[proofplan] Let $\kappa$ be a measurable cardinal, and let $U$ be a nonprincipal $\kappa$-complete ultrafilter on $\kappa$. First, $\kappa$ is uncountable by definition, and $U$ contains no subset of $\kappa$ of cardinality less than $\kappa$. This small-set exclusion forces regularity, because a cofinal sequence of length less than $\kappa$ would cover $\kappa$ by fewer than $\kappa$ small pieces. The same ultrafilter argument forces the strong limit property: if $\kappa$ injected into $\mathcal P(\lambda)$ for some $\lambda < \kappa$, then $\kappa$-completeness would concentrate the injection on a single subset of $\lambda$, contradicting injectivity and nonprincipality. [/proofplan] [step:Fix a measurable cardinal and record its ultrafilter] Let $\kappa$ be a measurable cardinal. By the adopted definition of measurable cardinal, $\kappa$ is uncountable, and there exists a nonprincipal $\kappa$-complete ultrafilter $U$ on $\kappa$. Here $U$ is a collection of subsets of $\kappa$ such that exactly one of $A$ and $\kappa \setminus A$ lies in $U$ for each $A \subseteq \kappa$, the intersection of fewer than $\kappa$ members of $U$ again lies in $U$, and no singleton belongs to $U$. [/step] [step:Show that the ultrafilter contains no small subset of $\kappa$] Let $A \subseteq \kappa$ satisfy $|A| < \kappa$. For each $\alpha \in A$, nonprincipality gives $\{\alpha\} \notin U$, so the ultrafilter property gives $\kappa \setminus \{\alpha\} \in U$. Since $|A| < \kappa$ and $U$ is $\kappa$-complete, the intersection of these complements belongs to $U$: \begin{align*} \bigcap_{\alpha \in A} (\kappa \setminus \{\alpha\}) = \kappa \setminus A \in U. \end{align*} Because $U$ is an ultrafilter, $A \notin U$. Thus no subset of $\kappa$ of cardinality less than $\kappa$ is $U$-large. [/step] [step:Use the small-set lemma to prove regularity] Suppose, for contradiction, that $\kappa$ is singular. Let $\mu := \operatorname{cf}(\kappa)$, where $\operatorname{cf}(\kappa)$ denotes the least cardinality of a cofinal subset of $\kappa$; since $\kappa$ is singular, $\mu < \kappa$. Choose a strictly increasing cofinal map $c: \mu \to \kappa$. For each $i < \mu$, define \begin{align*} D_i := \{\alpha < \kappa : \alpha < c(i)\}. \end{align*} Each $D_i$ has cardinality less than $\kappa$, since it is bounded below the cardinal $\kappa$. The family $\{D_i : i < \mu\}$ covers $\kappa$: for every $\alpha < \kappa$, cofinality of $c$ gives some $i < \mu$ with $\alpha < c(i)$, so $\alpha \in D_i$. By the preceding step, no $D_i$ lies in $U$. Therefore each complement $\kappa \setminus D_i$ lies in $U$. Since $\mu < \kappa$ and $U$ is $\kappa$-complete, the intersection of these fewer than $\kappa$ complements lies in $U$, but this intersection is empty because the $D_i$ cover $\kappa$. This contradicts the definition of an ultrafilter, which never contains $\varnothing$. Hence $\kappa$ is regular. [/step] [step:Use ultrafilter concentration to prove the strong limit property] Let $\lambda < \kappa$ be a cardinal, and let $\mathcal P(\lambda)$ denote the power set of $\lambda$. Suppose, for contradiction, that $2^\lambda \geq \kappa$. Then there is an injective map $F: \kappa \to \mathcal P(\lambda)$. For each $\beta < \lambda$, define the fiber set \begin{align*} B_\beta := \{\alpha < \kappa : \beta \in F(\alpha)\} \subseteq \kappa. \end{align*} Define a subset $S \subseteq \lambda$ by \begin{align*} S := \{\beta < \lambda : B_\beta \in U\}. \end{align*} For every $\beta < \lambda$, the set $C_\beta$ defined by $C_\beta := B_\beta$ if $\beta \in S$ and $C_\beta := \kappa \setminus B_\beta$ if $\beta \notin S$ belongs to $U$ by the definition of $S$ and the ultrafilter property. Since $\lambda < \kappa$ and $U$ is $\kappa$-complete, \begin{align*} C := \bigcap_{\beta < \lambda} C_\beta \in U. \end{align*} For every $\alpha \in C$ and every $\beta < \lambda$, the equivalence $\beta \in F(\alpha) \iff \beta \in S$ holds. Hence $F(\alpha) = S$ for every $\alpha \in C$. Since $C \in U$, the preceding small-set lemma gives $|C| = \kappa$, so $C$ contains at least two distinct ordinals. This contradicts the injectivity of $F$. Therefore $2^\lambda < \kappa$ for every cardinal $\lambda < \kappa$, so $\kappa$ is a strong limit cardinal. [guided] The goal is to show that no cardinal below $\kappa$ has a power set as large as $\kappa$. Fix a cardinal $\lambda < \kappa$ and assume toward a contradiction that $2^\lambda \geq \kappa$. This means that there is an injective map \begin{align*} F: \kappa &\to \mathcal P(\lambda), \end{align*} so the ordinals below $\kappa$ are assigned pairwise distinct subsets of $\lambda$. We now use the ultrafilter to decide each coordinate $\beta < \lambda$. For each $\beta < \lambda$, define \begin{align*} B_\beta := \{\alpha < \kappa : \beta \in F(\alpha)\}. \end{align*} Because $U$ is an ultrafilter, exactly one of $B_\beta$ and $\kappa \setminus B_\beta$ belongs to $U$. Define \begin{align*} S := \{\beta < \lambda : B_\beta \in U\}. \end{align*} Thus $S$ records the coordinates that occur on a $U$-large set of indices. For each $\beta < \lambda$, set $C_\beta := B_\beta$ when $\beta \in S$, and set $C_\beta := \kappa \setminus B_\beta$ when $\beta \notin S$. By construction, every $C_\beta$ lies in $U$. Since there are only $\lambda < \kappa$ many coordinates and $U$ is $\kappa$-complete, their intersection is still $U$-large: \begin{align*} C := \bigcap_{\beta < \lambda} C_\beta \in U. \end{align*} Now take any $\alpha \in C$. If $\beta \in S$, then $\alpha \in C_\beta = B_\beta$, so $\beta \in F(\alpha)$. If $\beta \notin S$, then $\alpha \in C_\beta = \kappa \setminus B_\beta$, so $\beta \notin F(\alpha)$. Therefore, for every $\beta < \lambda$, \begin{align*} \beta \in F(\alpha) \iff \beta \in S. \end{align*} Hence $F(\alpha) = S$ for every $\alpha \in C$. The set $C$ cannot be small. The earlier small-set lemma says that no subset of $\kappa$ of cardinality less than $\kappa$ belongs to $U$; since $C \in U$, it follows that $|C|$ is not less than $\kappa$. In particular, $C$ contains two distinct ordinals $\alpha_0$ and $\alpha_1$. But then $F(\alpha_0) = S = F(\alpha_1)$, contradicting that $F$ is injective. This contradiction proves $2^\lambda < \kappa$. Since $\lambda < \kappa$ was arbitrary, $\kappa$ is a strong limit cardinal. [/guided] [/step] [step:Combine the defining properties of strong inaccessibility] We have shown that $\kappa$ is uncountable, regular, and a strong limit cardinal. By the definition of a strongly inaccessible cardinal, these three properties imply that $\kappa$ is strongly inaccessible. Since $\kappa$ was an arbitrary measurable cardinal, every measurable cardinal is strongly inaccessible. [/step]