Measurable Cardinals are Mahlo Cardinals I am new to set theory and have been working through the proof that every measurable cardinal is Mahlo on page 135 of Jech's text. With the help of Asaf's comments (Measurable $\rightarrow$ Mahlo), I have been able to make sense of the first half of the proof. 
However, I found the second half (that argues by contradiction that $\{\alpha < \kappa : \alpha \text{ is regular} \} \in D$) quite terse, and cannot quite follow what is going on. 
Could someone please provide a detailed version of Jech's proof or perhaps a detailed alternative proof (that mimics the proof that every measurable cardinal is inaccessible, which I believe I understand better). 
Thank you in advance for your help. 
 A: Here is a proof leveraging ultrapowers. It generalizes well to all sorts of situations which is why I recommend learning it at some point:
In the following let $U$ be a normal ultrafilter on a measurable cardinal $\kappa$ and let
$$
\pi \colon V \to \mathrm{Ult}(V;U)
$$
be the canonical ultrapower embedding (we regard $\mathrm{Ult}(V;U)$ as transitive).
Claim. Let $X \subseteq \kappa$. Then $\pi(X) \cap \kappa = X$.
Proof. For $\xi < \kappa$ we have
$$
\xi \in X \iff \pi(\xi) = \xi \in \pi(X).
$$
Q.E.D.
Claim. Let $C \subseteq \kappa$ be a club. Then $\kappa \in \pi(C)$.
Proof. By elementarity
$$
\mathrm{Ult}(V;U) \models \pi(C) \text{ is a club in } \pi(\kappa)
$$
and $\pi(C) \cap \kappa = C$ is unbounded below $\kappa < \pi(\kappa)$.
Thus
$$
\mathrm{Ult}(V;U) \models \kappa \in \pi(C)
$$ 
and (by $\Sigma_0$-absoluteness) hence $\kappa \in \pi(C)$. Q.E.D.
Claim. $\mathrm{Ult}(V;U) \models \kappa \text{ is regular}$.
Proof. $\kappa$ is regular in $V$, $\mathrm{Ult}(V;U) \subseteq V$ and regularity is downward-absolute (a short cofinal sequence in $\mathrm{Ult}(V;U)$ would also witness in $V$ that $\kappa$ is singular). Q.E.D.
Now combine all of this:
Let $C \subseteq \kappa$ be a club. Then
$$
\mathrm{Ult}(V;U) \models \kappa \in \pi(C) \text{ and } \kappa \text{ is regular }.
$$
In particular
$$
\mathrm{Ult}(V;U) \models \pi(C) \text{ contains a regular cardinal}.
$$
By the elementarity of $\pi$ we obtain that
$$
V \models C \text{ contains a regular cardinal}.
$$
Since $C$ was an arbitrary club in $\kappa$, it follows that $\kappa$ is Mahlo.
A: Here is a detailed explanation of Jech's proof.
Let $D$ be a normal measure on $\kappa$. Suppose, towards a contradiction, that $\kappa$ is not Mahlo. Then there is some club $C \subseteq \kappa$ such that
$$
C \cap \{ \alpha < \kappa \mid \mathrm{cof}(\alpha) = \alpha \} = \emptyset.
$$
Since $D$ is normal, it contains all clubs. In particular $C \in D$. Since $D$ is closed under intersections, we therefore must have that $\{ \alpha < \kappa \mid \mathrm{cof}(\alpha) = \alpha \} \not \in D$ and hence that
$$
\{ \alpha < \kappa \mid \mathrm{cof}(\alpha) < \alpha \} = \kappa \setminus \{ \alpha < \kappa \mid \mathrm{cof}(\alpha) = \alpha \} \in D.
$$
By normality there is some $\lambda < \kappa$ such that
$$
E_\lambda = \{ \alpha < \kappa \mid \mathrm{cof}(\alpha) = \lambda \} \in D.
$$
By replacing $E_\lambda$ with $E_\lambda \setminus \lambda$ we may and shall assume that $E_\lambda \cap \lambda = \emptyset$
For each $\alpha \in E_\lambda$ fix a strictly increasing, cofinal function
$$
f_\alpha \colon \lambda \to \alpha
$$
Now, for each $\xi < \lambda$, the function
$$
g_\xi \colon E_\lambda \to \kappa, \ \alpha \mapsto f_\alpha(\xi)
$$
is decreasing. Hence there is some $A_\xi \in D$ and some $y_\xi < \kappa$ such that $f_\alpha(\xi) = y_\xi$ for all $\alpha \in A_\xi$.
Let 
$$
A = \bigcap_{\xi < \lambda} A_\xi.
$$
Since $\lambda < \kappa$ we have that $A \in D$.
Now let $\alpha \in A$. For all $\xi < \lambda$ we have $f_\alpha(\xi) = y_\xi$ is independent of $\alpha$ (by the construction of $A_\xi$). But
$$
\alpha = \sup_{\xi < \lambda} f_\alpha(\xi) = y_\xi
$$
is completely determined by the sequence $(y_\xi \mid \xi < \lambda)$.
Hence $A$ contains at most one element. This is a contradiction, since $D$ is non-principal.
It follows that $\kappa$ is Mahlo after all!
