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Your 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.

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 $M[j(G)][G\ast \bar g\ast\bar G_{\rm tail}]$, 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.

So a non-weakly compact cardinal can become supercompact 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.

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.

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.

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 $M[j(G)][G\ast \bar g\ast\bar G_{\rm tail}]$, 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.

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 $M[j(G)][G\ast \bar g\ast\bar G_{\rm tail}]$, 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.

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.

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 $M[j(G)][G\ast \bar g\ast\bar G_{\rm tail}]$, where $g'\subset\text{Add}(\kappa,1)^{M[G]}$ is $M[G]$-generic. But note that $P(\kappa)^V\subset M$ and so also $P(\kappa)^{V[G]}\subset M[G]$. Thus, $g'$ will have to be really $V[G]$-generic for this forcing, which is a contradiction if $g'\in V[G]$. So there can be no such embedding in $V[G]$, and so $\kappa$ is not measurable there.

Kunen observed that one can make a more extreme example as follows (Saturated ideals, {\em 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:

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 $\text{Add}(\kappa,1)$.

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 $M[j(G)][G\ast \bar g\ast\bar G_{\rm tail}]$, 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.

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: