Can measures be added by forcing? 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 $V$?
 A: Just some comments to complement Joel's answer: 
That forcing can destroy and then recreate measurability is due to Kunen: 

Kenneth Kunen. Saturated ideals, The Journal of Symbolic Logic 43 (1) (1978), 65–76. MR0495118 (80a:03068) 

The same is true for real valued measurability, this is due to Gitik: 

Moti Gitik, Saharon Shelah. More on Real-valued measurable cardinals and forcing with ideals, Israel Journal of Mathematics 124 (2001), 221–242 ([GiSh 582]). MR1856516 (2002g:03110)

For a while it was open whether we can destroy and then reconstruct measurability while still preserving weak compactness. In all arguments I was aware of, it is essential that weak compactness fails in the intermediate model, as what we accomplish is a (destructible) instance of failure of stationary set reflection. Joel points out that his argument actually preserves weak compactness, and can be carried out in other settings to preserve measurability or stronger properties. 
More generally, one can ask whether measurability can "appear spontaneously" by forcing, through some other means, just as we obtain saturation on the non-stationary ideal on $\omega_1$ by forcing MM, which cannot be traced back to some measure in some inner model that the forcing is reconstructing. 
For another proof of the specific result you are asking (but one that destroys weak compactness in the intermediate extension), see section 4 of my paper on RVM cardinals:

Real-valued measurable cardinals and well-orderings of the reals. In Set theory. Centre de Recerca Matemàtica, Barcelona, 2003–2004, Joan Bagaria, and Stevo Todorcevic, eds.; Trends in Mathematics, Birkhäuser Verlag, Basel, 2006, pp 83–120, MR2267147 (2007g:03064)

A: 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 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. 
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. 
