If $U$ is a $\kappa$-complete nonprincipal ultrafilter on $\kappa$ then every bounded subset of $\kappa$ has measure $0$. This is because if we had $X \in U$ with $X$ having size less than $\kappa$, then since $X$ is a union of less than $\kappa$ many singletons and since $U$ is $\kappa$-complete then one of the singletons has to be in $U$ which contradicts $U$ being non principal.
There must be a better way to explain this simple fact? Thanks.