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 $$ \varphi(K)=\varphi(T)? $$
For $m=1$ this is obvious, since in this case $K$ is a union of a countable set of intervals + a compact set of the Jordan measure 0, and on each interval the mapping $\varphi$ is constant. I wonder, if this is true for $m>1$?
I need this lemma for teaching, and I would appreciate any help.
I asked this also in MathOverflow.