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In the theory of Sobolev space, we have the following chain rule:

  • For a uniformly Lipschitz function $F : \mathbf{R}\to \mathbf{R}$ such that $F(0)=0$, and $u\in W^{1,1}(\mathbf{R}^n)$, then we have the following chain rule: $\partial_j F(u)=F'(u)\circ \partial_ju$.

But how to define the function $F'(u)$? It seems that we can't define $F'(u(.))$ a.e., and it may be not a measurable function.

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  • $\begingroup$ Does the fact that Lipschitz functions are differentiable a.e. not help you here? $\endgroup$
    – Daniel Shapero
    Commented Sep 13, 2022 at 3:16
  • $\begingroup$ @ Daniel Shapero, if $A$ is a null set in R, the set $u^{-1}(A)$ maybe not be a null set. $\endgroup$
    – sorrymaker
    Commented Sep 13, 2022 at 3:47

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The main difficulty in the proof of the rule is to prove that $\nabla u=0$ a.e. on the set $u^{-1}(\Sigma)$, where $\Sigma$ is the set where $F$ is not differentiable; and where $\nabla u=0$ one defines the product $F'(u)\nabla u$ to be 0, irrespective of the fact that $F'(u)$ is defined or not a such points. See e.g. Leoni, Morini: JEMS 9 pp 219-252

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    $\begingroup$ Also relevant in this context (though in the setting of vector-valued spaces): Arendt and Kreuter $\endgroup$
    – Jochen Glueck
    Commented Sep 13, 2022 at 7:18

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