From Jech, Set Theory:
Let $\kappa<\lambda$ be cardinals and suppose there exists an elementary embedding $j:V\longrightarrow M$ s.t.
(i) $j(\gamma)=\gamma$ for all $\gamma< \kappa$
(ii) $j(\kappa)>\lambda$
(iii)$M^\lambda\subset M$
Then there exists a fine normal measure $U$ on $P_{\kappa}(\lambda)=\{X\in P(\lambda)\mid |X|<\kappa\}$.
Proof:By (iii) the set $\{j(\gamma)\mid\gamma<\lambda\}$ belongs to $M$ and so the following defines an ultrafilter over $P_{\kappa}(\lambda)$: $$X\in U\iff\{j(\gamma)\mid \gamma<\lambda\}\in j(X)$$
A standard argument shows that $U$ is a $\kappa$-complete ultrafilter. $U$ is a fine measure because for every $\alpha\in \lambda, \{P\mid \alpha\in P\}$ is in $U$.
Now why is the "bold-type" sentence true? I suspect that the hypoteses (i)-(ii) are to be used, but how?
Thank you in advance.
By elementarity of $j$, you have $j(\{P:\alpha\in P\})=\{P:j(\alpha)\in P\}$, and this set contains $\{j(\gamma):\gamma<\lambda\}$ (because $\alpha<\lambda$). So $\{P:\alpha\in P\}\in U$ by definition of $U$.