[proofplan]
The proof is an unpacking of the semantics of the statement $W \models ``\kappa \text{ is measurable}"$. Measurability is expressed by an existential first-order formula asserting the existence of a nonprincipal $\kappa$-complete ultrafilter on $\kappa$. Since existential quantifiers in the structure $W$ range only over elements of $W$, any witness to this internal assertion must itself belong to $W$. Therefore a normal measure existing only externally in $V$ cannot serve as the internal witness for $W$, although the argument does not exclude the possibility of a different measure lying in $W$.
[/proofplan]
[step:Unpack the internal existential statement defining measurability]
Let $\Theta(X,\alpha)$ denote the first-order formula in the language of set theory which says that $X$ is a nonprincipal $\alpha$-complete ultrafilter on $\alpha$. With this notation, the assertion
$W \models ``\kappa \text{ is measurable}"$
means precisely that
\begin{align*}
W \models \exists X\, \Theta(X,\kappa).
\end{align*}
By the semantics of first-order satisfaction for the structure $W$, the displayed existential statement holds exactly when there is some set $U \in W$ such that
\begin{align*}
W \models \Theta(U,\kappa).
\end{align*}
Therefore there exists $U \in W$ such that $W$ satisfies that $U$ is a nonprincipal $\kappa$-complete ultrafilter on $\kappa$.
[guided]
We first separate the external notation from the internal assertion. Let $\Theta(X,\alpha)$ be the first-order formula which expresses: "$X$ is a nonprincipal $\alpha$-complete ultrafilter on $\alpha$." This formula has two free variables, $X$ and $\alpha$, and it is interpreted inside whatever model is being discussed.
The hypothesis is
\begin{align*}
W \models ``\kappa \text{ is measurable}".
\end{align*}
Using the ultrafilter formulation of measurability, this means
\begin{align*}
W \models \exists X\, \Theta(X,\kappa).
\end{align*}
Now we use the defining semantic rule for the existential quantifier in the structure $W$: an assertion of the form $W \models \exists X\, \Theta(X,\kappa)$ is true exactly when some element of the domain of $W$ can be assigned to $X$ and makes $\Theta(X,\kappa)$ true in $W$. Since the domain of the structure is $W$ itself, the witness must be an element of $W$. Thus there is a set $U \in W$ such that
\begin{align*}
W \models \Theta(U,\kappa).
\end{align*}
Expanding the meaning of $\Theta$, this says exactly that $W$ satisfies that $U$ is a nonprincipal $\kappa$-complete ultrafilter on $\kappa$. The important point is not any special property of ultrafilters; it is the existential quantifier. Internal existence in $W$ requires an internal witness belonging to $W$.
[/guided]
[/step]
[step:Exclude external measures from being internal witnesses]
Let $\mu \in V$ be a normal measure on $\kappa$ in the ambient universe $V$, and assume $\mu \notin W$. Since every witness to
\begin{align*}
W \models \exists X\, \Theta(X,\kappa)
\end{align*}
must be an element of $W$, the set $\mu$ cannot be assigned to the quantified variable $X$ in the satisfaction relation for $W$. Hence $\mu$ cannot witness, inside $W$, that $\kappa$ is measurable.
This conclusion is only a witness-membership obstruction. It does not imply that $W$ has no measure on $\kappa$; it says only that the particular external measure $\mu \notin W$ is unavailable as an internal witness.
[/step]
[step:Specialize the obstruction to the constructible universe]
Apply the preceding argument with $W = L$. If $\mu$ is a normal measure on $\kappa$ in $V$ and $\mu \notin L$, then $\mu$ is not among the sets over which the existential quantifier in $L \models \exists X\, \Theta(X,\kappa)$ ranges. Therefore this particular external normal measure cannot be reused to prove, inside $L$, that $\kappa$ is measurable.
Thus measurability in $V$ is not inherited by $L$ merely by carrying over a normal measure which lies outside $L$. This proves the stated claim.
[/step]
