Is my solution a typical solution? Any better solution? Exercise 8 on p.46 in Exercises 2C in "Measure, Integration & Real Analysis" by Sheldon Axler.

Question

I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
The following exercise is Exercise 8 on p.46 in Exercises 2C in this book.

Exercise 8
Give an example of a set $X$, a $\sigma$-algebra $\mathcal{S}$ of subsets of $X$, a set $\mathcal{A}$ of subsets of $X$ such that the smallest $\sigma$-algebra on $X$ containing $\mathcal{A}$ is $\mathcal{S}$, and two measures $\mu$ and $\nu$ on $(X,\mathcal{S})$ such that $\mu(A)=\nu(A)$ for all $A\in\mathcal{A}$ and $\mu(X)=\nu(X)<\infty,$ but $\mu\neq\nu.$

I solved this exercise.
I guess my solution is ok.
Is my solution a typical solution for this exercise?
Any better solution?

My solution is here:

Let $X:=\{1,2,3,4\}.$
Let $\mathcal{A}:=\{\{1,2\},\{1,3\}\}.$
Suppose $\mathcal{S}$ is the smallest $\sigma$-algebra on $X$ cotaining $\mathcal{A}.$
$\{1,2,3\}=\{1,2\}\cup\{1,3\}\in\mathcal{S}.$
$\{4\}=X\setminus\{1,2,3\}\in\mathcal{S}.$
$\{1,2,4\}=\{1,2\}\cup\{4\}\in\mathcal{S}.$
$\{1,3,4\}=\{1,3\}\cup\{4\}\in\mathcal{S}.$
$\{3\}=X\setminus\{1,2,4\}\in\mathcal{S}.$
$\{2\}=X\setminus\{1,3,4\}\in\mathcal{S}.$
$\{2,3,4\}=\{2\}\cup\{3\}\cup\{4\}\in\mathcal{S}.$
$\{1\}=X\setminus\{2,3,4\}\in\mathcal{S}.$
$\therefore\mathcal{S}=2^X.$
Let $x,y\in [0,\infty)$ and $x\neq y.$
Let $v:X\to [0,\infty]$ be the function such that $v(1)=x,v(2)=y,v(3)=y,v(4)=x.$
Let $w:X\to [0,\infty]$ be the function such that $w(1)=y,w(2)=x,w(3)=x,w(4)=y.$
Let $\mu:\mathcal{S}\to [0,\infty]$ be the function such that $\mu(E)=\sum_{x\in E} v(x)$ for $E\in\mathcal{S}.$
Let $\nu:\mathcal{S}\to [0,\infty]$ be the function such that $\nu(E)=\sum_{x\in E} w(x)$ for $E\in\mathcal{S}.$
Then, $\mu(\{1,2\})=v(1)+v(2)=x+y=y+x=w(1)+w(2)=\nu(\{1,2\}).$
Then, $\mu(\{1,3\})=v(1)+v(3)=x+y=y+x=w(1)+w(3)=\nu(\{1,3\}).$
Then, $\mu(X)=v(1)+v(2)+v(3)+v(4)=2x+2y=w(1)+w(2)+w(3)+w(4)=\nu(X)<\infty.$
Then, $\mu(\{1,4\})=v(1)+v(4)=2x\neq 2y=w(1)+w(4)=\nu(\{1,4\}).$
So, $\mu\neq\nu.$

Answer 1

Your example is correct, but there is a simpler way to present the answer.

Let $X=\{1,2,3,4\}$, $\mathcal{S}=2^X$ and $\mathcal{A}=\{\{1,2\},\{1,3\}\}$. It is easy to see that the smallest $\sigma$-algebra on $X$ containing $\mathcal{A}$ is $\mathcal{S}$.

Let us define the measure $\mu$ on $\mathcal{S}$ by $\mu(\{1\}) = \mu(\{2\}) = \mu(\{3\}) = \mu(\{4\})=1$ . Let us define $\nu$ on $\mathcal{S}$ by $\nu(\{1\}) = \nu(\{4\})=0$ and $\nu(\{2\}) = \nu(\{3\}) = 2$.

Then, $\mu(A)=\nu(A)$ for all $A\in\mathcal{A}$, that means $\mu(\{1,2\})= \nu(\{1,2\})= 2$ and $\mu(\{1,3\})= \nu(\{1,3\})= 2$, and $\mu(X)= \nu(X)=4<\infty$. Clearly, $\mu\neq\nu$.