Multivariable monotonic function

Question

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).

Answer 1

$f$ 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.

Answer 2

As for the second question, let $\hat f(x):=\inf_{y>x}f(y)$ on $[0,1)^n$, where $y>x$ means that $y_i>x_i$ for all $i$. One has $\hat f\ge f$ and it is easy to check that $\hat f$ is increasing and right-continuous in every variable.

Also, if you fix $x=(x_1,\dots,x_n)$ and $v:=(1,\dots,1)$, then $s\mapsto\hat f(x+sv)$ is bounded above by the right-continuous representative of the increasing function $s\mapsto f(x+sv)$, which is just $s\mapsto\inf_{t>s}f(x+sv)$. Since the two agree for a.e. $t$ (as in one dimension an increasing function has at most countably many "jumps"), Fubini's theorem implies $f=\hat f$ a.e.