# Proving that the $n$-simplex $\Delta^n$ is Jordan measurable

In our book we are using the following definitions:

A set $$U\subset \mathbb R^n$$ is said to be Jordan measurable if it is a bounded set and $$\partial U$$ is of content zero.

A set $$V\subset \mathbb R^n$$ is said to be of content zero if for every $$\epsilon>0$$ there are finite blocks $$W_1,\cdots, W_k$$ of the form $$W_i = (a_1,b_1)\times\cdots \times (a_n,b_n)$$ such that $$V\subset W_1\cup\cdots\cup W_k$$ and $$\sum$$Vol$$_n(W_i)<\epsilon$$

I want to prove that the $$n$$-simplex $$\Delta^n=\{x\in \mathbb R^n: x_i\geq 0, \sum x_i \leq 1\}$$ is a Jordan measurable set. I'm unsure about what its boundary is exactly. In our book, there's a tip to use the following result (which we already have proven):

Any $$K\subset a+E$$ is of content zero, where $$E$$ is subspace of $$\mathbb R^n$$ with dim $$E and $$K$$ is a compact subset

Attempt:

First of all, we prove that $$\Delta^n$$ is compact. Given the norm $$\|x\|_1= |x_1| + \cdots + |x_n|,$$ we notice that the n-simplex is the intersection of the unit closed ball with all the positive hyperplanes, that is, we have : $$\Delta_n = \overline B(0,1)_{\|.\|_1}\cap\{x\in \mathbb R^n: x_i\geq 0\}$$. Since the closed unit ball is bounded and closed and the hyperplanes are closed too, then $$\Delta^n$$ is compact. Now define the following functions $$f_1,\cdots, f_{n+1}$$ by $$f_1(x)=x_1,\cdots, f_n(x)=x_n$$ and $$f_{n+1}(x) = 1 - \sum x_i.$$ Now I think that the boundary is given by the union of the following sets $$F_i = f_i^{-1}(0)\cap_{j\neq i}\{x:f_j(x)\geq 0\}$$ for $$i=1,\cdots, n+1$$. Is this correct? Because if its, then all $$F_i$$ are compact sets in $$E_i=\{x: x_i = 0\}$$ for $$i=1,\cdots, n$$ and $$F_{i+1}$$ is compact in the affine plane $$x_1+\cdots+x_n=1$$.

Is this a correct reasoning? And is there any better way to prove that $$\Delta^n$$ is Jordan measurable? Thank you!

• It is not true that the simplex is the intersection of the unit ball with the half-spaces $\{x_i\ge 0\}$. For example, in dimension 2, the standard simplex is a triangle with vertices $(0,0)$, $(1,0)$ and $(0,1)$, in 3D is is a tetrahedron, etc. The boundary of a simplex are lower-dimensional simplices, contained in lower-dimensional affine subspaces and, as such, they have measure zero (as you mentioned). – GReyes Jun 12 at 5:45
• @GReyes I was considering the unit ball in the 1-norm $\|.\|_1$ above – math.h Jun 12 at 16:06
• I see. In any case I think this is not essential. A general simplex is a convex hull of points and their boundaries (or parts thereof) are not naturally related to any norm. It seems to me that it is all about the boundary being "thin", made up by lower-dimensional faces and thus having Jordan measure zero. – GReyes Jun 12 at 16:14