$\kappa$-complete ultrafilter and bounded subsets of $\kappa$ 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.
 A: It’s already pretty straightforward, but if you want to work even more directly from the usual definition of $\kappa$-completeness, note that since $U$ is non-principal, it contains $\kappa\setminus\{\alpha\}$ for each $\alpha\in\kappa$. In particular, if $B$ is a bounded subset of $\kappa$, then $\kappa\setminus\{\beta\}\in U$ for each $\beta\in B$, and $|B|<\kappa$, so
$$\kappa\setminus B=\bigcap_{\beta\in B}\Big(\kappa\setminus\{\beta\}\Big)\in U\;,$$
and therefore $B\notin U$.
Added: It’s not difficult to show that the usual definition is equivalent to the one in terms of partitions of the underlying set (as you probably already know), and I don’t see much to choose between them in getting this result.
A: Ideals and filters are dual. This means that any explanation that works for one should work for the other. 
To say that the union of less than $\kappa$ many sets of measure zero (read: in the dual ideal) is of measure zero, is the same as saying that the intersection of less than $\kappa$ many sets of measure one is of measure one. 
This is true simply by taking complements. 
Note that the fact that the filter is non-principal and $\kappa$-closed means that all the singletons are in the ideal, that is have measure zero. 
