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?
Best Answer
Your proof is correct. However it can be simplified.
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:
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.