Outer Measure on $\mathbb R$ with trivial measurable sets This question arises from wondering how bad an outer measure can be. Some authors prefer to work with general outer measures (e.g. Evans and Gariepy), and it would be nice to be able to think of outer measures as just extensions of measures. As in every outer measure induces a measure and every measure induces an outer measure, and it would be nice if these were inverse to each other, though in general they are not. There are of course conditions (e.g. regularity) that allow you to know that this is the case, but one can wonder how nasty a general outer measure can be.
One way an outer measure can be bad is not having enough measurable sets. Hence the question is whether there is an example of the most extreme version of this (on $\mathbb R$ and maybe $\mathbb R^n$ if it's interesting): does there exist an outer measure on $\mathbb R$ such that the only measurable sets (under the Caratheodory criterion) are $\mathbb R, \emptyset$?
 A: Yes there is! Consider the function $\mu^{*}:P(\mathbb{R})\to [0,\infty)$ given by
$$\mu^{*}(A)= \begin{cases} 1, & \text{if} \hspace{2mm} A\neq \emptyset, \\
0, & \text{if} \hspace{2mm} A=\emptyset .
  \end{cases}$$
It is easy to see that $\mu^{*}$ is an outer measure on $\mathbb{R}$. Now, I claim that the only two subsets of $\mathbb{R}$ satisfying the Carathéodory criterion are $\mathbb{R}$ and $\emptyset$. Clearly, if we let $E=\mathbb{R}$ or $E=\emptyset$ then, for every $A\subseteq \mathbb{R}$, the following equality is true $$\mu^{*}(A)= \mu^{*}(A\cap E) + \mu^{*}(A\setminus E).$$
Conversely, suppose that $\emptyset\neq E \subset \mathbb{R}$ (here the symbol $\subset$ means proper inclusion). Let $a\in E$, $b\in \mathbb{R}\setminus E$ and put $A=\{a,b\}$. Then, $$\mu^{*}(A)= 1 \neq 2 = 1+1 =\mu^{*}(A\cap E) + \mu^{*}(A\setminus E).$$
As a final comment, if we consider any set $X$ with at least two elements, then the corresponding $\mu^{*}$ produces an outer measure on $X$ such that the only $\mu^{*}$-measurable sets are the empty set and $X$ itself.
