Measurable Subsets + Caratheodory Measurability 1.) What can go wrong if one assigns a measure to more subsets, especially to all subsets? (I would like to understand the subtleties behind)
I imagine the first problem is to give the new subset some number in a unique way. There, however, one could just try to choose one of several possible choices. But what happens next? Is it that countable additivity cannot hold anymore?
2.) How come that Caratheodory's measurability argument works?
(I would like to understand the reason behind we restricting to these subsets make the outer measure become a measure)
When I think of this problem I can imagine that picking another sigma algebra could turn the outer measure into a measure as well. Clearly all subalgebras of the one chosen by Caratheodory will make it too. But are there more than these?
For example, pick a vitali set and all rational translates of it. Take their outer measure. This should constitute another sigma algebra different from the one by Caratheodory's construction.
 A: If $\bar \mu$ is the outer lebesgue measure, there is a subset $A \subset [0,1]$ such that $$
  \bar\mu(A) = \bar\mu([0,1] \setminus A) = 1 = \bar\mu([0,1]) \text{.}
$$
That breaks the requirement that if $\mu$ is a measure and $X,Y$ disjoint subsets you always have $$
  \mu(X) + \mu(Y) = \mu(X \cup Y) \text{.}
$$
(Just set $X=A$, $Y = [0,1] \setminus A$, then $X,Y$ are clearly disjoint and $X \cup Y = [0,1]$)
Thus, the outer measure $\bar\mu$ is not a measure. The problem is, as the above shows, that $\bar\mu$ is not additive. Caratheodory's avoids that by allows only those sets $A$ which split every set $S$ into two parts - $S \cap A$ and $S \cap A^C$, such that the sum of the outer measures of these two parts is equal to the outer measure of $S$. 
Pick two such allowed and disjoint sets $X,Y$. Set $S = X \cup Y$. Then $$
  \bar\mu(X\cup Y) = \bar\mu(S) = \bar\mu(\underbrace{S \cap X}_{=X}) + \bar\mu(\underbrace{S \cap X^C}_{=Y})
  = \bar\mu(X) + \bar\mu(Y) \text{,}
$$
which shows that for allowed sets, $\bar\mu$ is indeed additive.
A: Let $\Omega$ be a non- empty set. We define the Borel $\sigma$- algebra $B(\Omega)$ to be the intersection of all $\sigma$- algebras that contain the topology of $\Omega$.
Also let $M$ be the set of all measurables sets and $P(\Omega)$ be the power set of $\Omega$.
Then $B(\Omega)\subset M\subset P(\Omega)$ and also $B(\Omega)\neq M$,$M\neq P(\Omega)$.
So,for your first question we have Vitali's argument,that there exists a non measurable set http://en.wikipedia.org/wiki/Vitali_set.
As for the second question we wanted the additivity. For example in the reals (where you can imagine it) we want $m(A\cup B)=m(A)+m(B) $ for measurable sets such that $A\cap B=\emptyset$. 
First,we show that $m^{*}(X\cap(A\cup B))=m^{*}(Χ\cap A)+m^{*}(X\cap B) $. Then we show it for infinite many $A_i$ and after that we take $X=\Bbb R$ and we have the infinite additivity.
For me Caratheodory thought it vice versa. However it's hard to imagine it and somehow this definition came from the sky:P
