= Solution
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 ultrafilter>
$$
W=\{E\subseteq\mu:g^{-1}[E]\in U\}
$$
is 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.
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