= 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 $\beta<\alpha$, so
$$
j(\alpha)=\alpha\quad(\alpha<\kappa).
$$
The identity function $\operatorname{id}:\kappa\to\kappa$ represents an <ordinal> below $j(\kappa)$, and every constant $\alpha<\kappa$ is below it modulo $U$, since a tail is large. Consequently
$$
\boxed{\kappa\leq[\operatorname{id}]^{\mathrm{collapse}}<j(\kappa),\qquad
\operatorname{crit}(j)=\kappa.}
$$
Thus \b[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. Define
$$
U=\{A\subseteq\kappa:\kappa\in j(A)\}.
$$
Because $\kappa<j(\kappa)$, 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.
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