Usually if I have a outer measure $\mu^*:\mathbb R^n \to \mathbb R$ induced by a measure $\mu:\mathcal R \to \mathbb R$ where $\mathcal R$ denotes a ring over $\mathbb R^n$ we define the $\mu^*$-measurable sets as those sets $A$ for which $Q \subset \mathbb R^n \Rightarrow \mu^*(Q)= \mu^*(Q\cap A)+ \mu^*(Q\cap A^c)$. But I just read that if $\mu^*$ is finite, a set A is measurable if and only if $\mu^*(\mathbb R^n)=\mu^*(A)+\mu^*(A^c)$. How could I show this? (Any hints would also be welcome)
Equivalent definition of the measurable sets of a finite outer measure
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real-analysis
measure-theory
1 Answers
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Because R^n is of sigma-finite measure.
This definitions works on any space X on which there is an outer measure \mu^* induced from a sigma-finte premeasure.
For any E\subset X if \mu^(X) = \mu^(E) + \mu^*(E^C)
then E is \mu^* measurable