To be honest, I'm no sure if this consrtuction matches the criterion of non-boringness, but it is rather short (I promise you all this takes less than 40 pages ;) ) and works in every dimension. It is related to the answer of Andy Putman, and the subsequent comments. I should maybe have kept on commenting, but there was simply not room enough.
For my physics classes I introduce measuring the area of bounded open and closed sets (or volume in $\mathbb{R}^n$ ), by considering countable unions of closed rectangles, the edges of all of them being parallel to a given system of cartesian coordinates (let's call them acceptable). The union must be clean, in the sense that the rectangles overlap neatly to form a rectangulation (say). Because any nonempty intersection of two acceptable rectangles is again an acceptable rectangle, any countable union of acceptable rectangles clearly admits a clean sub-rectangulation.The same argument works also pretty well for describing a common sub-rectangulation of two others having the same image.
Define the area first for finite unions, starting by assigning the usual value to the area of a single rectangle (given as axiom), and extending the area functional using the usual additivity for disjointquasi-disjoint (intersect at most along a common edge) union of measurable setsacceptable rectangles (given as axiom). You retrieve all the usual properties of an area in that case.
Now a bounded countable and clean union $\bigcup_{n\in\mathbb{N}}R_n$ of acceptable rectangles can be assigned its area as the series $\sum_{n\in\mathbb{N}}\mathtt{Area}(R_n)$ (which always converges). As pointed out above, finding a common sub-rectangulation is fairly straightforward, and the limit does not depend on the choosen sub-rectangulation (commutatively summable series).
Since any open set is rectangulable (in the sense that it breaks in a countable clean union of acceptable rectangles), we can measure any open set. Any compact $K$ is included in the interior of an acceptable rectangle $R$ and $R\setminus{K}$ is a bounded rectangulable set $O$. We assign $\mathtt{Area}(R)-\mathtt{Area}(O)$ to the area of $K$.
By adding more natural axioms you can again extend this functional to all bounded elements of the borelian tribe, but that's a classical story.