Is anything known about the consistency strength of the statement:


"There is a normal measure (on a cardinal) that is not ordinal-definable"?


In particular, is it consistent relative to the existence of a measurable cardinal?
It looks like it's consistent relative to the existence of a supercompact cardinal.
If $\kappa$ is supercompact then we can force to make it Laver indestructible.


So assume that $\kappa$ is still $(\kappa+2)$-strong after we add $(2^{2^\kappa})^+$ many Cohen subsets of $\kappa^+$, more than the number of measures on $\kappa$ in $V$.
 Solovay proved that if $\kappa$ is $(\kappa+2)$-strong then for every set $X \in V\_{\kappa+2}$ there is a normal measure on $\kappa$ whose ultrapower contains $X$. So letting $X$ range over the Cohen subsets of $\kappa^+$ that we added, a counting argument shows that we must get some normal measures on $\kappa$ that are not in $V$. Cohen forcing is homogeneous, so these measures cannot be ordinal-definable.
I don't know how strong this kind of indestructibility is, or whether it's necessary.


I am also interested to know anything about countably complete measures on any set that are not ordinal-definable from that set.



Measures that are not OD
It was proved, long ago, that one can force over a model with a measurable cardinal $\kappa$ and get a model with lots ($2^{2^\kappa}$, I think) of normal measures. I believe that (1) most of those measures won't be OD in that model and (2) the relevant paper is 


Kenneth Kunen and Jeff Paris, Boolean extensions and measurable cardinals, Annals Math. Log., Vol. 2 (1971), pp. 359-377.


Unfortunately, I'm traveling and would find it difficult to check these things right now.


