Is there an outer measure on $\mathbb R$ whose only measurable sets are $\mathbb R, \emptyset$?

Question

I.e. an outer measure $\mu^*$ on the reals such that if for some $A \subset \mathbb R$, we have that for all $B \subset R$ $\mu^*(B) = \mu^*(B \cap A) + \mu^*(B \setminus A)$, then $A=\mathbb R$ or $A = \emptyset$.

Answer 1

Define $\mu^*$ by $\mu^*(\emptyset) = 0$ and $\mu^*(A) = 1$ for all $A \ne \emptyset$. It is trivial to check that this is an outer measure.

Now if $A$ is a set which is neither $\emptyset$ nor $\mathbb{R}$, taking $B = \mathbb{R}$, we have that $B, B\cap A$ and $B \setminus A$ are all nonempty, so $\mu^*(B) = 1$ and $\mu^*(B \cap A) + \mu^*(B \setminus A) = 2$. Hence $A$ is not measurable.

The same works if you replace $\mathbb{R}$ by any other set.