Suppose $\kappa$ is measurable, witnessed by a $\kappa$-complete non-principal ultrafilter $\mathcal{U}$ on $\kappa$. Form the ultrapower $j: V \to M = \mathrm{Ult}(V, \mathcal{U})$.

By Los's theorem, $j$ is an elementary embedding: $V \models \varphi(\bar{a}) \iff M \models \varphi(j(\bar{a}))$.

Key properties of $j$:

• $j(\alpha) = \alpha$ for all $\alpha < \kappa$ (since for $\alpha < \kappa$, the constant function $c_\alpha$ represents $\alpha$ in the ultrapower, and non-principality ensures no identification).
• $j(\kappa) > \kappa$ (since the identity function $\mathrm{id}: \kappa \to \kappa$ represents an element $[\mathrm{id}]$ in the ultrapower with $\alpha < [\mathrm{id}] < j(\kappa)$ for all $\alpha < \kappa$).

So $\kappa$ is the critical point of $j$: the least ordinal moved.

Assuming $V = L$: Then $M = L$ as well (since $j: L \to L$ by elementarity). In $L$, $\kappa$ is measurable, so $L \models$ "there exists a measurable cardinal." Let $\kappa$ be the least measurable cardinal in $V = L$.

Since $j$ is elementary and $j(\kappa) > \kappa$: $M \models$ "$j(\kappa)$ is the least measurable cardinal." But $M = L = V$, so $j(\kappa)$ is the least measurable cardinal in $V$. But we assumed $\kappa$ is the least measurable cardinal, and $j(\kappa) > \kappa$. Contradiction. $\blacksquare$