 Solution

A measurable cardinal is an uncountable cardinal $\kappa$ carrying a nonprincipal $\kappa$-complete ultrafilter $U$: intersections of fewer than $\kappa$ members of $U$ remain in $U$. Such an ultrafilter contains no set of size less than $\kappa$. Indeed each singleton is absent, and intersecting fewer than $\kappa$ singleton complements shows that every small set's complement is in $U$. In particular every tail of $\kappa$ is in $U$.

Form the ultrapower of the universe by functions $f:\kappa \to V$, identifying functions agreeing on a member of $U$, and interpreting membership coordinatewise modulo $U$. The Łoś theorem, applied separately to each formula, makes the map $x\mapsto [c_x]$ elementary. Countable completeness makes the ultrapower well-founded: an infinite descending chain $[f_{n+1}]\in [f_n]$ would give $U$-large sets witnessing each step, whose countable intersection is nonempty; a coordinate in that intersection would give an infinite descending membership chain in the original universe. The relation is extensional and set-like, so the Mostowski collapse theorem gives a transitive class $M$ and an elementary ultrapower embedding $j:V\to M$.

For each $\alpha <\kappa$, every function $\kappa \to \alpha$ is constant modulo $U$. If no fibre were in $U$, intersecting the fewer than $\kappa$ fibre complements would give the empty set in $U$. Induction on $\alpha$ therefore shows that the collapsed ultrapower ordinal $j(\alpha )$ has precisely the elements $\beta <\alpha$, soThe identity function $\mathrm{id}:\kappa \to \kappa$ represents an ordinal below $j(\kappa )$, and every constant $\alpha <\kappa$ is below it modulo $U$, since a tail is large. ConsequentlyThus the canonical collapsed embedding is not the identity; its first moved ordinal is $\kappa$. For a general measure the identity-function class need not collapse to exactly $\kappa$; that stronger equality holds for a normal measure. The argument above needs only its lower bound.

Conversely, suppose $j:V\to M$ is elementary, $M$ is transitive, its critical point is $\kappa$, and its derived measure is a set, as for the usual definable or amenable class embeddings. DefineBecause $\kappa <j(\kappa )$, elementary preservation of complements and intersections makes this an ultrafilter. It is nonprincipal since $j(\{\alpha \})=\{\alpha \}$ for $\alpha <\kappa$. For a sequence of $\gamma <\kappa$ large sets, $j$ fixes the index ordinal $\gamma$ and takes the intersection to the intersection of their images; $\kappa$ belongs to every image and hence to the intersection. Thus $U$ is $\kappa$-complete. The set-existence qualification excludes treating an arbitrary external class map as a set parameter without justification.

For context, measurability already implies regularity and the strong-limit property. A cofinal sequence of length below $\kappa$ partitions $\kappa$ into that many bounded, hence small, pieces; completeness and their absent fibres contradict covering all of $\kappa$. If there were $\kappa$ distinct subsets of some $\lambda <\kappa$, for each coordinate choose the $U$-large side deciding its membership bit. The intersection of these $\lambda$ large sets would still be large, but all its indices would label the same subset, so it would have at most one member. This contradiction gives $2^{\lambda }<\kappa$. Thus the first moved ordinal is an uncountable strongly inaccessible cardinal, in addition to being measurable.