Problem 10.5 from Set Theory, Jech: A measure $U$ on $\kappa$ is normal if and only if the diagonal function $d(\alpha)=\alpha$ is the least function $f$ with the property that for all $\gamma<\kappa$, $\{\alpha:f(\alpha)>\gamma\}\in U$.
(Here, for $f$ and $g$ in $\kappa^\kappa$, let $f\equiv g$ if and only if $\{\alpha<\kappa:f(\alpha)=g(\alpha)\}\in U$. We define $<$ on the equivalence classes by letting $f<g$ if and only if $\{\alpha<\kappa:f(\alpha)<g(\alpha)\}\in U$. A normal measure is a nonprincipal $\kappa$-complete ultrafilter closed under diagonal intersections.)
For the $\Rightarrow$ direction, to show that $d$ is the least function with such a property, we need to show $d<f$ for all such functions $f$, i.e. $\{\alpha<\kappa:d(\alpha)<f(\alpha)\}=\{\alpha<\kappa:\alpha<f(\alpha)\}\in U$. How do I show this?
Otherwise the complement $\{\alpha<\kappa:f(\alpha)\le\alpha\}\in U$. Since we have $\{\alpha<\kappa:f(\alpha)=d(\alpha)=\alpha\}\not\in U$, it follows that $\{\alpha<\kappa:f(\alpha)<\alpha\}\in U$. Now you can apply normality and get that $f$ must be constant (and hence less than $d$) on some measure one set.
Edit:
It's easy to get lost in all these set definitions, but really the line of reasoning is as follows: $d$ is almost always bigger than each individual ordinal. If $f$ is less than $d$, then $f$ is almost always regressive and hence almost always constant. But $d$ is almost always bigger than that constant value, so $d$ is almost always bigger than $f$.
Explicitly, once we have $d>f$, we have $\{\alpha<\kappa:d(\alpha )>f(\alpha )\}\in U$. This is the same set as $\{\alpha <\kappa:\alpha >f(\alpha )\}\in U$. What this means is that $f$ is regressive on a set in $U$. Normality implies $f$ is constant on a(nother) set in $U$, i.e. for some $\beta $, $\{\alpha <\kappa:f(\alpha )=\beta \}\in U$. This will imply $f<d$, a contradiction.
Consider all the ordinals after $\beta$, $\{\alpha <\kappa:\beta <\alpha \}$, i.e. $\kappa\setminus (\beta +1)$. This set is in $U$ by $\kappa$-completeness (and non-principality: each $\{\alpha\}\not\in U$ so that the union $\beta+1=\bigcup_{\alpha\le\beta}\{\alpha\}\not\in U$ and hence the complement $\kappa\setminus(\beta+1)\in U$). This set is equal to $\{\alpha <\kappa:\beta <d(\alpha )\}\in U$. Thus we can intersect to get another set in $U$: $$\{\alpha <\kappa:f(\alpha )=\beta \}\cap\{\alpha <\kappa:\beta <d(\alpha )\}\in U\text{.}$$ The set above is a subset of $\{\alpha <\kappa:f(\alpha )<d(\alpha )\}\in U$ so that $f<d$.