*} c_\alpha:\lambda \to \alpha. \end{align*} For each $i<\lambda$, define \begin{align*} g_i:S_\lambda &\to \kappa \end{align*} \begin{align*} \alpha &\mapsto c_\alpha(i). \end{align*} Since $c_\alpha(i)<\alpha$, the map $g_i$ is regressive. Normality gives an ordinal $\gamma_i<\kappa$ and a set $S_i\in U$ with $S_i\subseteq S_\lambda$ such that $c_\alpha(i)=\gamma_i$ for every $\alpha\in S_i$. Because $\lambda<\kappa$, $\kappa$-completeness gives \begin{align*} S_*=\bigcap_{i<\lambda}S_i\in U. \end{align*} Since $\kappa$ is regular and each $\gamma_i<\kappa$, the ordinal \begin{align*} \gamma=\sup_{i<\lambda}\gamma_i \end{align*} satisfies $\gamma<\kappa$. The tail $\kappa\setminus(\gamma+1)$ belongs to $U$, so \begin{align*} S_*\cap(\kappa\setminus(\gamma+1))\in U. \end{align*} Choose $\alpha\in S_*\cap(\kappa\setminus(\gamma+1))$. Then for every $i<\lambda$, \begin{align*} c_\alpha(i)=\gamma_i\leq\gamma<\alpha. \end{align*} Thus the range of $c_\alpha$ is bounded by $\gamma$ in $\alpha$, contradicting that $c_\alpha:\lambda\to\alpha$ is cofinal. Hence $N\notin U$. Since $U$ is an ultrafilter, the complement \begin{align*} R=\{\alpha<\kappa:\operatorname{cf}(\alpha)=\alpha\} \end{align*} belongs to $U$. Every nonzero regular ordinal is a cardinal: if a nonzero ordinal $\alpha$ with $\operatorname{cf}(\alpha)=\alpha$ were equipotent with a cardinal $\mu<\alpha$, then a bijection $e:\mu\to\alpha$ would have cofinal range in $\alpha$, giving $\operatorname{cf}(\alpha)\leq\mu<\alpha$, a contradiction. Hence $R$ is the $U$-large set of regular cardinals below $\kappa$, up to the harmless bounded point $0$.[/step]

[guided]The goal is to prove that $U$ concentrates on regular ordinals. The obstruction is the set \begin{align*} N=\{\alpha<\kappa:\alpha\neq 0 \text{ and } \operatorname{cf}(\alpha)<\alpha\}. \end{align*} If $N$ were $U$-large, then the cofinality function would be a regressive function on a $U$-large set. Define \begin{align*} f:N &\to \kappa \end{align*} \begin{align*} \alpha &\mapsto \operatorname{cf}(\alpha). \end{align*} For each $\alpha\in N$, the defining condition of $N$ gives $f(\alpha)=\operatorname{cf}(\alpha)<\alpha$, so $f$ is regressive. Normality of $U$ therefore gives a fixed $\lambda<\kappa$ and a set $S_\lambda\in U$, with $S_\lambda\subseteq N$, such that $\operatorname{cf}(\alpha)=\lambda$ for every $\alpha\in S_\lambda$. Now every $\alpha\in S_\lambda$ has the same cofinality $\lambda$. For each such $\alpha$, choose an increasing cofinal map \begin{align*} c_\alpha:\lambda \to \alpha. \end{align*} The key idea is to freeze not only the cofinality, but every coordinate of these cofinal sequences. Fix $i<\lambda$ and define \begin{align*} g_i:S_\lambda &\to \kappa \end{align*} \begin{align*} \alpha &\mapsto c_\alpha(i). \end{align*} Because $c_\alpha$ maps into $\alpha$, we have $c_\alpha(i)<\alpha$ for every $\alpha\in S_\lambda$. Thus $g_i$ is regressive, and normality gives an ordinal $\gamma_i<\kappa$ and a set $S_i\in U$ such that $S_i\subseteq S_\lambda$ and $c_\alpha(i)=\gamma_i$ for every $\alpha\in S_i$. We have one $U$-large set $S_i$ for each coordinate $i<\lambda$. Since $\lambda<\kappa$ and $U$ is $\kappa$-complete, their intersection remains $U$-large: \begin{align*} S_*=\bigcap_{i<\lambda}S_i\in U. \end{align*} For every $\alpha\in S_*$ and every $i<\lambda$, the $i$th coordinate of the chosen cofinal sequence is forced to be the fixed ordinal $\gamma_i$. Now use regularity of $\kappa$. The family $(\gamma_i)_{i<\lambda}$ has length $\lambda<\kappa$, and each $\gamma_i<\kappa$, so \begin{align*} \gamma=\sup_{i<\lambda}\gamma_i \end{align*} is still below $\kappa$. Since bounded subsets are $U$-null, the tail $\kappa\setminus(\gamma+1)$ belongs to $U$. Therefore \begin{align*} S_*\cap(\kappa\setminus(\gamma+1))\in U, \end{align*} so this intersection is nonempty. Choose $\alpha$ in it. Then $\gamma<\alpha$, and for every $i<\lambda$, \begin{align*} c_\alpha(i)=\gamma_i\leq\gamma<\alpha. \end{align*} Thus the entire range of $c_\alpha$ is bounded below $\alpha$ by $\gamma$. This contradicts the fact that $c_\alpha:\lambda\to\alpha$ was chosen cofinal. Hence $N$ cannot belong to $U$. Since $U$ is an ultrafilter, the complement of $N$ belongs to $U$. That complement is the set \begin{align*} R=\{\alpha<\kappa:\operatorname{cf}(\alpha)=\alpha\}, \end{align*} up to the harmless convention at $0$. A nonzero ordinal satisfying $\operatorname{cf}(\alpha)=\alpha$ must be an initial ordinal, hence a cardinal. Indeed, if such an ordinal $\alpha$ had cardinality $\mu<\alpha$, then a bijection $e:\mu\to\alpha$ would have range all of $\alpha$, hence cofinal in $\alpha$, so $\operatorname{cf}(\alpha)\leq\mu<\alpha$, contradicting $\operatorname{cf}(\alpha)=\alpha$. Therefore $R$ is the desired $U$-large set of regular cardinals below $\kappa$.[/guided]

[step:Show that strong-limit cardinals below $\kappa$ form a measure-one set on the regulars] Let \begin{align*} T=\{\alpha\in R:\alpha \text{ is not a strong limit cardinal}\}. \end{align*} We prove that $T\notin U$. Suppose toward a contradiction that $T\in U$. For each $\alpha\in T$, since $\alpha$ is a cardinal but not strong limit, choose the least cardinal $\lambda(\alpha)<\alpha$ such that \begin{align*} 2^{\lambda(\alpha)}\geq \alpha. \end{align*} This defines a regressive map \begin{align*} \lambda:T &\to \kappa \end{align*} \begin{align*} \alpha &\mapsto \lambda(\alpha). \end{align*} By normality, there are a cardinal $\lambda_0<\kappa$ and a set $T_0\in U$ with $T_0\subseteq T$ such that $\lambda(\alpha)=\lambda_0$ for every $\alpha\in T_0$. For every $\alpha\in T_0$, \begin{align*} \alpha\leq 2^{\lambda_0}. \end{align*} Since $\kappa$ is strong limit and $\lambda_0<\kappa$, we have $2^{\lambda_0}<\kappa$. Hence $T_0\subseteq 2^{\lambda_0}+1$, so $T_0$ is bounded in $\kappa$. This contradicts $T_0\in U$, because bounded subsets of $\kappa$ are $U$-null. Therefore $T\notin U$. Since $R\in U$ and $U$ is an ultrafilter, the set \begin{align*} H=\{\alpha\in R:\alpha \text{ is a strong limit cardinal}\} \end{align*} belongs to $U$. [/step]

[step:Intersect the measure-one sets and conclude Mahloness] Let \begin{align*} I=\{\alpha<\kappa:\alpha \text{ is an uncountable regular strong limit cardinal}\}. \end{align*} The set $H$ from the previous step consists of regular strong-limit cardinals, and removing the bounded initial segment $\omega+1$ preserves membership in $U$. Therefore \begin{align*} I=H\cap(\kappa\setminus(\omega+1)) \end{align*} belongs to $U$. By the first step, every $U$-large subset of $\kappa$ is stationary in $\kappa$. Hence $I$ is stationary in $\kappa$. Thus the inaccessible cardinals below $\kappa$ form a stationary subset of $\kappa$. By the definition of a Mahlo cardinal, $\kappa$ is Mahlo. [/step]