Proof of every measurable cardinal carries a normal measure I'm reading the proof of Theorem 10.20 in Set Theory by Jech and I don't understand the last argument.
The theorem says every measurable cardinal carries a normal measure. The proof goes: Let $U$ be a nonprincipal $\kappa$-complete ultrafilter on $\kappa$. For $f$ and $g$ in $\kappa^\kappa$, define $f<g$ if and only if $\{\alpha<\kappa:f(\alpha)<g(\alpha)\}\in U$. Now let $f:\kappa\rightarrow\kappa$ be the least function with the property that for all $\gamma<\kappa$, $\{\alpha:f(\alpha)>\gamma\}\in U$. Let $D=\{X\subset\kappa: f^{-1}(X)\in U\}$.
Then we claim $D$ is a normal measure: Let $h$ be a regressiv efunction on a set $X\in D$. Let $g$ be the function defined by $g(\alpha)=h(f(\alpha))$. As $g(\alpha)<f(\alpha)$ for all $\alpha\in f^{-1}(X)$, we have $g<f$. Here is the part I don't understand: It follows by minimality of $f$ that $g$ is constant on some $Y\in U$. Hence $h$ is constant on $f(Y)$ and $f(Y)\in D$.
Why does it follow from the minimality of $f$ that $g$ is constant on some $Y\in U$? And $h$ is constant on $f(Y)$ and $f(Y)\in D$?

By minimality of $f$, we have $$\{\alpha:g(\alpha)\le\gamma\}\in U $$ for some $\gamma,$ so by $\kappa$-completeness, there is a $\beta \le \gamma$ such that $\{\alpha:g(\alpha)=\beta\}\in U.$ 
(Your other questions can be answered by straightforward definition chasing. For instance that $h$ is constant on $f(Y)$ is a trivial consequence of $g$ being constant on $Y.$)
