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Let $u$ be a convex function defined on a bounded open set $\Omega$ in $\mathbb{R}^n$. Then $u$ is twice differentiable a.e. Let $E_u$ be the set on which $u$ is not twice differentiable. Then $E_u$ has zero Lebesgue measure. Let $T_u$ be the normal mapping of $u$.

Question: is $T_u(E_u)$ Lebesgue measurable?

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  • $\begingroup$ How is the normal mapping defined? $\endgroup$
    – Saúl RM
    Commented Oct 28, 2022 at 4:17
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    $\begingroup$ $T_u(x):=\{p\in\mathbb{R}^n:u(z)\geq u(x)+\langle p,z-x\rangle,\forall\,z\in\Omega\}$. $\endgroup$
    – User1999
    Commented Oct 28, 2022 at 4:46
  • $\begingroup$ Do you mean $T_u(x)=\partial u(x)$? i.e., $T_u(x)$ is the subgradient of $u$ at $x$? I am sure your set is Borel and that seems obvious, but a detailed proof might be annoying like all proofs that something nontrivial is Borel meausurable. $\endgroup$
    – Piotr Hajlasz
    Commented Nov 13, 2022 at 3:57
  • $\begingroup$ Yes. Thanks for your comment. $\endgroup$
    – User1999
    Commented Nov 14, 2022 at 5:18

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