How to show that $D_{f}$ is a Borel function

Question

How to show that $D_{f}$ is a Borel function.

Well I have one Lipschitz function $f:\Bbb{R}^{n}\to \Bbb{R}$ and I want to proof that

$D_{f}:D\to L(\Bbb{R}^{n},\Bbb{R})$ is Borel function, where $D=\{ x\in \Bbb{R}^{n}: f'(x) \text{ exists in the Fréchet sense }\}$

I try with the definition to show that $\forall U$ Borel set in $ L(\Bbb{R}^{n},\Bbb{R})$ imply $D_{f}^{-1}(U)$ is Borel set.

Then let $U$ Borel set in $L(\Bbb{R}^{n},\Bbb{R})$ hence

$D_{f}^{-1}(U)=\{x\in \Bbb{R}^{n}: D_{f}(x)\in U\}$ but $D_{f}(x)$ is one linear transformation (using or not using Fréchet sense) so

$D_{f}^{-1}(U)=\{x\in \Bbb{R}^{n}: T(x)\in U \}$ can to say : Like $T$ is continuous because is linear transformation then $D_{f}^{-1}(U)$ is measurable imply is Borel set?, can somebody help me please or give me one hint...thank you

Answer 1

Let us prove the following result:

Let $f:\Bbb{R}^{n}\to \Bbb{R}$ be a Lipschitz function and $D=\{ x\in \Bbb{R}^{n}: f'(x)\: \text{exists }\}$. Let $D_{f}:D\to L(\Bbb{R}^{n},\Bbb{R})$ be the function that associates to each point $x\in D$ the value $f'(x)$. Then $D_f$ is a Borel measurable function.

(Note: $f'(x)$ is the derivative of $f$ at $x$ and $L(\Bbb{R}^{n},\Bbb{R})$ is the space of linear transformations from $\Bbb R^n$ to $\Bbb R$).

Since $f:\Bbb{R}^{n}\to \Bbb{R}$ is a Lipschitz function, there is $K>0$, such that for all $x, y \in \Bbb{R}^n$, $|f(x)-f(y)| < K \|x - y\|$. Note that $f$ is continuous and so it is a Borel measurable function.

First, note that Borel sets in $L(\Bbb{R}^{n},\Bbb{R})$ are the elements of the $\sigma$-algebra generate by the open sets of the norm topology of $L(\Bbb{R}^{n},\Bbb{R})$. Since $L(\Bbb{R}^{n},\Bbb{R})$ is isomorphic to $\Bbb{R}^{n}$, the borel sets in $L(\Bbb{R}^{n},\Bbb{R})$ can be thought of as the Borel sets in $\Bbb R^n$.

Now, to prove that $D_{f}:D\to L(\Bbb{R}^{n},\Bbb{R})$ is Borel function, note that, by using the canonical basis in $\Bbb R^n$, we have that $$D_f= \left (\frac {\partial f}{\partial x_1}, \dots , \frac {\partial f}{\partial x_n} \right)$$

So it is enough to prove that any $i\in \{1,...,n\}$, the function
$\frac {\partial f}{\partial x_i}: D\to \Bbb R $ is a Borel function.

Let $(t_j)_{j \in \Bbb N}$ be a sequence of positive real numbers converging to $0$. Now, for any $i\in \{1,...,n\}$ and any $j \in \Bbb N $, let us define $d_{i,j} : D\to \Bbb R $ by $$d_{i,j}(x) = \frac{f(x+t_je_i) -f(x)}{t_j}$$

Since $f$ is a Borel measurable function, clearly $d_{i,j}$ are Borel measurable functions and, as $j\to \infty$, they converge pointwise to $\frac {\partial f}{\partial x_i}$ in $D$. So $\frac {\partial f}{\partial x_i}$ are Borel measurable functions. So $D_f$ is a Borel measurable function.

Remark: It is also possible to show that $D$ is a Borel measurable set.