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Let $f$ be a function from $[0,1]\times[0,1]$ to $\mathbb{R}$ that is nondecreasing in both variables, i.e. $f(x_1,y_1)\le f(x_2,y_2)$ whenever $x_1\le x_2$ and $y_1\le y_2$. It is known that the partial derivatives $\partial f/\partial x$ and $\partial f/\partial y$ exist almost everywhere, but are they Lebesgue measurable?

The one-dimensional version of this problem is well-studied, and it is known that the restriction of $\partial f/\partial x$ to the set $[0,1]\times\{y\}$ is measurable for any $y$. It is also known that $\partial f/\partial x$ is measurable when restricted to the subset of $[0,1]\times[0,1]$ where $f$ is differentiable.

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  • $\begingroup$ If $f$ is measurable, then this follows from Theorem 1 of Measurability of partial derivatives by Moshe and Mizel. How do you establish the existence of partial derivatives without measurability anyways? Is such a function automatically measurable? $\endgroup$
    – Cameron Zwarich
    Commented Dec 7, 2016 at 3:38
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    $\begingroup$ @CameronZwarich: I think $f$ is automatically measurable. Let $a \in \mathbb{R}$; we must show that $f^{-1}((-\infty,a])$ is measurable. For each $x \in [0,1]$ define $g(x) = \sup\{y: f(x,y) \leq a\}$. Then $g$ is decreasing, which surely implies that its graph has measure zero, and $f^{-1}((-\infty,a])$ is the region below this graph (measurable) plus some subset of the graph itself (null). $\endgroup$
    – Nik Weaver
    Commented Dec 7, 2016 at 5:15
  • $\begingroup$ @NikWeaver Seems like that works. Thanks. $\endgroup$
    – Cameron Zwarich
    Commented Dec 7, 2016 at 6:20
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    $\begingroup$ @CameronZwarich: $\partial _1 f$ and $\partial_2 f$ are also automatically measurable, since they are a.e. limits of incremental quotients, $n(f(x+1/n,y)-f(x,y))$ resp. $n(f(x,y+1/n)-f(x,y))$, which are measurable too. $\endgroup$
    – Pietro Majer
    Commented Dec 7, 2016 at 12:07
  • $\begingroup$ @PietroMajer: Ah, it's that simple! Thanks, all of you! $\endgroup$
    – Jonas Sjöstrand
    Commented Dec 7, 2016 at 23:56

1 Answer 1

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Yes. As Nik Weaver's comment points out, your function is automatically measurable. Therefore you can apply Theorem 1 of Measurability of partial derivatives by Moshe and Mizel.

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