Proof that the general polytope is measurable So I have proved that a compact, convex polytope is Jordan measurable at the heed of Terrance Tao's "Introduction to measure theory" Exercise 1.1.9. My current question is how to expand this to the general compact polytope?
I could imagine a proof going something like let $P$ be a polytope and let $B=\{E\subseteq P|E $ is convex and open$ \}$. $B$ covers $int(P)$, and it feels like we could somehow $int(P)$ had a finite cover due to compactness, but I can't see how?
If you could expand on my attempt or provide there own I would be thrilled, thanks.
 A: Jump to the bottom for the proof.
Half-space background
A half-space is all the points on one side of a line in $\mathbb{R}^2$ or all the points on one side of a plane in $\mathbb{R}^3$, and so forth. For lines/planes that cut through the origin, you can represent the half-space as all vectors $x$ such that $x \cdot v \ge 0$, for some $v$ perpendicular to the line/plane. To cut the space somewhere other than the origin, you can displace the line/plane along $v$ by defining the half-space as all vectors $x$ such that $x \cdot v \le c$, for some $c \in \mathbb{R}$.
If the half-space includes the line/plane that does the cutting, it will be a closed subspace (convince yourself).
Polytope background
Polytopes are the higher dimensional versions of polygons. Closed convex polygons can be expressed as the intersection of closed half-planes (convince yourself). Closed convex polytopes can be expressed as the intersection of closed half-spaces.
A closed convex polytope in $\mathbb{R}^d$ is bounded if it is contained in a box in $\mathbb{R}^d$.
Lemma 1
A half space $H \subseteq \mathbb{R}^d$ can be represented in the form $\{(x', t) : x' \in \mathbb{R}^{d-1}; t \le f(x')\}$, where $f : \mathbb{R}^{d-1} \to \mathbb{R}$ is continuous.
Proof.
By definition, a half-space $H$ in $\mathbb{R}^d$ can be expressed as $H = \{x \in \mathbb{R}^d : x \cdot v \le c \}$ for some $v \in \mathbb{R}^d$ and $c \in \mathbb{R}$. We can reword this as follows.
$$
\begin{align}
H &= \{x \in \mathbb{R}^d : x \cdot v \le c \} \\
&= \{x \in \mathbb{R}^d : x_1v_1 + ... x_dv_d \le c \} \\
&= \{x \in \mathbb{R}^d : x_dv_d \le c - x_1v_1 + ... x_{d-1}v_{d-1} \} \\
&= \{x \in \mathbb{R}^d : x_d \le \frac{c - x_1v_1 + ... x_{d-1}v_{d-1}}{v_d} \} \\ 
\end{align}
$$
Without loss of generality, assume that $v_d$ is non-zero (one of the dimensions must be non-zero, and to avoid messy indexing, I'll order the dimensions such that the last one is the non-zero one).
To tidy this expression up, let $x' := (x_1, x_2, ... x_{d-1})$ be the vector $x$ without the last dimension and let $f : \mathbb{R}^{d-1} \to \mathbb{R}$ be the function $f(x') := \frac{c - x'_1v_1 + x'_2v_2 + ... x'_{d-1}v_{d-1}}{v_d} $, and observe that it is continuous.
Then continue like so:
$$
\begin{align}
H &= \{(x', x_d) \in \mathbb{R}^d : x_d \le f'(x') \} \\
&= \{(x', x_d) : x' \in \mathbb{R}^{d-1}  \text { and } x_d \le f'(x') \}
\end{align}
$$
q.e.d.
Lemma 2
A half-space intersected with a box is Jordan measurable.
Proof.
Let $H$ be a half-space in $\mathbb{R}^d$, let $B$ be a box in $\mathbb{R}^d$ and let Let $x \in \mathbb{R}^d$.
Pause for some notation
$H = \{x \in \mathbb{R}^d : x \cdot v \le c \}$ for some $v \in \mathbb{R}^d$ and constant $c \in \mathbb{R}$. $B$ can be expressed as $B = I_1 \times I_2 \times ... \times I_d$ where $I_1 ... I_d$ are intervals. Let $B' = I_1 \times I_2 \times ... \times I_{d-1}$. So then $B = B' \times I_d$. Let $x' = (x_1, ... x_{d-1})$, all but the last dimension of $x$.
We can express set membership of $H \cap B$ as follows.
$$
\begin{align}
x \in H \cap B &\iff x \cdot v \le c \text{ and } x \in B' \times I_d \\
\end{align}
$$
By lemma 1 (along with that definition of $f$ and our assumption of $v_d$ being non-zero):
$$
\begin{align}
x \in H \cap B &\iff x_d \le f(x') \text{ and } x \in B' \times I_d \\
&\iff x \in B' \text{ and } a \le x_d \le\min(f(x'), b) \quad \text{where } I_d = [a, b] \\
\end{align}
$$
Thus, $H \cap B = \{(x', x_d) : x' \in B' \text{ and } a \le x_d \le \min(f(x'), b) \}$.
Now, translate $H \cap B$ by the vector $k = (0, 0, ..., -a) \in \mathbb{R}^d$ to get:
$$(H \cap B) + k = \{(x', x_d) : x' \in B' \text{ and } 0 \le x_d \le \min(f(x'), b) - a \}$$
This is the exact form we have in 1.1.7 (2) in the book, we know that this is measurable, so $m((H \cap B) + k) \in \mathbb{R}$ exists. By translation invariance $m((H \cap B) + k) = m(H \cap B)$.
q.e.d.
Proof
Let $P = H_1 \cap H_2 \cap ... \cap H_n$ be a bounded polytope in $\mathbb{R}^d$ expressed as the intersection of $n$ closed half-spaces $H_1 ... H_n$. Such a polytope is necessarily convex. As $P$ is bounded, $P = B \cap P$ for some box $B \in \mathbb{R}^d$. So we have:
$$
\begin{align}
P &= B \cap P \\
&= B \cap (H_1 \cap H_2 \cap ... \cap H_n) \\
&= (B \cap H_1) \cap (B \cap H_2) \cap ... \cap (B \cap H_n) \\
\end{align}
$$
By Lemma 2, $B \cap H_i$ is Jordan measurable for all $1 \le i \le n$, and by boolean closure (1.1.6 (1) in the book) it follows that $(B \cap H_1) \cap (B \cap H_2) \cap ... \cap (B \cap H_n)$ is Jordan measurable. Thus, $P$ is Jordan measurable.
q.e.d.
