Show that Polygons are Jordan Measurable in $\mathbb R^2$
where we were just introduced to Jordan-Measurability that is defined as:
$M \subset \mathbb R^{d}$ is Jordan measurable if for any $\epsilon > 0$ figures $F, F^{'}\in \mathcal{F}^{d}$ exists (where $\mathcal{F}^{d}$ is the ring generated by the product of right-open intervals) where $F \subset M \subset F^{'}$ so that $\lambda^{d}(F)-\lambda^{d}(F^{'})< \epsilon$
My idea: let $\epsilon >0$ I assume $M$ as a polygon can be written as: $M:=\{ x \in \mathbb R^{2}: Ax \leq b\}$
$M$ is further bounded and closed and hence for every inequality we can find a maximising $x^{i+}\in M$ as well as a minimizing $x^{i-}$ so that $A_{i.}x^{i+}=b_{i}$ where $A_{i.}$ is the i-th row vector. I would thus define my $F:=\bigcap\limits_{i=1}^{m}[x^{i-}+\frac{1}{2n},x^{i+}-\frac{1}{2n}[$ and $F^{'}:=\bigcup\limits_{i=1}^{m}[x^{i-}-\frac{1}{2n},x^{i+}+\frac{1}{2n}[$ where we chose $n \in \mathbb N$ so that $\frac{1}{n} < \epsilon$ it is clear that $F, F^{'}\in \mathcal{F}^{d}$ as finite intersection as well as finite unions of half open intervals. Furthermore $F \subset M \subset F^{'}$ (I am unsure whether this is true) and note that: $\lambda^{d}(F^{'})-\lambda^{d}(F)< \epsilon$
It is clearly not correct, I need some help.