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 $\mathbb{R}^n$. Let $f \in \mathcal{U}_1$. Prove that for all $x \in \mathbb{R}$ the limit of $f(y)$ as $y$ tends to $x$ while satisfying $y > x$, denoted $\lim_{y \rightarrow x^+} f(y)$, exists and that there exist finitely many reals $x \in \mathbb{R}$ such that $f(x) \neq \lim_{y \rightarrow x^+} f(y)$.
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 $\mathbb{R}^n$.
Let $f \in \mathcal{U}_1$. Prove that for all $x \in \mathbb{R}$ the limit of $f(y)$ as $y$ tends to $x$ while satisfying $y > x$, denoted $\lim_{y \rightarrow x^+} f(y)$, exists and that there exist finitely many reals $x \in \mathbb{R}$ such that $f(x) \neq \lim_{y \rightarrow x^+} f(y)$.