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 disjoint union of measurable sets (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.