I know that there are many similar questions to mine. But this is more a question concerning a certain "logic".
I want to show, that every monotone function $f: \mathbb{R} \rightarrow \mathbb{R}$ is Borel-measurable.
Via definition of Borel-measurability, it suffices to show, that $$\forall y \in \mathbb{R}: \{f \geq y\} \in \mathfrak{B}(\mathbb{R}). $$
Proof:
Let wlog $f$ be a non-increasing function and $y \in \mathbb{R}$ fix.
(1) If $\{f \geq y \} = \emptyset$ than $\{f \geq y \} \in \mathfrak{B}(\mathbb{R}) $
(2) If $\{f \geq y \} \neq \emptyset $ define $x^* :=\sup\{x \in \mathbb{R}: f(x) \geq y \} $. Than we have:
-
$x^* = \infty \Rightarrow \{ f \geq y \} = \mathbb{R} \in \mathfrak{B}(\mathbb{R}) $
-
$x^* < \infty$:
- $x^* \in \{ f \geq y \} \Rightarrow \{f \geq y \} = (- \infty, x^*] \in \mathfrak{B}(\mathbb{R}) $
- $x^* \notin \{ f \geq y \} \Rightarrow \{f \geq y \} = (- \infty, x^*) \in \mathfrak{B}(\mathbb{R}) $
Since all these sets are in $\mathfrak{B}(\mathbb{R})$ and $y \in \mathbb{R}$ arbitrary $\Rightarrow$ $f$ is measurable.
I hope that this is so far okay. But I am not really happy with the last distinction: How is it even possible, that $x^* \notin \{ f \geq y \}$, since $f$ is a non-increasing function? I have a hard time thinking about this trivial fact; some help would be awesome!
Best Answer
Your proof is fine. Let $f(x)=0$ for $x<0$ and $f(x)=-1$ if $x \geq 0$. Take $y=0$. Then $\{x: f(x) \geq y\}=(-\infty, 0)$ and $x^{*}=0$.