I know that there are two very similar questions here, one asking if the condition of $\kappa$ being a regular cardinal in the definition of $\kappa$-complete filter is really needed and another asking about the definition of measurable cardinals, but my doubt is when I put those definitions together.
According to Wikipedia and Jech, in the definition of measurable cardinal, $\kappa$ doesn't need to be regular, just uncountable, but the definition of $\kappa$-complete filter ask for it. So what am I missing here? Should $\kappa$ be regular or not? Sorry for the noob question and thanks in advance.
Let $U$ be a nonprincipal, $\kappa$-complete ultrafilter on $\kappa$. Then $\kappa$ is regular.
Proof. First note that $U$ is uniform, i.e. every member of $A \in U$ has size $\kappa$. (Otherwise $A = \bigcup_{\xi \in A } \{\xi \}$ is a union of length $< \kappa$ so that by $\kappa$-completeness we must have $\{\xi\} \in U$ for some $\xi \in A$.)
Suppose $\kappa$ is singular. Write $\kappa = \bigcup_{\xi < \eta} A_\xi$, with $A_\xi$ of cardinality $< \kappa$ and $\eta < \kappa$. Since $U$ is uniform, we have $A_\xi \not \in U$ for all $\xi < \eta$. But then, by $\kappa$-completeness, $$ \emptyset = \bigcap_{\xi < \eta} \kappa \setminus A_\xi \in U. $$ Contradiction. $\dashv$