I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
The following exercise is Exercise 6 on p.46 in Exercises 2C in this book.
Exercise 6
Find all $c\in [3,\infty)$ such that there exists a measure space $(X,\mathcal{S},\mu)$ with $$\{\mu(E):E\in\mathcal{S}\}=[0,1]\cup [3,c].$$
I solved Exercise 6, but I am not sure if my solution is ok or not.
Is my solution ok?
My solution is here:
$$T:=\{c\in [3,\infty):\text{ there exists a measure space }(X,\mathcal{S},\mu)\text{ with }\{\mu(E):E\in\mathcal{S}\}=[0,1]\cup [3,c]\}.$$
Then, $T=\{4\}.$
Proof:
Let $c\in T.$
Then, there exists a measure space $(X,\mathcal{S},\mu)$ with $$\{\mu(E):E\in\mathcal{S}\}=[0,1]\cup [3,c].$$
Since $\mu(E)\leq\mu(X)$ for $E\in\mathcal{S}$, $\mu(X)=c.$
There exists $E_3\in\mathcal{S}$ such that $\mu(E_3)=3.$
There exists $E_1\in\mathcal{S}$ such that $\mu(E_1)=1.$
Since $E_3=(E_3\setminus E_1)\cup (E_3\cap E_1)$ and $(E_3\setminus E_1)\cap (E_3\cap E_1)=\emptyset$, $\mu(E_3)=\mu(E_3\setminus E_1)+\mu(E_3\cap E_1).$
Since $E_3\setminus E_1\in\mathcal{S}$ and $E_3\setminus E_1\subset E_3$, $\mu(E_3\setminus E_1)\leq 3.$
So, $\mu(E_3\setminus E_1)\in [0,1]\cup\{3\}.$
If $\mu(E_3\setminus E_1)\in [0,1]$, then $0\leq 3-\mu(E_3\cap E_1)\leq 1.$
So, $2\leq\mu(E_3\cap E_1)\leq 3.$
Since $E_3\cap E_1\in\mathcal{S}$ and $E_3\cap E_1\subset E_1$, $\mu(E_3\cap E_1)\leq\mu(E_1)=1.$
This is a contradiction.
So, $\mu(E_3\setminus E_1)=3.$
So, $3=\mu(E_3)=\mu(E_3\setminus E_1)+\mu(E_3\cap E_1)=3+\mu(E_3\cap E_1).$
So, $\mu(E_3\cap E_1)=0.$
Since $E_1=(E_1\setminus E_3)\cup (E_1\cap E_3)$ and $(E_1\setminus E_3)\cap (E_1\cap E_3)=\emptyset$,$1=\mu(E_1)=\mu(E_1\setminus E_3)+\mu(E_1\cap E_3)=\mu(E_1\setminus E_3)+0=\mu(E_1\setminus E_3).$
Therefore, $\mu(E_1\cup E_3)=\mu(E_1\setminus E_3)+\mu(E_1\cap E_3)+\mu(E_3\setminus E_1)=1+0+3=4.$
So, $4\leq c.$
Assume that $4<c.$
- If $6\leq c$, let $b\in (c-3,c-1)$.
Then, $3\leq c-3<b<c-1.$
So, $b\in [3,c].$
There exists $E_b\in\mathcal{S}$ such that $\mu(E_b)=b.$
Then, $1<\mu(X\setminus E_b)=c-b<3.$
This is a contradiction.- If $4<c<6$, then $1<\mu(X\setminus E_3)=c-3<3.$
This is a contradiction.Therefore, $c$ must be equal to $4.$
So, $T\subset\{4\}.$
Suppose $X:=\mathbb{Z}^+\cup\{0\}$, $\mathcal{S}:=2^X$, and $w:X\to [0,\infty]$ is a function such that $w(i):=\frac{1}{2^i}$ for $i\in\mathbb{Z}^+$ and $w(0):=3.$
Define a measure $\mu$ on $(X,\mathcal{S})$ by $$\mu(E)=\sum_{x\in E} w(x)$$ for $E\in\mathcal{S}.$
Then, $$\{\mu(E):E\in\mathcal{S}\}=[0,1]\cup [3,4].$$
So, $\{4\}\subset T.$
Therefore, $T=\{4\}.$
Best Answer
Your solution is correct. Here is a shorter solution.
Proof: Suppose $(X,\mathcal{S},\mu)$ is a measure space and $\{\mu(E):E\in\mathcal{S}\}=[0,1]\cup [3,c]$.
Clearly $\mu(X)=c$ and there is $E_1 \in \mathcal{S}$ such that $\mu(E_1)=1$. Then, $\mu(X \setminus E_1)= c-1$. So $c-1 \in [0,1]\cup [3,c]$. Since $c \geqslant 3$, we have $c-1 \geqslant 2$ and so $c-1 \in [3,c]$.
Suppose $3 < c-1$, then there is $\varepsilon$, $0< \varepsilon <1$, such that $3< c-1-\varepsilon <c$. It follows that there is $E_2 \in \mathcal{S}$ such that $\mu(E_2)=c-1-\varepsilon$. So, $\mu(X \setminus E_2) = c-(c-1-\varepsilon)=1+\varepsilon$. It means $1 < \mu(X \setminus E_2) <2$. Contradiction, because $\{\mu(E):E\in\mathcal{S}\}=[0,1]\cup [3,c]$. So, we have $c-1 \leqslant 3$. But since $c-1 \in [3,c]$, we have that $c-1=3$, that is, $c=4$.
Now, let us show that, for $c=4$, there is, in fact, a measure space $(X,\mathcal{S},\mu)$ with $\{\mu(E):E\in\mathcal{S}\}=[0,1]\cup [3,4]$. Consider $X=[0,1]$, $\mathcal{S}$ the Borel $\sigma$-algebra and $\lambda$ the Lebesgue measure. Let $\delta$ be the measure defined in $\mathcal{S}$ by, for all $B \in \mathcal{S}$, $\delta(B)=3$ if $0 \in B$, and $\delta(B)=0$ if $0 \notin B$. It is easy to see that $\delta$ is a measure. Take $\mu = \lambda + \delta$. It is easy to see that $\{\mu(E):E\in\mathcal{S}\}=[0,1]\cup [3,4]$.