Lebesgue measurability of singular set

Question

Let $\Omega$ be a bounded domain in $\mathbb{R}^{d}$ and $f:Q\to\mathbb{R}$ be continuous function. Define a superdifferential of $f$ at $x\in Q$ by $$ D^{+}f(x)=\{p\in\mathbb{R}^{d} \mid \text{$f(y)\le f(x)+p\cdot(y-x)+o(|x-y|)$ as $Q\ni y\to x$}\}. $$

Is a set $$ \Sigma(f)=\{x\in Q \mid D^{+}f(x)=\emptyset\} $$ Lebesgue measurable?

I'm facing the above problem, but am not familiar with Lebesgue measure theory and so I don't know how to verify. Of course, I know that if $f\in C^{1}(Q)$, then $\Sigma(f)=\emptyset$ thus Lebesgue measurable set, and it is clear even if $f\in Lip(Q)$. However, I'm wondering if this is really true for general continuous functions.

This problem may be fundamental, but I'm glad if you teach me how to discuss. In addition, I want to consider the same problem for upper semicontinuous functions and so it is great pleasure if you give some comments.

Thank you in advance.

Answer 1

Seems so. Let's describe the existence of superdifferential as a measurable condition. We denote by $\mathbb{Q}_+$ the set of positive rational numbers, $Z:=(\mathbb{Q}\cap [-1,1])^d$. The existence of supperdifferential at a point $x$ is equivalent to the following condition $$ \exists R\in \mathbb{Q}_+\, \forall \delta\in \mathbb{Q}_+\, \exists \varepsilon\in \mathbb{Q}_+\,\exists p\in R\cdot Z\,\forall y\in \varepsilon\cdot Z\\ f(x+y)\leqslant f(x)+p\cdot y+\delta\|y\|. $$ It is clear that if superdifferential exists, this condition holds. To prove the opposite, taking such $p$ for $\delta=1,1/2,1/3,\dots$ and choosing a converging subsequence of $p$'s gives you a genuine superdifferential.