# Which of the following is an outer measure?

by 1256   Last Updated January 16, 2018 06:20 AM

Which of the following is an outer measure

Let $X$ be any set and $\mu ^* : P(X)\rightarrow [0, \infty)$ be given by

1) $$\mu ^*(A)= \begin{cases} 0,& \text{if } A=\phi\\ 1, & \text{otherwise} \\ \end{cases}$$

2) $$\mu ^*(A)= \begin{cases} 0,& \text{if } A & \text{is countable}\\ 1, & \text{otherwise} \\ \end{cases}$$

3) $$\mu ^*(A)= \begin{cases} 0,& \text{if } A & \text{is finite}\\ 1, & \text{otherwise} \\ \end{cases}$$

I checked first two properties namely $\mu^*(A)\geq0$ for all $A$ and $\mu^*(\phi)=0$ and $A \subset B$ then $\mu^*(A) \leq \mu^*(B)$. My problem on showing countable subaditivity for all three case. Please help

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For #1, note that if $\bigcup_{i=1}^{\infty} A_{i} = \emptyset$, then each $A_{i} = \emptyset$. Clearly, $\mu^{*} \left( \bigcup_{i=1}^{\infty} A_{i} \right) = 0$ and $\sum_{i=1}^{\infty} \emptyset = 0$. If your countable union is not empty, then the measure of the union will be 1, whereas the sum of the measures will be at least 1.

For #2, you can use a similar argument, but replace "empty" with "countable".

For #3, one can easily union a bunch of finite sets to get an infinite set, no? So, you can get that the measure of the union of these finite sets is 1, whereas the sum of the measures of these finite sets will be 0.

Dan
January 16, 2018 06:17 AM