# Outer measure, Caratheodory measurability

Let X be a nonempty set. If $m^* : \mathcal{P}(X) \rightarrow [0, + \infty]$ is an outer measure, we say that $B \subseteq X$ is $m^*$-measurable if:

$$m^*(A) = m^*(A \cap B) + m^*(A \cap B^c), \forall A \in \mathcal{P}(X)$$

I can't think of an outer measure and a set where this property fails. Can you show me an example? (We defined outer measure as a monotonous, $\sigma$-subadditive function $m^* : \mathcal{P}(X) \rightarrow [0, + \infty]$ which satisfies $m^*(\emptyset) = 0$)

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How about we take $m^*$ to be Lebesgue outer measure and consider some nonmeasurable set? – anonymous Feb 25 '13 at 16:26
You could take $X$ to be a set with more than one element and define $m^*(S)=1$ for any nonempty subset $S$ of $X$. Then no proper subset of $X$ is $m^*$-measurable. – David Mitra Feb 25 '13 at 18:17

For subsets $A$ of $\mathbb R$, define $$m^*(A) = \inf \sum_{k=1}^\infty |I_k|^{1/2}$$ where the infimum is over all countable covers of $A$ by open intervals $I_k$. Here, $|I_k|$ denotes the length of the interval. This is an outer measure.
Then we can show $m^*([0,1]) = m^*((1,2]) = 1$ and $m^*([0,2]) = \sqrt{2}$ so that $[0,1]$ is not measurable.
You can check out chapter 17 (Nonmeasurable and Non-Borel Sets) in the book "The Elements of Integration and Lebesgue Measure" by Robert G. Bartle and/or section 4.6 (Nonmeasurable subsets of $\mathbb{R}$) in "An Introduction to Measure and Integration" by Inder K. Rana. For some examples.