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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? $\endgroup$ – Noah Schweber Apr 2 '14 at 6:29
  • $\begingroup$ No, it is not. If it seems so trivial for you, then please tell me the solution. $\endgroup$ – Kurisuto Asutora Apr 3 '14 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. $\endgroup$ – Noah Schweber Apr 3 '14 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. $\endgroup$ – Kurisuto Asutora Apr 4 '14 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 $\endgroup$ – Joseph Van Name Jun 5 '14 at 8:28
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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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