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This is crossposted from MSE, I hope this is suitable here, since there is no reaction there. I need this lemma for teaching, and I would appreciate any help.

Briefly:

Is the image of a Jordan compact set $K$ under a degenerate smooth map $\varphi$ equal to an image of a compact subset $T\subseteq K$ of zero measure, $\mu(T)=0$: $$ \varphi(K)=\varphi(T)? $$

In detail: let $K$ be a compact set in ${\mathbb R}^m$, and suppose it is measurable in the sence of Jordan. Suppose $U$ is an open set in ${\mathbb R}^m$, and $K\subseteq U$. Let $\varphi:U\to{\mathbb R}^n$ be an infinitely smooth mapping with $n\ge m$ such that its differential $$ d\varphi(x)(p)=\lim_{t\to 0}\frac{\varphi(x+tp)-\varphi(x)}{t},\qquad x\in U,\ p\in{\mathbb R}^m $$ is not injective everywhere on $K$: $$ \forall x\in K\quad \exists p\ne 0\quad d\varphi(x)(p)=0. $$ Question:

Does there exist a compact subset $T\subseteq K$ of zero measure, $\mu(T)=0$, such that $$ \varphi(K)=\varphi(T)? $$

Examples:

  1. For $m=1$ this is obvious, since in this case $K$ is a union of a countable set of closed intervals + a compact set of Jordan measure 0, and on each interval the mapping $\varphi$ is constant. So I wonder, if this is true for $m>1$.

  2. As far as I understand, it is sufficient to clarify the situation in the case of $m=n=2$, where the question becomes the following:

Suppose $U\subseteq{\mathbb R}^2$ is an open set, $K\subseteq U$ a compact set, measurable in the sense of Jordan, and $\varphi:U\to{\mathbb R}^2$ an infinitely smooth map with the Jacobian determinant vanishing everywhere on $K$: $$ \forall x\in K\qquad J\varphi(x)=0. $$ Does there exist a compact subset $T\subseteq K$ of zero measure, $\mu(T)=0$, such that $$ \varphi(K)=\varphi(T)? $$

Thank you.

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1 Answer 1

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I think this is Theorem 1.5 in Hajlasz's UPitt notes.

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  • $\begingroup$ I would not recognize. And I am afraid, I need somebody to explain this. Is there a simple way to prove this proposition, without entering this deep theory? $\endgroup$ Aug 6, 2015 at 14:37
  • $\begingroup$ Igor, as far as I understand, Hajlasz proves that the $m$-th Hausdorff measure of $\varphi(K)$ vanishes, but why does this mean, that $T$ exists? $\endgroup$ Aug 6, 2015 at 17:33
  • $\begingroup$ @SergeiAkbarov I think that if you look at the argument, you can back out what you want (you look at ball covers, which go to skinny ellipsoids, so middle parts (along the major axis) of the balls, in the limit, give you your $K.$ Obviously, I did not work out the details, but I think this is the way to go. $\endgroup$
    – Igor Rivin
    Aug 6, 2015 at 19:32

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