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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
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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

2 Answers 2

up vote 3 down vote accepted

An example I put in my book ... (p. 155 of 2nd edition, Measure, Topology, and Fractal Geometry Springer 2008)

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.

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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.

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