# Multivariate monotonic function

Let $f(x_1, \dots, x_n)$ be a real function on the $n$-dimensional unit cube (that is, mapping $[0,1]^n \mapsto \mathbb{R}$). Assume furthermore that $f$ is monotonic in every coordinate, and that $f$ is bounded.

I have the following questions:

1. Is it clear that $f$ is measurable (with respect to Borel sets on $[0,1]^n$)?

2. Is it true that there exists a function $\hat{f}$ which is right-continous (at every point, in every coordinate) such that $f=\hat{f}$ except on a set of Lebesgue measure zero?

If you know the answer, please also provide a reference (if possible).

• Is this a homework problem? – Noah Schweber Apr 2 '14 at 6:29
• No, it is not. If it seems so trivial for you, then please tell me the solution. – Kurisuto Asutora Apr 3 '14 at 10:42
• It doesn't seem trivial to me (although I know barely any analysis :P); I was only worried because there is no clear motivation given, and this form of question is common in real analysis courses. – Noah Schweber Apr 3 '14 at 16:27
• The motivation is that I am proving something about the integral of bounded, monotonic (multivariate) functions. And the question is, if I know that the function is bounded and monotonic, is it then clear that it is also integrable, or do I also have to assume that it is measurable. – Kurisuto Asutora Apr 4 '14 at 2:18
• While there are functions monotonic which are not Borel measurable, every function which is monotonic in all variables is Lebesgue measurable. See this question. mathoverflow.net/q/134304/22277 – Joseph Van Name Jun 5 '14 at 8:28

$f$ need not be Borel (!) measurable: Let $f(x,y)=0$ on $x+y<1$ and $f=1$ on $x+y>1$, and on the diagonal $x+y=1$, set $f=1/2$ for $x\in E$ and $f=0$ otherwise, where $E\subset [0,1]$ is not Borel. Then $f^{-1}(\{ 1/2 \})$ is not a Borel set in the square.