The following argument appears in the proof of Theorem 4.7. in Bagaria-Magidor's paper Group radicals and strongly compact cardinals.
Let $\delta<\kappa$ be uncountable cardinals which may be singular and let $\alpha$ be an ordinal such that $\alpha\geq\kappa$. Suppose there exists a $\delta$-complete fine measure $\mathcal{U}$ on $\mathcal{P}_{\kappa}(\alpha)$, that is, a $\delta$-complete ultrafilter $\mathcal{U}$ on $\mathcal{P}_{\kappa}(\alpha)=\{x\subseteq\alpha:|x|<\kappa\}$ such that $\{x\in\mathcal{P}_{\kappa}(\alpha):a\in x\}\in\mathcal{U}$ for every $a\in\alpha$. Let $j_{\mathcal{U}}:V\longrightarrow Ult(V,\mathcal{U})$ be the corresponding ultrapower embedding. Since $\mathcal{U}$ is $\delta$-complete, then $Ult(V,\mathcal{U})$ is well-founded. Moreover, also by $\delta$-completeness, the critical point of $j_{\mathcal{U}}$ is greater than or equal to $\delta$. Now my question:
Why $Ult(V,\mathcal{U})\vDash|[id]_{\mathcal{U}}|<j_{\mathcal{U}}(\kappa)$?
Thanks in advance.
(I would have added the tag ultrapowers if it existed, but it does not and I have no reputation to create it).
Best Answer
Remember that $j_{\mathcal U}(\kappa)$ is (the image under transitive collapse of) the equivalence class in the ultrapower of the constant function $c$ with value $\kappa$. So, by Los's theorem, what needs to be proved is that $|id_{\mathcal U}(a)|<c(a)$ for $\mathcal U$-almost all $a\in\mathcal P_\kappa(\alpha)$. That is, $|a|<\kappa$ for almost all $a$. But this inequality is in fact true for all $a\in\mathcal P_\kappa(\alpha)$, by definition of $\mathcal P_\kappa(\alpha)$.