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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).

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  • $\begingroup$ Is this a homework problem? $\endgroup$ Apr 2, 2014 at 6:29
  • $\begingroup$ No, it is not. If it seems so trivial for you, then please tell me the solution. $\endgroup$ Apr 3, 2014 at 10:42
  • $\begingroup$ 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. $\endgroup$ Apr 3, 2014 at 16:27
  • $\begingroup$ 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. $\endgroup$ Apr 4, 2014 at 2:18
  • $\begingroup$ 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 $\endgroup$ Jun 5, 2014 at 8:28

2 Answers 2

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$f$ 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.

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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.

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  • $\begingroup$ I just realized that this question was posted 6 years ago. I have no clue why it appeared on the homepage. $\endgroup$
    – Mizar
    Sep 7, 2020 at 15:45
  • $\begingroup$ No worries. Thanks all the same. $\endgroup$ Sep 8, 2020 at 7:29

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