Gauging the "size" of measurable cardinals and inaccessible cardinals I have been told that measurable cardinals are much "larger" than ordinary strongly inaccessible cardinals. I have even heard that the former make the latter look "tiny" in comparison. This is difficult for me to visualize.
Is there some way to gauge, intuitively, how much larger the smallest measurable cardinal is than the smallest strongly inaccessible cardinal?
Furthermore, can it be proven that the smallest strongly inaccessible cardinal is not measurable?
 A: Here's one sense in which the least measurable is much larger than the least inaccessible: Let $\kappa$ be the smallest measurable cardinal. Then the set of inaccessible cardinals $<\kappa$ has size $\kappa$. Jech's book has details on this (and much stronger) results.

Let me sketch why the least measurable has to be so big. Let $\kappa$ be measurable, with a $\kappa$-complete measure $\mathcal{U}$. We can consider the ultrapower of the entire universe by this measure, $\prod V/\mathcal{U}$. This might seem iffy, but by Scott's trick this structure can naturally be identified with a definable class, so that the "$\in$"-relation is again definable; so we may talk about this ultrapower in the universe of sets.
The structure $\prod V/\mathcal{U}$ is well-founded, since $\mathcal{U}$ is $\kappa$- (specifically, countably-)complete, so we may take its Mostowski collapse; that is, we have some transitive proper class $M$ such that $(M, \in)$ is isomorphic to $\prod V/\mathcal{U}$ via a map $m$. (Again, I'm sweeping a bunch of technical details under the table, but this all works.) The usual embedding of a structure into its own ultrapower now yields a map $j: V\rightarrow M$; this map is an elementary embedding.
Moreover, this map is nontrivial: $j(\kappa)$ is an ordinal strictly larger than $\kappa$! And $\kappa$ is the least such cardinal moved (the critical point of $j$). To see that $\kappa$ is moved, note that by transfinite induction, for each $\alpha<\kappa$ we have $j(\alpha)=\alpha$; meanwhile, consider the element of the ultrapower $x=m([(\alpha)_{\alpha<\kappa}]_{\mathcal{U}})$, the (Mostowski collapse of) the equivalence class of the identity sequence. By elementarity, $x$ is an ordinal $<j(\kappa)$; but also $x>j(\alpha)$ for every $\alpha<x$, since $\mathcal{U}$ is $\kappa$-complete. So $j(\kappa)>\kappa$.
So what? Well, lots of "small" large cardinal properties - inaccessible, weakly compact, etc. - are $\Pi_1$. If $\mu$ has some $\Pi_1$ property in $V$, then $\mu$ also has that property in any inner model; $\Pi_1$ properties are downwards absolute. So let $\kappa$ be the least measurable. If $\kappa$ were also the least inaccessible, then there would be a contradiction: by elementarity, $M\models$"$j(\kappa$) is the least inaccessible," but by absoluteness $M\models$"$\kappa$ is inaccessible."
Similarly, the least measurable can't be the second weakly compact, . . . , the (least inaccessible)th weakly compact,  . . . measurables are freakin' HUGE! (Except not nearly as huge as, well, huge cardinals. :P)
See Jech's book for more details.
