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Let $\mu^{\ast}$ be the outer measure on $R$.A collection $\left\{A_i\right\}$ is a partition of $R$ if $A_i \cap A_j=\phi$ if $i\neq j$ and $\bigcup^\infty_{i=1} A_i=R$. Prove that all sets on the partition $\left\{A_i\right\}$ are measurable if and only if $\mu^\ast (B)=\sum^\infty_{i=1} \mu^\ast (B\cap A_i)$ for all subset $B$ in $R$.

I think the relationship can be proved from the definition of measurable sets directly. But I don't know how to do it. It seems not that trivial.

Michael Hardy
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Jack
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    In the most simple case where the partition is ${ A, A^c }$ for some set $A$, this statement is precisely the Caratheodory criterion for measurability (which is sometimes taken as the definition of measurability). See also this post for a proof. Maybe your more general statement can be deduced from this particular case. – Derived Cats Jul 24 '16 at 05:09

1 Answers1

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Forward implication:

First,

$$\mu^* (B) = \mu^*\left( B \cap \bigcup_{i=1}^{\infty} A_i \right) = \mu^* \left( \bigcup_{i=1}^{\infty} (B \cap A_i) \right) \le \sum_{i=1}^{\infty} \mu^*(B \cap A_i)$$

Now, by the definition of measurability, we can show by induction that for all $n$,

$$\mu^*(B) = \sum_{i=1}^n \mu^*(B \cap A_i) + \mu^* \left( \bigcup_{i=n+1}^{\infty} B \cap A_i \right) \ge \sum_{i=1}^n \mu^*(B \cap A_i)$$

Hence $\mu^*(B) \ge \sum_{i=1}^{\infty} \mu^*(B \cap A_i)$.

Backward implication:

Let $i \in \Bbb N$. Let $B \subset \Bbb R$. It suffices to show that $\mu^*(B) \ge \mu^*(B \cap A_i) + \mu^*(B \setminus A_i)$. We have:

$$\mu^*(B) = \sum_{j=1}^{\infty} \mu^*(B \cap A_j) = \mu^*(B \cap A_i) + \sum_{j \neq i} \mu^*(B \cap A_j) \\ \ge \mu^*(B \cap A_i) + \mu^* \left( \bigcup_{j\neq i} (B\cap A_j) \right) = \mu^*(B \cap A_i) + \mu^*\left(B \cap \left(\bigcup_{j \neq i} A_j \right) \right) \\ = \mu^*(B \cap A_i) + \mu^*(B \cap (\Bbb R \setminus A_i)) = \mu^*(B \cap A_i) + \mu^*(B \setminus A_i)$$