# Exercise 21 on p.39 in Exercises 2B in "Measure, Integration & Real Analysis" by Sheldon Axler. Is my proof ok?

I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
The following exercise is Exercise 21 on p.39 in Exercises 2B in this book.

Exercise 21
Prove 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.

2.39 condition for measurable function
Suppose $$(X,\mathcal{S})$$ is a measurable space and $$f:X\to\mathbb{R}$$ 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.

2.25 $$\sigma$$-algebras are closed under countable intersection
Suppose $$\mathcal{S}$$ is a $$\sigma$$-algebra on a set $$X$$. Then
(a) $$X\in\mathcal{S}$$;
(b) if $$D,E\in\mathcal{S}$$, then $$D\cup E\in\mathcal{S}$$ and $$D\cap E\in\mathcal{S}$$ and $$D\setminus E\in\mathcal{S}$$;
(c) if $$E_1,E_2,\dots$$ is a sequence of elements of $$\mathcal{S}$$, then $$\bigcap_{k=1}^\infty E_k\in\mathcal{S}.$$

I mimicked the proof of 2.39.

My proof is here:

Let $$\mathcal{T}=\{A\subset [-\infty,\infty]:f^{-1}(A)\in\mathcal{S}\}.$$
We want to show that every Borel subsets of $$[-\infty,\infty]$$ is in $$\mathcal{T}$$. To do this, we will first show that $$\mathcal{T}$$ is a $$\sigma$$-algebra on $$[-\infty,\infty]$$.
Certainly $$\emptyset\in\mathcal{T}$$, because $$f^{-1}(\emptyset)=\emptyset\in\mathcal{S}.$$
If $$A\in\mathcal{T}$$, then $$f^{-1}(A)\in\mathcal{S}$$; hence $$f^{-1}([-\infty,\infty]\setminus A)=X\setminus f^{-1}(A)\in\mathcal{S}$$ by 2.33(a), and thus $$[-\infty,\infty]\setminus A\in\mathcal{T}.$$ In other words, $$\mathcal{T}$$ is closed under complementation.
If $$A_1,A_2,\dots\in\mathcal{T}$$, then $$f^{-1}(A_1),f^{-1}(A_2),\dots\in\mathcal{S}$$; hence $$f^{-1}(\bigcup_{k=1}^\infty A_k)=\bigcup_{k=1}^\infty f^{-1}(A_k)\in\mathcal{S}$$ by 2.33(b), and thus $$\bigcup_{k=1}^\infty A_k\in\mathcal{T}.$$ In other words, $$\mathcal{T}$$ is closed under countable unions. Thus $$\mathcal{T}$$ is a $$\sigma$$-algebra on $$[-\infty,\infty].$$
By hypothesis, $$\mathcal{T}$$ contains $$\{(a,\infty]:a\in\mathbb{R}\}.$$ Because $$\mathcal{T}$$ is closed under complementation, $$\mathcal{T}$$ also contains $$\{[-\infty,b]:b\in\mathbb{R}\}.$$ Because the $$\sigma$$-algebra $$\mathcal{T}$$ is closed under finite intersections (by 2.25), we see that $$\mathcal{T}$$ contains $$\{(a,b]:a,b\in\mathbb{R}\}.$$ Because $$(a,b)=\bigcup_{k=1}^\infty (a,b-\frac{1}{k}]$$ and $$(-\infty,b)=\bigcup_{k=1}^\infty (-k,b-\frac{1}{k}]$$ and $$\mathcal{T}$$ is closed under countable unions, we can conclude that $$\mathcal{T}$$ contains every open subset of $$\mathbb{R}.$$
So, $$\mathbb{R}\in\mathcal{T}$$ and $$(a,\infty)\in\mathcal{T}$$ for all $$a\in\mathbb{R}.$$
So, $$\{-\infty,\infty\}=[-\infty,\infty]\setminus\mathbb{R}\in\mathcal{T}$$ and $$\{\infty\}=(a,\infty]\setminus (a,\infty)\in\mathcal{T}$$ by 2.25(c).
So, $$\{-\infty\}=\{-\infty,\infty\}\setminus\{\infty\}\in\mathcal{T}$$ by 2.25(c).
Thus the $$\sigma$$-algebra $$\mathcal{T}$$ contains the smallest $$\sigma$$-algebra on $$\mathbb{R}$$ that contains all open subsets of $$\mathbb{R}.$$
And the $$\sigma$$-algebra $$\mathcal{T}$$ contains $$\{-\infty\}$$ and $$\{\infty\}$$.
In other words, $$\mathcal{T}$$ contains every Borel subset of $$[-\infty,\infty]$$. Thus $$f$$ is an $$\mathcal{S}$$-measurable function.

Is my proof ok?

• Your proof is correct. Mar 19, 2023 at 13:12

Your proof is correct. However it can be simplified.

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.

Proof: Let $$\Sigma$$ be the Borel $$\sigma$$-algebra of $$[-\infty,\infty]$$. Since, any open set of $$[-\infty,\infty]$$ is a countable union of open intervals of the form:

1. $$(a,b)$$ where $$a,b \in \Bbb R$$
2. $$(a, \infty]$$ where $$a \in \Bbb R$$
3. $$[-\infty,b)$$ where $$b \in \Bbb R$$

it is easy to see that $$\Sigma$$ is the $$\sigma$$-algebra generated by $$\{ (a,+\infty] \: : \: a \in \Bbb R\}$$. It follows that, if $$f^{-1}((a,\infty])\in\mathcal{S}$$, for all $$a\in\mathbb{R}$$, then $$f$$ is an $$\mathcal{S}$$-measurable function.

• Ramiro, Thank you very much for your answer. Mar 19, 2023 at 23:07