The question is in the title. For Easton's theorem see [Wikipedia](https://en.wikipedia.org/wiki/Easton%27s_theorem). Loosely speaking we can use forcing to manipulate the powerset function on regular cardinals as much as we like given we satisfy the following restrictions: (a) $\kappa<\lambda$ implies $2^\kappa\le 2^\lambda$ and (b) $\kappa<cf(2^\kappa)$.


In addition, Easton's forcing preserves cardinals.


**Question**: Does Easton forcing preserve measurable cardinals? In particular, if $\mu$ is the first measurable, does it remain the first measurable in the extension (measurable here means "there is an $\aleph\_1$-complete non principal ultrafilter on $\mu$).
If the answer is "no" does the restriction of Easton's theorem below the first measurable $\mu$ preserve the measurability of $\mu$?



Does Easton forcing preserve measurable cardinals?
There are a variety of positive and negative results on this topic, depending on the Easton function and the set-theoretic background. 


The Kunen-Paris theorem, for example, provides a large number of positive instances.


**Theorem.** If $\kappa$ is a measurable cardinal and $2^\kappa=\kappa^+$, then for any Easton function $E:A\to\kappa$ defined on a set $A\subset\kappa$ of regular cardinals below $\kappa$, with $A\notin\mu$ for some normal measure $\mu$ on $\kappa$, then in the Easton forcing extension $V[G]$, the cardinal $\kappa$ remains measurable. 


**Proof.** Let $\newcommand\P{\mathbb{P}}\P$ be the Easton forcing for $E$, and suppose that $G\subset\P$ is $V$-generic. Let $j:V\to M$ be the ultrapower by $\mu$. If you consider $j(\P)$, you can factor it at $\kappa$ and see that $j(\P)=\P\times\P^\*$, where $\P^\*$ is the forcing beyond stage $\kappa$. There is no forcing at coordinate $\kappa$ precisely because $\kappa\notin j(A)$. It follows that $\P^\*$ is $\leq\kappa$-closed in $M$. Furthermore, the size of $\P^\*$ is $|j(\kappa)|^M$, and so the number of subsets of it in $M$ is $|j(2^\kappa)|^M=|j(\kappa^+)|^M$, and this is $\kappa^+$ in $V$, since $(\kappa^+)^\kappa=\kappa^+$. So in $V$ we may line up the dense subsets of $\P^\*$ in $M$ into a $\kappa^+$ sequence. Since $M^\kappa\subset M$, we may therefore perform a diagonalization to build in $V$ an $M$-generic filter $G^\*\subset\P^\*$, simply by meeting the dense sets one-by-one. Since $G$ is $V$-generic, it is also $M[G^\*]$-generic, and so $G\times G^\*$ is $M$-generic for $j(\P)$. This filter fulfills the lifting criterion, and so we may lift the embedding to $j:V[G]\to M[j(G)]$, with $j(G)=G\times G^\*$. Thus, $\kappa$ remains measurable in $V[G]$, as desired. 
$\Box$


Note that the assumption $2^\kappa=\kappa^+$ can be easily forced without adding subsets to $\kappa$ and thefore while preserving the measurability of $\kappa$. So the argument is applicable to any measurable cardinal, whether or not the GCH holds at $\kappa$, provided one also forces $2^\kappa=\kappa^+$. 


Since the construction of $G^\*$ has wide flexibility, there are many such $G^\*$ and consequently it follows that in $V[G]$ there are $2^{2^\kappa}$ many normal measures on $\kappa$. This is how Kunen and Paris proved the consistency of having many normal measures on a measurable cardinal. 


There are many similar further such lifting arguments in many of my papers, but let me point you to: 


* *Hamkins, Joel David*, [**Destruction or preservation as you like it**](http://jdh.hamkins.org/asyoulikeit/), Ann. Pure Appl. Logic 91, No. 2-3, 191-229 (1998). [DOI:10.1016/S0168-0072(97)00044-4](http://dx.doi.org/10.1016/S0168-0072(97)00044-4), [ZBL0949.03047](https://zbmath.org/?q=an:0949.03047), [blog post](http://jdh.hamkins.org/asyoulikeit/).


Using results in that paper, one can show that $\kappa$ is not necessarily the least measurable cardinal in $V[G]$, since it could be that the measurability of some smaller cardinal is turned on by the forcing. 


The general case of $A\in\mu$ is not always possible, since if you control some pattern of the continuum function on a set $A$ of measure one, this will have consequences for the continuum function at $\kappa$ itself in $M$. And so the general rule is that the function $E$ must exhibit certain kinds of self-reference in order to be able to preserve the large cardinal property at $\kappa$. 


Another limitation or negative result is that the consistency strength of $\kappa$ is measurable plus $2^\kappa>\kappa^+$ is strictly stronger than a measurable cardinal. One needs $\kappa$ to be $\kappa^{++}$-tall. (See my paper, 
*Hamkins, Joel D.*, [**Tall cardinals**](http://jdh.hamkins.org/tallcardinals/), [Doi:10.1002/malq.200710084](http://dx.doi.org/10.1002/malq.200710084), Math. Log. Q. 55, No. 1, 68-86 (2009). [ZBL1165.03044](https://zbmath.org/?q=an:1165.03044), [blog post](http://jdh.hamkins.org/tallcardinals/).) So it will be necessary to start with a larger large cardinal situation if this is desired.


Meanwhile, one can achieve a large number of positive instances when one begins with a supercompact cardinal. In many of these cases it is often easier to use the Easton-support iteration rather than the Easton product forcing. 


There is a whole literature on combining Easton's theorem with large cardinals, and perhaps people can post further references. Let me mention a few papers of Brent Cody, who first looked at this topic in his disseration (under my supervision). 


* *Cody, Brent*, [**Easton’s theorem in the presence of Woodin cardinals**](http://dx.doi.org/10.1007/s00153-013-0332-0), Arch. Math. Logic 52, No. 5-6, 569-591 (2013). [ZBL1305.03039](https://zbmath.org/?q=an:1305.03039).
* *Cody, Brent; Friedman, Sy-David; Honzik, Radek*, [**Easton functions and supercompactness**](http://dx.doi.org/10.4064/fm226-3-6), Fundam. Math. 226, No. 3, 279-296 (2014). [ZBL1341.03069](https://zbmath.org/?q=an:1341.03069).
* *Cody, Brent; Gitman, Victoria*, [**Easton’s theorem for Ramsey and strongly Ramsey cardinals**](http://dx.doi.org/10.1016/j.apal.2015.04.006), Ann. Pure Appl. Logic 166, No. 9, 934-952 (2015). [ZBL1371.03064](https://zbmath.org/?q=an:1371.03064).


