The Lévy-Solovay theorem says that small forcings do not create measures. J.D. Hamkins has generalized this to a larger class of forcings called gap forcings. I would assume this cannot be generalized to *all* forcings, but I cannot think of a counterexample.
Is there a forcing notion that creates a $\kappa$-complete (or even countably complete) measure $\mu$ on some uncountable cardinal $\kappa$ such that $\mu \cap V$ is not in $V$?



Can measures be added by forcing?
Your title question is asking whether the measurability of a measurable
cardinal is downwards absolute to ground models: if $\kappa$ is
measurable in a forcing extension $V[G]$, must it be measurable in
$V$? This is a question that makes sense for any of the large
cardinal notions.


The answer is that, although the smaller large cardinal notions
such as inaccessibility and Mahlo-ness are downward absolute, this
phenomenon does not generally hold for the much larger large
cardinals. Specifically, a non-measurable cardinal $\kappa$ can become measurable after forcing with $\text{Add}(\kappa,1)$. 


Here is one way to see it. Suppose $\kappa$ is a measurable
cardinal in $V$. We may force $2^\kappa=\kappa^+$ while preserving
the measurability of $\kappa$, since this adds no new subsets to
$\kappa$; so suppose $2^\kappa=\kappa^+$ already in $V$. Let
$\mathbb{P}$ be the Easton support forcing iteration of length
$\kappa$, which forces with $\mathbb{Q}\_\gamma=\text{Add}(\gamma,1)$
to add a Cohen subset at
every inaccessible cardinal $\gamma$. Let
$\mathbb{Q}\_\kappa=\text{Add}(\kappa,1)$ be the stage $\kappa$
forcing, to do so at the top. Suppose
$G\ast g\subset\mathbb{P}\ast\mathbb{Q}\_\kappa$ is $V$-generic. First, I
claim that $\kappa$ is measurable in $V[G][g]$. This follows from
the usual lifting arguments, which appear in many of my papers.
Start with $j:V\to M$, an ultrapower embedding by a measure on
$\kappa$. The forcing $j(\mathbb{P})$ is isomorphic to
$\mathbb{P}\ast\mathbb{Q}\ast\mathbb{P}\_{tail}$, where the tail
forcing is $\leq\kappa$-closed in $M[G][g]$. Since
$|j(\kappa^+)|^V=\kappa^+$, we may enumerate the dense subsets of
$\kappa$ in $M[G][g]$ in a $\kappa^+$ sequence in $V[G][g]$. And
since $M[G][g]^\kappa\subset M[G][g]$ in $V[G][g]$, we may thereby
diagonalize to produce an $M[G][g]$-generic filter $G\_{tail}\subset\mathbb{P}\_{tail}$, and thus lift $j$ to
$j:V[G]\to M[j(G)]$, where $j(G)=G\ast g\ast G\_{tail}$. Similarly, the
object $g$ is essentially a condition in $j(\mathbb{Q}\_\kappa)$,
and so we may diagonalize again to produce an $M[j(G)]$-generic
filter $h\subset j(\mathbb{Q}\_\kappa)$ containing it, and thus
lift $j$ fully to $j:V[G][g]\to M[j(G)][j(g)]$, with $j(g)=h$.
Thus, $\kappa$ is measurable in $V[G][g]$.


Meanwhile, and this is the main point, I claim that $\kappa$ is
*not* measurable in $V[G]$. Suppose it were, with embedding
$j:V[G]\to \bar M$ in $V[G]$. By elementarity, since $V[G]$ thinks
it is a forcing extension by $\mathbb{P}$, it follows that $\bar
M=M[j(G)]$ for some inner model $M$. (My [theorems on
approximation and covering](http://jdh.hamkins.org/approximation-and-cover-properties/) show that in fact $M\subset V$ and $j\upharpoonright V:V\to M$ is definable in $V$, but we don't need that here.)
We may factor the forcing as $M[j(G)]=M[G\ast \bar g\ast\bar G\_{\rm
tail}]$, where $\bar g\subset\text{Add}(\kappa,1)^{M[G]}$ is $M[G]$-generic.
But note that $P(\kappa)^V\subset M$ and so also
$P(\kappa)^{V[G]}\subset M[G]$. Thus, $\bar g$ will have to be really
$V[G]$-generic for this forcing, which is a contradiction if
$\bar g\in V[G]$. So there can be no such embedding in $V[G]$, and so
$\kappa$ is not measurable there. This kind of argument is used many times in my paper [Destruction or preservation as you like it](http://jdh.hamkins.org/asyoulikeit/). 


Thus, we have a model $\bar V$, namely $\bar V=V[G]$, where
$\kappa$ is not measurable, but it becomes measurable after
forcing with $\text{Add}(\kappa,1)$.


The same essential argument works with all the stronger large
cardinals. If we had started with $\kappa$ supercompact, for
example, then we could make a model where it is not measurable,
but becomes supercompact after forcing with
$\text{Add}(\kappa,1)$.


Kunen observed that one can make a more extreme example as follows
(Saturated ideals, *Journal of Symbolic Logic*, 43(1):65--76,
March 1978). He relies on the fact that if one first adds a
homogeneous Souslin $\kappa$-tree, and then forces with that tree,
the combined forcing is equivalent to $\text{Add}(\kappa,1)$. But
the first step kills the weak compactness of $\kappa$. So
combining this observation with the previous, one arrives at the
conclusion:


If $\kappa$ is a measurable cardinal, then there is a forcing
extension $\bar V$ in which $\kappa$ is no longer weakly compact,
but forcing with a $\kappa$-Souslin tree in $\bar V$ makes
$\kappa$ suddenly measurable in the forcing extension $\bar V[g]$.
Furthermore, if $\kappa$ was tall, strong or supercompact in th
original ground model, then it will also retain those stronger
properties in $\bar V[g]$.


So a non-weakly compact cardinal can become supercompact by
forcing.


*Addition.* Consider forcing $\mathbb{P}$ which is an Easton support iteration of length $\kappa$, which at inaccessible $\gamma$ forces with the lottery sum of either doing nothing, or using $\text{Add}(\gamma,1)$. The argument above shows that if $\kappa$ is measurable and $G$ is $V$-generic for $\mathbb{P}$, then $\kappa$ is measurable in $V[G]$. We don't need the stage $\kappa$ forcing, now, since we may opt for trivial forcing in the stage $\kappa$ lottery. But also, if we force to add $V[G]$-generic $g\subset\kappa$, then $\kappa$ remains measurable in $V[G][g]$, since we may instead opt for the nontrivial forcing at stage $\kappa$. But the key argument above show that every normal measure in $V[G]$ must concentrate on $\gamma$ for which we did trivial forcing, and every normal measure $\mu$ on $\kappa$ in $V[G][g]$ concentrates on the $\gamma$ for which we did nontrivial forcing. It follows that such a $\mu$ in $V[G][g]$ must have $\mu\notin V[G]$. So we can preserve measurability, while preventing normal measures from extending ground model measures. 


