I was recently reading an article called An Elementary Proof of Rademacher's Theorem. In the text below $U$ is a ball and $ f: U \to \mathbb{R}$ is a lipschitz function. Let $S$ the set of $x\in U$ for which $\partial_v f(x)$ does not exist.
My question. How can we prove that the continuity of $ f $ implies that the set $ S $ is measurable?
I'm trying to look at $ S $ as the following set. $$ S= \left\{ x\in U \left| \begin{matrix} \forall T:\mathbb{R}^n\to \mathbb{R} \mbox{ linear }\exists \epsilon>0 \\ \mbox{ such that } \forall \delta>0 \\ |t|<\delta \mbox{ and } \frac{|f(x+tv)-f(x)-Tv|}{|v|}>\epsilon \end{matrix} \right. \right\} $$ Then, prove that such a set is measurable. But something tells me that $ S $ can not be described in this way. So the question seems to be this. How to describe $ S $ properly and then prove that it is a measurable set?
Update. Examining the text in the image better, I realized that I can express the set $ S $ in the form below. I believe it's the correct expression for $ S $. Recall that $|v|=1$.
$S=\{x\in U: \partial_{v}f(x) \mbox{ does not exist }\}$
$S=\left\{x\in U: \lim_{t\to 0} \frac{f(x+tv)-f(x)}{t}\mbox{ does not exist }\right\}$
$S=\left\{x\in U\left| \begin{matrix} \mbox{ logical negation of} \\ (\exists \partial_vf(x)\in\mathbb{R}) (\forall \epsilon>0)(\exists \delta >0) \\ 0<|t|<\delta\implies\left|\frac{f(x+tv)-f(x)}{t}-\partial_vf(x)\right|<\epsilon \end{matrix}\right.\right\}$
$S=\left\{x\in U\left| \begin{matrix} (\forall \partial_vf(x)\in\mathbb{R}) (\exists \epsilon>0)(\forall \delta >0) \\ 0<|t|<\delta\mbox{ and } \left|\frac{f(x+tv)-f(x)}{t}-\partial_vf(x)\right|\geq \epsilon \end{matrix}\right.\right\}$
$S=\displaystyle\bigcap_{L\in\mathbb{R}} \displaystyle\bigcap_{\delta>0} \big( S_{L,\delta}^+\cup S_{L,\delta}^+) $ for $ S_{L,\delta}^+=\left\{x\in U\left| \frac{f(x+tv)-f(x)}{t}\leq L+ \epsilon, 0<|t|<\delta \right.\right\} $ and $ S_{L,\delta}^-=\left\{x\in U\left| \frac{f(x+tv)-f(x)}{t}\geq L - \epsilon, 0<|t|<\delta \right.\right\} $