# Multivariate monotonic function

MathOverflow Asked by Kurisuto Asutora on October 25, 2020

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).

## 2 Answers

$$f$$ 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 $$xin E$$ and $$f=0$$ otherwise, where $$Esubset [0,1]$$ is not Borel. Then $$f^{-1}({ 1/2 })$$ is not a Borel set in the square.

Correct answer by Christian Remling on October 25, 2020

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 $$hat fge f$$ 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 $$smapstohat f(x+sv)$$ is bounded above by the right-continuous representative of the increasing function $$smapsto f(x+sv)$$, which is just $$smapstoinf_{t>s}f(x+sv)$$. 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.

Answered by Mizar on October 25, 2020

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