Suppose $\kappa$ is a measurable cardinal and let $\mathcal{F}\subset\wp(\kappa)$ be a $\kappa$-complete non-principal filter. Can we extend $\mathcal{F}$ to a $\kappa$-complete ultrafilter?


My motivation comes from a problem I have just encountered. I need a $\kappa$-complete ultrafilter whereas the best I can do is to construct only some filter. 


Of course, the standard Zorn argument does not work here by a simple meta-argument. Take the filter on $\omega\_1$ consisting of sets with countable complement. It is $\omega\_1$-complete but there is no $\omega\_1$-complete ultrafilter on $\omega\_1$, since this cardinal is not measurable (in ZFC). 



Extending complete filters
If your filter is generated by $\kappa$ many sets, then indeed the
conclusion you seek can be made, by a direct argument that does
not go through strong compactness.


**Theorem**. The following are equivalent, for any uncountable
regular cardinal $\kappa$.


1. $\kappa$ is a measurable cardinal.
2. Every $\kappa$ complete filter $F$, generated by at most
$\kappa$-many sets, extends to a $\kappa$-complete ultrafilter.


Proof: It is easy to see that $2$ implies $1$, since the filter of
co-bounded sets in $\kappa$ is $\kappa$-complete and generated by
the tails, so there is a $\kappa$-complete non-principal
ultrafilter on $\kappa$.


For the main direction, assume $\kappa$ is measurable and $F$ is a
$\kappa$-complete filter on a set $D$ with $F$ generated by at
most $\kappa$ many sets $X\_\alpha$, for $\alpha\lt\kappa$. Let
$j:V\to M$ be an elementary embedding with critical point
$\kappa$. By applying $j$ to $\vec X=\langle
X\_\alpha\lt\kappa\rangle$ and restricting to $\kappa$, we see that
$\langle j(X\_\alpha)\mid\alpha\lt\kappa\rangle$ is in $M$. And
since this is fewer than $j(\kappa)$ many elements of $j(F)$,
which is $j(\kappa)$-complete in $M$, it follows that
$\bigcap\_{\alpha\lt\kappa}j(X\_\alpha)\in j(F)$, and in particular,
there is some $a\in \bigcap\_\alpha j(X\_\alpha)$. Define
$U=\{X\subset D\mid a\in j(X)\}$. It is easy to verify that $U$
is a $\kappa$-complete ultrafilter on $D$ and $F\subset U$, as
desired. QED


For $\theta$-generated filters, one generally needs
$\theta$-strong compactness, as mentioned in the comments, and this is in fact equivalent to $\theta$-strong compactness. The essence of the argument above, then, is that a cardinal $\kappa$ is measurable if and only if it is $\kappa$-strongly compact. 


That said, if you want this filter extension property, then I encourage you to go ahead and make the strong compactness assumption. There are many beautiful theorems using strongly compact cardinals. 


