Exercise 27 on p.40 in Exercises 2B in “Measure, Integration & Real Analysis” by Sheldon Axler.

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I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
The following exercise is Exercise 27 on p.40 in Exercises 2B in this book.

Exercise 27
Prove or give a counterexample: If $(X,\mathcal{S})$ is a measurable space and $$f:X\to [-\infty,\infty]$$ is a function such that $f^{-1}((a,\infty))\in\mathcal{S}$ for every $a\in\mathbb{R}$, then $f$ is an $\mathcal{S}$-measurable function.

The author wrote the following theorem.

2.52 condition for measurable function
Suppose $(X,\mathcal{S})$ is a measurable space and $f:X\to[-\infty,\infty]$ is a function such that $$f^{-1}((a,\infty])\in\mathcal{S}$$ for all $a\in\mathbb{R}.$ Then $f$ is an $\mathcal{S}$-measurable function.

If the statement in Exercise 27 were true, the author would have written the following theorem instead of 2.52:

2.52' condition for measurable function
Suppose $(X,\mathcal{S})$ is a measurable space and $f:X\to[-\infty,\infty]$ is a function such that $$f^{-1}((a,\infty))\in\mathcal{S}$$ for all $a\in\mathbb{R}.$ Then $f$ is an $\mathcal{S}$-measurable function.

So, I guess the statement in Exercise 27 is false.

My attempt is here:

Let $\mathcal{B}$ be the collection of Borel subsets of $\mathbb{R}.$
Let $$f:\mathbb{R}\to\mathbb{R}$$ be a function such that $f^{-1}((a,\infty))\in\mathcal{B}$ for every $a\in\mathbb{R}.$
The author wrote "We will see later that there exist subsets of $\mathbb{R}$ that are not Borel sets." on p.29 in the book.
Let $C_0$ be a subset of $\mathbb{R}$ which is not a Borel set.
Let $g$ be a function such that $g(x):=\infty$ for $x\in C_0$ and $g(x):=f(x)$ for $x\in\mathbb{R}\setminus C_0.$
Then, $\mathbb{R}\setminus C_0$ is not a Borel subset of $\mathbb{R}$ because $C_0=\mathbb{R}\setminus(\mathbb{R}\setminus C_0)$ is not a Borel subset of $\mathbb{R}.$
$g^{-1}((a,\infty))=f^{-1}((a,\infty))\cap (\mathbb{R}\setminus C_0)$ is an intersection of a Borel subset of $\mathbb{R}$ and a non-Borel subset of $\mathbb{R}$.
So, in general, $g^{-1}((a,\infty))\in\mathcal{B}$ doesn't hold.
$\dots$

Please tell me how to solve Exercise 27.

Best Answer

It depends on $\mathcal{S}$.

  1. If $\mathcal{S}=2^X$, then all functions $f:X\to [-\infty,\infty]$ are $\mathcal{S}$-measurable functions.

  2. Let us suppose that $\mathcal{S}\neq 2^X$. Then there is $C \subset X$, such that $C \notin \mathcal{S}$. Define $f:X\to [-\infty,\infty]$ by $f(x)=+\infty$ if $x \in C$ and $f(x)=-\infty$ if $x \in X \setminus C$. Then, for every $a\in\mathbb{R}$, $$f^{-1}((a,\infty))=\emptyset \in\mathcal{S}$$
    But, clearly $f$ is not a $\mathcal{S}$-measurable function, because $f^{-1}(\{+\infty\})=C \notin \mathcal{S}$.

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