I am doing Measure theory from the book by p.k jain . The definition of Measurable function is given that: An extended real valued function defined on a measurable set E is said to be measurable if the set of those x in E for which f(x) > a (for all real a) is measurable.I am stuck In the proof ' A continuous function defined on a measurable set is measurable .' I know that inverse image of an open set is open under a continuous function .In the book it is proved that the set { x in E : f(x) > a } is open being the inverse image of an open set ( a , b) ,b= infinity.
My question is that to show that f is measurable ,one should prove that { x in E : f(x) > a} is measurable being the inverse image of (a ,b] , b = infinity .Why the extended real number line is ignored here . Is (a, b] ,b= infinity is open in extended real number line?
Sets of the form $(a,\infty ]$ are open in the standard topology of the extended real line. Also the order in the extended real line is given such that $a< \infty $ for all $a\in[-\infty ,\infty )$, therefore
$$ \{x\in \Omega : f(x)>a\}=\begin{cases} f^{-1}((a,\infty]),& a<\infty \\ \emptyset ,& a=\infty \end{cases} $$
To show that $f$ is measurable it is enough to show that $f^{-1}((a,\infty ])$ is measurable for each $a\in \mathbb{R}$ because
$$ f^{-1}(\infty )=f^{-1}\left(\bigcap_{n\in \mathbb{N}}(n,\infty ]\right)=\bigcap_{n\in \mathbb{N}}f^{-1}((n,\infty ])\\ f^{-1}(-\infty )=f^{-1}\left(\left(\bigcup_{n\in \mathbb{Z}}(n,\infty ]\right)^\complement \right)=\left(f^{-1}\left(\bigcup_{n\in \mathbb{Z}}(n,\infty ]\right)\right)^\complement =\left(\bigcup_{n\in \mathbb{Z}}f^{-1}((n,\infty ])\right)^\complement $$
therefore $f^{-1}(\infty )$ and $f^{-1}(-\infty )$ are measurable if all $f^{-1}((a,\infty ])$, with $a\in \mathbb{R}$, are measurable.