I am trying to show this using the theorem:
A function $f: \Omega \to \mathbb{R}$ is measurable if and only if $f^{-1}(E) \in \mathcal{F}$ for all borel sets $E$.
The proof to show a continuous function is measurable says:
$\{x : f(x) > c \}$ is open, and hence Borel and hence Lebesgue measurable.
Now, I am having troubles with this because I don't think I understand exactly what measurable means. We have a sigma algebra $\mathcal{F}$, and we say $f$ is measurable if $\{x : f(x) > c \} \in \mathcal{F}$ for every $ c \in \mathbb{R}$ – what does $\{x : f(x) > c \} \in \mathcal{F}$ mean exactly?
In the example showing a continuous function is measurable, I understand that $\{x: f(x) > c\} = f^{-1}((c,\infty])$ and since $f$ is continuous, $f^{-1}((c,\infty])$ is open, so it is Borel – but how does this included that $f^{-1}((c,\infty])\in \mathcal{F}?$ surely this is only true if $\mathcal{F}$ is the Borel sigma algebra, but the question has not specified it was, and I am assuming $\mathcal{F}$ is a general $\sigma-$algebra.
Thanks.
Best Answer
Continuous functions are not always measurable. It depends on the $\sigma$-algebra! Actually, continuous functions are measurable for an algebra $\mathcal{F}$ if and only if $\mathcal{F}$ contains the Borel algebra.
I think you can see why it is sufficient. To see that it is necessary, suppose that all continuous functions are $\mathcal{F}$-measurable, then $Id:\Bbb{R} \to \Bbb{R}:x\mapsto x$ is measeurable (because continuous). But then $Id^{-1}(c,\infty)=(c,\infty)$ and $Id^{-1}(-\infty,c)=(-\infty,c)$ both belong to $\mathcal{F}$ for all $c$. Since $\mathcal{F}$ is an algebra, it therefore contains all open sets and thus contains the Borel algebra.