I am looking at the proof that $\kappa$ a cardinal is measurable iff it is the critical point of an elementary embedding from $V$ to an inner model $M$, specifically the forward direction.
I understand the broad strokes of the proof – one uses the ultrafilter on $\kappa$ to create an ultrapower of $V$ itself and use the transitive collapse to find the relevant inner model.
We have that $f\sim g \iff \{\lambda<\kappa:f(\lambda)=g(\lambda)\} \in U$, which we then apply Scott's Trick to, to get the equivalence class $[f]=\{g: g\sim f \land \forall h(h \sim f \implies rank(g) \leq rank(h)\}$. We then define a relation $E$ on the equivalence classes, where $[f]E[g] \iff \{\lambda<\kappa:f(\lambda) \in g(\lambda)\} \in U$.
In order to use the collapse, we require that $E$ be well-founded, extensional, and set-like. I understand why it is well-founded, and why it is set-like. My question is, why is it extensional?
Best Answer
In my opinion it's easiest to think contrapositively rather than trying to prove that $\{[h]: [h]E[f]\}=\{[h]: [h]E[g]\}$ implies $[f]=[g]$ directly.
Suppose $[f]\not=[g]$. WLOG, assume $f(x)\not=g(x)$ for all $x\in\kappa$. Then we can find a sequence $a\in V^\kappa$ such that for all $x$ we have $a(x)\in f(x)\triangle g(x)$. Either $\mathcal{U}$-many $x$ have $a(x)\in f(x)\setminus g(x)$ or $\mathcal{U}$-many $x$ have $a(x)\in g(x)\setminus f(x)$; in the former case we get $[a]E[f]$ but $[a]\not E[g]$, and in the latter case we get $[a]E[g]$ but $[a]\not E[f]$.
NOTE: as per the comments below, we can also prove this via Los' theorem. However, note that the above argument really is the atomic clause of Los' theorem already, so that would be overkill.