# Are the partial derivatives of a function increasing in both variables measurable?

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.

• 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? – Cameron Zwarich Dec 7 '16 at 3:38
• @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). – Nik Weaver Dec 7 '16 at 5:15
• @NikWeaver Seems like that works. Thanks. – Cameron Zwarich Dec 7 '16 at 6:20
• @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. – Pietro Majer Dec 7 '16 at 12:07
• @PietroMajer: Ah, it's that simple! Thanks, all of you! – Jonas Sjöstrand Dec 7 '16 at 23:56