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).



Multivariable monotonic function
$f$ 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.


