(1) I am searching for a non-technical explanation linking one or more of the properties a measurable cardinal has by definition with that fact that there is no second-order property which holds for the first time at a measurable cardinal.
For example, the definition that looks most promising is the one in terms of elementary embeddings: $\kappa$ is measurable if it is the critical point of a non-trivial elementary embedding of the universe V into a transitive class M.
However, the embedding only guarantees a mapping of $first$-order formulas, so I can't see how second-order reflection properties could enter the picture.
(2) Is a measurable the smallest of the large cardinals with this property?
Best Answer
The reason is that if $j\colon V\to M$ is an elementary embedding with critical point $\kappa$, then $\mathcal P(\kappa)^V=\mathcal P(\kappa)^M$. They have the same power sets.
Now. A second-order property of $\kappa$, in fact of $V_\kappa$, is really just a first-order property when you have the entire power set. And the fact $V_{\kappa+1}=M_{\kappa+1}$, or $\mathcal P(\kappa)^V=\mathcal P(\kappa)^M$, means that the two models agree on the second-order properties of $\kappa$ and $V_\kappa$.
But now, being the least cardinal with a second-order property is a first-order property in $V$, so $j(\kappa)$ is the least one in $M$. But oh-uh, $V$ and $M$ agree on second-order properties of $\kappa$!