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I failed to give an appropriate title to the question, so any suggestion for a better title is welcome: Here's the question: I would like to prove the following result: Given $\varepsilon>0$, a Lipschitz map $u:\mathbb R^n \to \mathbb R^m$ and two distinct points $p,q\in \mathbb R^n$, there exists a curve $\gamma:[0,1]\to\mathbb R^n$ with $\gamma(0)=p, \gamma(1)=q$ and such that $$ \int_0^1|\gamma'(t)| \mathrm dt < |p-q|+\varepsilon $$ and such that $(u\circ \gamma)'=Du (\gamma')$ $\mathcal L^1$-almost everywhere in $[0,1]$, and $\mathcal L^1$ denotes the one-dimensional Lebesgue-measure on $[0,1]$.

EDIT: The curve $\gamma$ should be Lipschitz, but if the result holds for smooth $\gamma$ as well this is of course fine.

EDIT for a better explanation: $(u\circ \gamma)$ is a.e. differentiable by Rademachers Theorem. However, the equality $(u\circ \gamma)'=Du (\gamma')$ need a priori not to hold a.e. in $[0,1]$ since $Du$ is only defined $\mathcal L^n$-a.e. a priori. So the result says essentially that this can be fixed with a small perturbation.

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The tool you should be using here is the coarea formula, a kind of curvilinear generalization of Fubini's theorem. For simplicity take $n=2$, $p=(0,0),q=(1,0)$. Let $E$ be the set where the derivative of $u$ exists. Let $v: \mathbb R^2 \to \mathbb R^2$ be given by $v(x,y)=(x,(y)(x)(x-1))$. We have $$\int_0^\epsilon H^1(v([0,1] \times \{y\}))dy=m(v([0,1] \times [0,\epsilon]))=m(E \cap v([0,1] \times [0,\epsilon]))=\int_0^\epsilon H^1(E \cap v([0,1] \times \{y\}))dy.$$ In particular, for almost every $y$ with $y<\epsilon$, $H^1(v([0,1] \times \{ y \}) \cap E) = H^1(v([0,1] \times \{y\}))$. Here $H^1$ denotes the one-dimensional Hausdorff measure, which coincides with the usual notion of length for rectifiable curves. For $y$ small enough, this curve has length close to $1$.

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  • $\begingroup$ Thank you for your great answer. $\endgroup$
    – frog
    Dec 17, 2014 at 14:11
  • $\begingroup$ I'm sorry but there is some confusion again: Say for example $\varepsilon=1$. In this case (since $H^1(\nu([0,1]\times\{y\}))≥1$ for all $\{y\}$, one could conclude that the first of your integrals must be strictly larger than 1. On the other hand, a direct computation shows that $m(\nu([0,1]\times[0,1]))=\frac{1}{6}$. What am I missing? $\endgroup$
    – frog
    Jan 7, 2015 at 13:55

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