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

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Dec 7, 2016 at 23:56	comment	added	Jonas Sjöstrand		@PietroMajer: Ah, it's that simple! Thanks, all of you!
Dec 7, 2016 at 13:19	vote	accept	Jonas Sjöstrand
Dec 7, 2016 at 12:24	review	Close votes
Dec 7, 2016 at 13:37
Dec 7, 2016 at 12:07	comment	added	Pietro Majer		@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.
Dec 7, 2016 at 6:20	comment	added	Cameron Zwarich		@NikWeaver Seems like that works. Thanks.
Dec 7, 2016 at 5:15	comment	added	Nik Weaver		@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).
Dec 7, 2016 at 3:38	comment	added	Cameron Zwarich		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?
Dec 7, 2016 at 3:25	answer	added	Cameron Zwarich		timeline score: 1
Dec 7, 2016 at 2:45	review	First posts
Dec 7, 2016 at 3:05
Dec 7, 2016 at 2:42	history	asked	Jonas Sjöstrand	CC BY-SA 3.0