I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
The following exercise is Exercise 19 on p.39 in Exercises 2B in this book.
Exercise 19
Suppose $X$ is a nonempty set and $\mathcal{S}$ is the $\sigma$-algebra on $X$ consisting of all subsets of $X$ that are either countable or have a countable complement in $X$. Give a characterization of the $\mathcal{S}$-measurable real-valued functions on $X$.
My solution is here:
Let $f$ be an $\mathcal{S}$-measurable real-valued function on $X.$
Let $A:=\{a\in\mathbb{R}:f^{-1}((a,\infty))\text{ is countable}\}.$
We prove that $A$ is not empty.
Assume that $A=\emptyset.$
Then, $f^{-1}((a,\infty))$ is uncountable for any $a\in\mathbb{R}.$
Then, $X\setminus f^{-1}((a,\infty))=f^{-1}((-\infty,a])$ is countable for any $a\in\mathbb{R}.$
Since $X=f^{-1}(\mathbb{R})=f^{-1}(\bigcup_{n\in\mathbb{Z}} (-\infty,n])=\bigcup_{n\in\mathbb{Z}}f^{-1}((-\infty,n])$, $X$ is countable.
Let $a\in\mathbb{R}.$
Since $f^{-1}((a,\infty))\subset X$, $f^{-1}((a,\infty))$ is countable.
This is a contradiction.
So, $A$ is not empty.
Let $a_1\in A.$
Let $a_2$ be any element of $\mathbb{R}$ such that $a_1<a_2.$
Then, $f^{-1}((a_2,\infty))\subset f^{-1}((a_1,\infty)).$
Since $f^{-1}((a_1,\infty))$ is countable, $f^{-1}((a_2,\infty))$ is countable.
So, if $a_1\in A$, then $a_2\in A$ for any $a_2\in\mathbb{R}$ such that $a_1<a_2.$
So,
- if $A$ is not bounded below, $A=\mathbb{R},$
- if $A$ is bounded below, $A=(b,\infty)$ or $A=[b,\infty)$ for some $b\in\mathbb{R}.$
First, we consider the case $A=(b,\infty)$ for some $b\in\mathbb{R}.$
$f^{-1}((b,\infty))$ is uncountable since $b\notin A.$
So, $f^{-1}((-\infty,b])=X\setminus f^{-1}((b,\infty))$ is countable.
Let $\epsilon$ be any positive real number.
Then, $f^{-1}((b+\epsilon,\infty))$ is countable since $b+\epsilon\in A.$
Therefore, $f^{-1}((b,b+\epsilon])=f^{-1}((b,\infty))\setminus f^{-1}((b+\epsilon,\infty))$ is uncountable.Next, we consider the case $A=[b,\infty)$ for some $b\in\mathbb{R}.$
$f^{-1}((b,\infty))$ is countable since $b\in A.$
Let $\epsilon$ be any positive real number.
$f^{-1}((b-\epsilon,\infty))$ is uncountable since $b-\epsilon\notin A.$
$f^{-1}((-\infty,b-\epsilon])=X\setminus f^{-1}((b-\epsilon,\infty))$ is countable.
$f^{-1}((b-\epsilon,b])=f^{-1}((b-\epsilon,\infty))\setminus f^{-1}((b,\infty))$ is uncountable.Conversely, let $f$ be a real-valued function on $X$ which satisfies one of the following conditions:
- $f^{-1}((a,\infty))$ is countable for any $a\in\mathbb{R}.$
- There exists $b\in\mathbb{R}$ such that for any positive real number $\epsilon$, $f^{-1}((b+\epsilon,\infty))$ is countable and $f^{-1}((b,b+\epsilon])$ is uncountable and $f^{-1}((-\infty,b])$ is countable.
- There exists $b\in\mathbb{R}$ such that for any positive real number $\epsilon$, $f^{-1}((b,\infty))$ is countable and $f^{-1}((b-\epsilon,b])$ is uncountable and $f^{-1}((-\infty,b-\epsilon])$ is countable.
If $f$ satisfies 1, then $f$ is an $\mathcal{S}$-measurable real-valued function on $X.$
Suppose that $f$ satisfies 2.
Then, $f^{-1}((c,\infty))$ is countable if $b<c$ since there exists a positive real number $\epsilon$ such that $b+\epsilon<c.$
Then, $X\setminus f^{-1}((c,\infty))=f^{-1}((-\infty,c])$ is countable if $c\leq b.$
So, $f$ is an $\mathcal{S}$-measurable real-valued function on $X.$Suppose that $f$ satisfies 3.
Then, $f^{-1}((c,\infty))$ is countable if $b\leq c.$
Then, $X\setminus f^{-1}((c,\infty))=f^{-1}((-\infty,c])$ is countable if $c<b$ since there exists a positive real number $\epsilon$ such that $c<b-\epsilon.$
So, $f$ is an $\mathcal{S}$-measurable real-valued function on $X.$
Is my solution ok?
Even if my solution is ok, the conditions 1, 2 and 3 are not simple.
How to improve this?
Best Answer
The case $A = \Bbb R$. Then for all $a \in \Bbb R$, $f^{-1}((a,\infty))$ is countable. Since $$ X = f^{-1}(\Bbb R) = f^{-1}\left(\bigcup_n (-n,\infty)\right)=\bigcup_n f^{-1}\left((-n,\infty)\right)$$ we have that $X$ is countable.
The case $A=(b,\infty)$ for some $b\in\mathbb{R}$. In this case, $f^{-1}((b,\infty))$ is uncountable since $b\notin A.$
On the other hand, let $\epsilon$ be any positive real number. Then, $f^{-1}((b+\epsilon,\infty))$ is countable since $b+\epsilon\in A$. In particular, for any $n \in \Bbb N$, $n>0$, $f^{-1}((b+\tfrac{1}{n},\infty))$ is countable. Since, $$f^{-1}((b,\infty))= f^{-1}\left(\bigcup_n (b+\tfrac{1}{n},\infty)\right)=\bigcup_n f^{-1}\left((b+\tfrac{1}{n},\infty)\right)$$
we have that $f^{-1}((b,\infty))$ is countable. Contradiction. So it is not possible to have $A=(b,\infty)$.
The case $A=[b,\infty)$ for some $b\in\mathbb{R}$. In this case, $f^{-1}((b,\infty))$ is countable since $b\in A$.
Let $\epsilon$ be any positive real number. Then $f^{-1}((b-\epsilon,\infty))$ is uncountable since $b-\epsilon\notin A$. In particular $f^{-1}(\Bbb R)$ is uncountable. Note that $f^{-1}((-\infty,b-\epsilon])=X\setminus f^{-1}((b-\epsilon,\infty))$ is countable. In particular, for any $n \in \Bbb N$, $n>0$, $f^{-1}((-\infty,b-\tfrac{1}{n}])$ is countable. Since, $$f^{-1}((-\infty,b))= f^{-1}\left(\bigcup_n (-\infty, b-\tfrac{1}{n} )\right)=\bigcup_n f^{-1}\left((-\infty, b-\tfrac{1}{n})\right)$$ we have that $f^{-1}((-\infty, b))$ is countable. Since $f^{-1}((b,\infty))$ is also countable, we have that $f^{-1}(\Bbb R \setminus \{b\})$ is countable and, since $f^{-1}(\Bbb R)$ is uncountable, we have that $f^{-1}(\{b\})$ is uncountable.
Conclusion: We have only two cases.
Remark: As pointed out by Prof. Axler, the conditions for $\mathcal{S}$-measurability in the two cases above ($X$ countable and $X$ uncountable) can be combined in a single elegant statement: