Hello there,


This is probably very easy but I can't find an argument.


Call a function $f: R^n \to R$ monotone increasing if $x\_i \le y\_i$ for each $1 \le i \le n$ implies $f(x) \le f(y)$.


I'd like to show that such a function is measurable; I'd be very surprised if this is not the case.


If $n=1$ it's ok, for the set $f^{-1}(-\infty,c)$ is either empty or an interval. I thought about studying each section of $f$, that is, fix e.g. $\hat{x}\_1=(x\_2,\ldots,x\_n)$, then $f\_{\hat{x}\_1}(x) := f(x,x\_2,\ldots,x\_n)$ is monotone, hence measurable. Similarly any section is measurable. But it seems this is not sufficient to conclude...


Thanks!



Monotone functions are measurable
We use induction on $n$. For $n=1$ this is trivial. Otherwise, suppose $f : \mathbb{R}^{n+1} \to \mathbb{R}$ is monotone increasing, fix $a \in \mathbb{R}$, and define $g : \mathbb{R}^n \to \mathbb{R}$ by $g(x) = \inf\{t \in \mathbb{R} : f(x,t) \ge a\}$. Then $g$ is monotone decreasing. By the induction hypothesis, $g$ is Lebesgue measurable. So the epigraph $E = \{(x,t) : t > g(x)\}$ is Lebesgue measurable, and the graph $G = \{(x,t) : g(x) = t\}$ is Lebesgue null (Fubini's theorem). However, $E \triangle\ \{f \ge a\} \subset G$ so $\{f \ge a\}$ is Lebesgue measurable.


