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If $\kappa$ is the least omega-measurable cardinal and $U$ witnesses it, failure of $\kappa$-completeness partitions a member of $U$ into fewer than $\kappa$ fibres, none belonging to $U$. The pushforward ultrafilter under the fibre-index map is then a nonprincipal countably complete measure on a smaller cardinal, a contradiction. Thus $U$ is $\kappa$-complete and $\kappa$ is a measurable cardinal.

Fix a regular uncountable cardinal number $\kappa$. A club set $C\subseteq \kappa$ is unbounded in $\kappa$ and closed under suprema below $\kappa$ of nonempty increasing sequences with no last term. Equivalently, every limit ordinal $\delta <\kappa$ with $\sup (C\cap \delta )=\delta$ belongs to $C$. The club filter isIt consists of all sets containing a club set, not only the club sets themselves.

An omega-measurable cardinal is a cardinal number carrying a nonprincipal ultrafilter closed under countable intersections, that is, a countably complete ultrafilter. Such a cardinal must be uncountable: on a countable set the intersection of the complements of all singletons would be empty.

A measurable cardinal is an uncountable cardinal number $\kappa$ carrying a nonprincipal $\kappa$-complete ultrafilter. A kappa-complete filter is closed under intersections of every family of size strictly below $\kappa$. Thus for an omega-measurable cardinal the completeness required is countable completeness, not merely closure under finitely many intersections.

Take the least omega-measurable cardinal $\kappa$, with a countably complete ultrafilter $U$ on $\kappa$ that is nonprincipal. We prove that $U$ is $\kappa$-complete. This will give $\boxed{\kappa \text{ is measurable}}$.

Suppose $\mu <\kappa$, $A_i\in U$ for $i<\mu$, and ${\bigcap }_{i<\mu }{A}_i\notin U$. Then $B=\kappa \setminus {\bigcap }_{i<\mu }{A}_i$ belongs to $U$. For $x\in B$ let $g(x)$ be the least $i<\mu$ with $x\notin A_i$. The pushforward ultrafilteris a countably complete ultrafilter on $\mu$; this uses $B\in U$, so $U$ restricts to an ultrafilter on $B$. For each $i$, the fibre $g^{-1}[\{i\}]$ is contained in $\kappa \setminus A_i$, hence is not in $U$. Thus $W$ is nonprincipal. If $\mu$ is finite or countable, countable completeness already rules this out. Otherwise $|\mu |<\kappa$ would be an omega-measurable cardinal, contradicting minimality of $\kappa$.

Therefore every intersection of fewer than $\kappa$ members of $U$ lies in $U$. Since $\kappa$ is uncountable, it is a measurable cardinal. This proves that the least omega-measurable cardinal is measurable, even if the originally supplied cardinal carried only a countably complete ultrafilter.