In the book [1], Clarke et al. define the generalized gradient for a Lipschitz function $f:\mathbb{R}^n\to\mathbb{R}$ as follows.
8.1. Theorem (Generalized Gradient Formula). Let $x\in\mathbb{R}^n$, and let $f:\mathbb{R}^n\to\mathbb{R}$ be Lipschitz near $x$. Let $\Omega$ be any subset of zero measure in $\mathbb{R}^n$, and let $\Omega_f$ be the set of points in $\mathbb{R}^n$ at which $f$ fails to be differentiable. Then, \begin{equation*} \partial f(x):=co\{\lim\nabla f(x_i):x_i\to x, x_i\notin\Omega,x_i\notin\Omega_f\} \end{equation*}
In other words, the generalized gradient of $f$ at $x$ is the convex hull whose elements are the limiting points of the gradient of $f$ computed at the elements of sequences converging to $x$. Moreover, these elements of sequences do not belong to any set of measure zero nor to the set of points where $f$ fails to be differentiable.
Here follows my question. If $f$ is defined on any open subset $S$ of $\mathbb{R}^n$, i.e., $f:S\to\mathbb{R}$, does the above formula hold as below
Let $x\in S$, and $\Omega(S)$ be any subset of $S$ with measure zero with respect to $\mathbb{R}^n$, and let $\Omega_f(S)$ be the set of points in $S$ at which $f$ fails to be differentiable. Then, \begin{equation*} \partial f(x):=co\{\lim\nabla f(x_i):x_i\to x, x_i\notin\Omega(S),x_i\notin\Omega_f(S)\} ? \end{equation*}
My answer is no, because $S$ is open. Consequently, $S$ does not contains the limit $\lim\nabla f(x_i)$. Thus, $\partial f(x)$ may not be defined in $S$. Is this reasoning correct?
References
[1] Clarke et al, "Nonsmooth Analysis and Control Theory", Springer 1998
It is true that $S$ does not contain $\lim \nabla f(x_i)$, but I fail to see any significance to this fact as I can see no point in demanding that the range of a gradient must be contained within its domain. Would you demand that $f\ :\ (0, \infty) \to \Bbb R\ :\ x \mapsto \frac 1x$ has no derivative because $f'(x) < 0$?
Your definition only works for $x \in \overline{(S - \Omega_f(S) - \Omega(S))}$ (unless you are happy with the empty set as a value), but this is only to be expected.