I will be restricting myself to $O(2)$ here to make things a bit easier. I will also be using Jordan measure.
I know that showing a set is measurable can be done by showing that its boundary has measure $0$. In the case of $O(2)$, I can show that every point in it is a boundary point. Meaning $O(2)=\partial\: O(2)$. So I am now trying to show that $O(2)$ is a nullset.
I had an idea to try to write $O(2)$ as the graph of an integrable function $f: \mathbb{R}^3\to \mathbb{R}$. Then we have for $x \in \mathbb{R}^3$, $$ \begin{pmatrix}x_1 & x_3\\x_2& f(x)\end{pmatrix} $$ and we want to build an integrable $f$ so that $\{(x,f(x)): \: x\in \mathbb{R}^3\} \cong O(2)$. The graph would be a nullset and so $O(2)$ as well.
However this doesn't seem to work. I don't know how to get the graph to cover the entirety of $O(2)$.
I would be grateful if I could get a tip to continue or maybe if there is a better way of doing this.