Let $\mathcal{F}_n$ be the $\mathbb{R}$-vector space of functions $f : \mathbb{R}^n \rightarrow \mathbb{R}$. For all $X \subset \mathbb{R}^n$, we denote $\mathbb{1}_X$ the indicator function of $X$. Let $\mathcal{U}_n$ be the vector subspace of $\mathcal{F}_n$ generated by the functions $\mathbb{1}_P$ where $P$ is a polytope of $\mathbb{R}^n$. Prove that the following definition allows us to define a linear form $\chi_n : \mathcal{U}_n \rightarrow \mathbb{R}$. We define $\chi_1(f)$ by the sum $\chi_1(f) = \sum_{x \in \mathbb{R}} \left(f(x) - \lim_{y \rightarrow x^+} f(y)\right)$, then for $f \in \mathcal{U}_n$ with $n > 1$, we set $$\chi_n(f) = \chi_1(g) \text{ with } g \text{ defined by } g(z) = \chi_{n-1}(f_z) \text{ for } z \in \mathbb{R}.$$ We will show at the same time the formula $\chi_n(\mathbb{1}_P) = 1$ for every polytope $P$ of $\mathbb{R}^n$ and we will justify that $\chi_n$ is independent of the coordinate system, namely that for every invertible linear map $A : \mathbb{R}^n \rightarrow \mathbb{R}^n$ we have $\chi_n(f \circ A) = \chi_n(f)$ for all $f \in \mathcal{U}_n$.
Let $\mathcal{F}_n$ be the $\mathbb{R}$-vector space of functions $f : \mathbb{R}^n \rightarrow \mathbb{R}$. For all $X \subset \mathbb{R}^n$, we denote $\mathbb{1}_X$ the indicator function of $X$. Let $\mathcal{U}_n$ be the vector subspace of $\mathcal{F}_n$ generated by the functions $\mathbb{1}_P$ where $P$ is a polytope of $\mathbb{R}^n$.
Prove that the following definition allows us to define a linear form $\chi_n : \mathcal{U}_n \rightarrow \mathbb{R}$. We define $\chi_1(f)$ by the sum $\chi_1(f) = \sum_{x \in \mathbb{R}} \left(f(x) - \lim_{y \rightarrow x^+} f(y)\right)$, then for $f \in \mathcal{U}_n$ with $n > 1$, we set
$$\chi_n(f) = \chi_1(g) \text{ with } g \text{ defined by } g(z) = \chi_{n-1}(f_z) \text{ for } z \in \mathbb{R}.$$
We will show at the same time the formula $\chi_n(\mathbb{1}_P) = 1$ for every polytope $P$ of $\mathbb{R}^n$ and we will justify that $\chi_n$ is independent of the coordinate system, namely that for every invertible linear map $A : \mathbb{R}^n \rightarrow \mathbb{R}^n$ we have $\chi_n(f \circ A) = \chi_n(f)$ for all $f \in \mathcal{U}_n$.