I want to prove the following
We have $\kappa$ an infinite regular cardinal such that every $\kappa$-complete filter on $\kappa$ can be extended to an $\omega_1$-complete ultrafilter on $\kappa$.
Show that there exists a measurable cardinal $\lambda\leq\kappa$.
Moreover, if $\lambda$ is the least measurable cardinal less than or equal to $\kappa$, then in fact every $\kappa$-complete filter on $\kappa$ can be extended to a $\lambda$-complete ultrafilter on $\kappa$.
Attempt For the first part, I need to prove that there exist a $\lambda\leq \kappa$ such that there exist a $\lambda$-complete non-principal ultrafilter on $\lambda$. I'm stomped... I tried to the partition equivalence of a $\kappa$-complete ultrafilter, trying to fit in some Ramsey theory and weakly compact cardinal, but i can't reach any conclusion. I also looked at strongly compact cardinal with no solution. I'm now trying a more constructive approach. Using the property of my cardinal to build a $\lambda$-complete non-principal ultrafilter on $\lambda$.
Maybe there are some equivalent definition of measurable that I could use here? Any help is welcomed, thanks