Differential of a Lipschitz map is Borelian

Question

Let $f: \mathbb R^n \to \mathbb R$ be a Lipschitz map. We say that $f$ is differentiable in $x \in \mathbb R^n$ in the Frechet sense if exists a linear operator $A: \mathbb R^n \to \mathbb R$ such that $$ \lim_{h \to 0} \frac{f(x+h) - f(x) - A\cdot h}{\|h\|} = 0 \quad (I) $$ Denoting by $D_f$ the set of points where $f$ is differentiable, we define the function $Df: D_f \to \mathbb R$. I'm tring to proof the following statements:

1. $D_f$ is a Borelian subset of $\mathbb R^n$.
2. $Df: D_f \to \mathbb L(\mathbb R^n, \mathbb R)$ is a Borelian map.

My attempt: Using that $(I)$ is equivalent to existing a linear map $A$ such that $$ \lim_{t \to 0^+} \frac{f(x+tv) - f(x)}{t} = A\cdot v $$ uniformly in $\|v\|=1$. We have that $x \in D_f$ iff $$ \exists A \in L(\mathbb R^n, \mathbb R), \, \forall \epsilon > 0, \, \exists \delta > 0, \, \left | F(x) - A\cdot v \right | < \epsilon, $$ where $F_t(x) = \frac{f(x+tv) - f(x)}{t}$. Hence $$ x \in \bigcup_{\epsilon > 0} \bigcap_{0<t<\delta} F_t ^{-1}(|A \cdot v| - \epsilon, |A \cdot v| + \epsilon), $$ which is Borelian, since $F$ is continuous.

Is this sufficient to show that $D_f$ is a Borelian set? I don't know how to start the argument to see that the map $Df$ is Borelian. Help?

Answer 1

Partial answer.

For $n=1$

$$g_n(x):= \frac{f(x+\frac{1}{n})-f(x)}{\frac{1}{n}}$$

So $f'(x)=\lim_ng_n(x)$

Since $g_n$ are Borel measurable,then $\limsup_ng_n,\liminf_ng_n$ are Borel measurable.

So $D_f=\{x:\limsup_ng_n(x)=\liminf_ng_n(x)\}$ is Borel measurable.

Also $f':D_f \to \Bbb{R}$ is Borel measurable as a pointwise limit of Borel measurable functions

There is a fact in measure theory that says that a Lipschitz continuous function $f: \Bbb{R} \to \Bbb{R}$ is differentiable almost everywhere.