I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
The following exercise is Exercise 3 on p.45 in Exercises 2C in this book.
Exercise 3
Give an example of a measure $\mu$ on $(\mathbb{Z}^+,2^{\mathbb{Z}^+})$ such that $$\{\mu(E):E\subset\mathbb{Z}^+\}=[0,1].$$
I solved Exercise 3.
But I am not sure if my solution is ok or not.
I used the following Example 2.55 on p.42 in section 2C in this book to solve Exercise 3.
- Suppose $X$ is a set, $\mathcal{S}$ is a $\sigma$-algebra on $X$, and $w:X\to [0,\infty]$ is a function.
Define a measure $\mu$ on $(X,\mathcal{S})$ by $$\mu(E)=\sum_{x\in E} w(x)$$ for $E\in\mathcal{S}.$ [Here the sum is defined as the supremum of all finite subsums $\sum_{x\in D} w(x)$ as $D$ ranges over all finite subsets of $E.$]
My solution is here:
Let $w:\mathbb{Z}^+\to [0,\infty]$ be a function such that $w(i):=\frac{1}{2^i}$ for each $i\in\mathbb{Z}^+.$
Define a measure $\mu$ on $(\mathbb{Z}^+,2^{\mathbb{Z}^+})$ by $$\mu(E)=\sum_{x\in E} w(x)$$ for $E\in 2^{\mathbb{Z}^+}.$
Then, by Example 2.55 on p.42 in section 2C in the book, $\mu$ is really a measure on $(\mathbb{Z}^+,2^{\mathbb{Z}^+}).$
$\mu(\mathbb{Z}^+)=\sum_{x\in \mathbb{Z}^+} w(x)=1.$
So, $0\leq\mu(E)\leq\mu(\mathbb{Z}^+)=1$ for $E\subset\mathbb{Z}^+.$
Let $a\in [0,1].$
Then we can write $a=a_1w(1)+a_2w(2)+\cdots$, where $a_i\in\{0,1\}$ for each $i\in\{1,2,\dots\}.$
Let $E_a:=\{i\in\{1,2,\dots\}:a_i=1\}.$
Then, $E_a\subset\mathbb{Z}^+$ and $\mu(E_a)=a.$
Therefore, $\{\mu(E):E\subset\mathbb{Z}^+\}=[0,1]$ holds.