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.
On
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.
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.