Let $\mathcal{I}$ be a proper [ideal](http://en.wikipedia.org/wiki/Ideal_%28set_theory%29) on $\omega$. If $\mathcal{I}$ is Borel as a subset of $2^\omega$, does it follow that $\mathcal{I}$ is meager?


**Edit:** What if $\mathcal{I}$ contains all finite subsets of $\omega$?



Borel ideals on $\omega$ are meager?
Let us consider a prime ideal J containing I. Suppose I is non meager. Then J would have non empty interior (modulo meager). Since complementing a set of integers is a meager preserving operation so the dual ultrafilter U of J has the same property. But both J and U are closed under rational translations and they both have interiors (modulo meager) so they are both comeager which is impossible.


