Does the Carathéodory definition of measurable set encompass the most sets possible?
By this I mean, are all the sets that are well behaved enough with pre-measure that they have all the properties that we want completely given by the Carathéodory criterion of measurability?
It's reasonable to require that measurable sets cut up measurable sets into components that sum up to the measure of the original set but it's not obvious why all sets need to be cut up in this way or why this is the most general definition of measurable set.
Why couldn't lebesgue measurable sets be some proper subset of an even less restrictive definition?
In some sense, you are asking the wrong question. We don't want the $\sigma$-algebra of Lebesgue measurable sets to encompass too many sets - in fact, we prefer as few sets as reasonably permissible to be Lebesgue measurable. The Lebesgue measurability criterion is designed to exclude bad sets. The real question should be, "Does the definition of Lebesgue measurability exclude enough bad sets to be useful?"
The most important properties that we want out of Lebesgue measurability are the following:
A product of bounded intervals should be measurable. Makes sense, since it's obvious how to define the volume of a $n$-dimensional box.
There should be a reasonable definition of measure so that if we translate a measurable set, it shouldn't change the measure.
The Lebesgue measurable sets encompass all products of intervals but not too many more (they form a minimal good $\sigma$-algebra containing products of intervals in some sense), and Lebesgue measure is a complete translation-invariant measure on the Lebesgue measurable sets. In fact, even better: one can prove it is the unique such measure. And as it turns out, it has several nice intuitive properties in addition: see http://en.wikipedia.org/wiki/Lebesgue_measure#Properties. This is what I would consider a positive answer to the question I raised in the first paragraph.
In principle, one could easily come up with a less restrictive definition of measurable set. Take any non-measurable set, and use it and the Lebesgue $\sigma$-algebra to generate a new $\sigma$-algebra. But then, of course, the Carathéodory criterion no longer holds by fiat, and that's something we'd really like.
The best reason I see to assume the Carathéodory criterion is that the subsets in a measurable space that satisfy the criterion automatically form a $\sigma$-algebra, so it actually makes sense to use it to define a $\sigma$-algebra. (In particular, you use it to obtain countable additivity.) There are probably measure spaces that don't use the Carathéodory criterion to define measurability, but at least for Euclidean space the Carathéodory criterion makes too much sense to replace it with a weaker criterion for measurability.