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:
Is it clear that $f$ is measurable (with respect to Borel sets on $[0,1]^n$)?
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).
$f$ 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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