I am looking for an attractive, but rigorous definition of area; say in Euclidean plane. Probably there is no short definition. It is OK to make it even longer, but can it be built from useful parts in a not boring way? Say in such a way that not all the students will sleep on the lecture?


  • The real problem is to prove existence, the uniqueness is easy.

  • Using integral does not seem to be a good idea.

  • There is an approach where you write the formula for area and then proving its properties. I do not like it since it moves you to discrete geometry which is completely irrelevant and the ideas used nearly useless anywhere else (so no reason to learn this stuff). [See for example "Geometry: A Metric Approach with Models" by Millman and Parker.]

  • The method with measuring grid (cutting everything into small squares and counting) looks much better. One can consider this method as an introduction to integral. There is only one technical statement which has to be proved --- if you rotate square then its area does not change. The only problem is that it is not generalizable to say absolute plane or sphere...

  • One may define the area as a limit of $\varepsilon^2\cdot N_\varepsilon$, where $N_\varepsilon$ denotes the maximal number of points in the figure one distance $>\varepsilon$ from each other. The only hard part is to prove existence of the limit $\varepsilon^2\cdot N_\varepsilon$ for say polygons. (You can exchange the limit to ultralimit --- this way everything works smoothly, but I do not want to sell my soul just to get a def of area...)

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    $\begingroup$ How general a notion of area are you looking for? Eg do you want something very general like Lebesgue measure, or do you just want to be able to find the areas of simple figures in the plane (and if so, how simple? For instance, would subspaces with polygonal boundary be enough?) $\endgroup$ Jan 26, 2013 at 18:18
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    $\begingroup$ Can you be a bit more specific on the type of spaces and structures you are considering ? $\endgroup$ Jan 26, 2013 at 18:23
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    $\begingroup$ Maybe it's just the reference: Chapter IV of Hilbert's "Foundations of Geometry" treats precisely on this topic in about 10 pages. If you add Supplement III that gives another 7 pages. Boltyankii's little book "Equivalent and Equidecomposable figures" is also nice. I don't think any math student would be bored with it. Of course this equidecomposability approach does not work in dimensions higher than 2. That was the subject of Hilbert's third problem. $\endgroup$ Jan 27, 2013 at 6:49
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    $\begingroup$ What's wrong with Euclid's approach? Area is a primitive, so he doesn't define it. It has the properties given in common notions 1-5. If area were non-unique, it would contradict common notion 1, and the system would be inconsistent. Existence isn't an issue, because he isn't trying to construct a map onto the reals. Non-measurable sets don't arise, because the approach is constructive, and the methods of construction are incapable of creating non-measurable sets. Calculus and measure theory are topics for courses in calculus and measure theory. $\endgroup$
    – user21349
    Jan 27, 2013 at 16:02
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    $\begingroup$ My feeling is that if this is aimed at undergraduate students, you might have to bite the bullet and work with a grid...it's boring but easy. $\endgroup$ Jan 30, 2013 at 6:34

5 Answers 5


There is an intuitive approach to area, based on the fact that polygons $P, P'$ have the same area if and only if they are equidecomposable (that is, one may be cut into pieces and reassembled to form the other).

The first three pages of this note sketch a "motivic" approach to the definition of area, for polygons. Namely, one defines $K(\text{Poly})$ to be the free Abelian group generated by plane polygons $P$, subject to the following two relations:

  1. $[P]=[P']$ if $P$ is congruent to $P'$
  2. $[P]=[P_1]+[P_2]$ if $P$ may be cut into polygons $P_1$ and $P_2$.

An easy exercise (sketched in the note I link to) shows that $[P]=[P']$ if and only if $P$ and $P'$ have the same area, so $K(Poly)\simeq \mathbb{R}$. But even better, one can define the area of a polygon $P$ as its class $[P]$ in $K(\text{Poly})$.

Indeed, for many reasonable classes of subsets of the plane, one may extend this definition to assign to such a set a class in $K(\text{Poly})$. For example, say a sequence of classes $[P_i]$ in $K(\text{Poly})$ converges to $[P]$ if there exists a representative $[A]-[B]$ for $[P-P_i]$ with both $A$ and $B$ contained in $[0, \epsilon_i]\times [0, \epsilon_i]$, with $\epsilon_i\to 0$.

Suppose $X$ is a subset of the plane so that there is a sequence of polygons $P_i$, such that the symmetric difference $(X\cup P_i)-(X\cap P_i)$ is contained in a polygon $Q_i$. Suppose further that $[Q_i]=[Q'_i]$ and $Q'_i\subset [0, \epsilon_i]\times [0,\epsilon_i]$, with $\epsilon_i\to 0$. Then we assign to $X$ the class $$\lim_{i\to \infty} [P_i]$$ if it exists. It's not hard to check (geometrically!) that this assignment is well-defined.

NB: This approach does not work to define volume in $\mathbb{R}^n, n>2$. Indeed, Dehn showed that there are many polyhedra with the same volume that are not equidecomposable.

  • $\begingroup$ For your limit procedure to work as expected, one need a bit of extra structure (for example, to trace the action of dilatations on the k-group) for otherwise, as the isomorphism $K(poly)\cong R$ is only an isomorphism of abelian groups and R has many automorphisms, the limits you are taking might do weird things. $\endgroup$ Jan 26, 2013 at 18:48
  • $\begingroup$ I don't think so, Mariano---I define a topology on K(Poly) in the 3rd paragraph, which agrees with the usual topology on R. $\endgroup$ Jan 26, 2013 at 18:53
  • $\begingroup$ This is the same as in Millman and Parker --- with proofs it takes 40 pages, not sexy at all. $\endgroup$ Jan 26, 2013 at 18:59
  • $\begingroup$ Ah, I haven't seen the book. What could possibly take them so long? $\endgroup$ Jan 26, 2013 at 19:02
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    $\begingroup$ @JeffStrom: This is actually well-studied. The $3$-dimensional version gives $\mathbb{R}\oplus (\mathbb{R}\otimes \mathbb{R}/\mathbb{Z})$, where the first factor is ordinary volume and the second factor is the Dehn invariant. IIRC we don't know a complete description of the $4$-dimensional version. But lots of people (e.g. Bloch, Goncharov, and others) have worked on this sort of thing--the hyperbolic version has relations to periods and other things; also some people (e.g. Inna Zakharevich) have studied "higher $K$-groups" in this setting. $\endgroup$ Nov 19, 2015 at 15:21

If you are content with areas of polygons, you can define it à la K-theory.

Let $\mathcal A$ be the free abelian group generated by polygons in the plane modulo the relations that identify pairs of congruent polygons and such that $P=P_1+P_2$ if $P_1$ and $P_2$ are the result of cutting the polygon $P$ into two pieces with a line.

One can show that $\mathcal A\cong\mathbb R$ as a group, and that the canonical map $\mathrm{Polygons}\to\mathcal A$ sending a polygon to its class is the area.

  • $\begingroup$ This approach has issues in higher dimensions, though... $\endgroup$ Jan 26, 2013 at 18:43
  • $\begingroup$ Daniel wrote the same thing as I typed this... $\endgroup$ Jan 26, 2013 at 18:43
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    $\begingroup$ See also geometric measure theory and Hadwiger's theorem, which is very nicely explained in a series of posts on the n-Category Café starting here: golem.ph.utexas.edu/category/2011/06/hadwigers_theorem_part_1.html $\endgroup$ Jan 27, 2013 at 2:21

For students, probably the most elementary thing to do would be to restrict yourself to regions in the plane that can be triangulated with finitely many triangles (with straight sides). Accepting the area of a triangle as known, you then define the area of a region by adding up the areas of the triangles in a finite triangulation. The one thing you have to check is that this is well-defined. For this, I would first prove that any two finite triangulations of a region in the plane have a common subdivision (if you're reasonably clever about it, this can be done very quickly; certainly in 3-5 pages), and then prove that your notion of area is invariant under subdivisions.

The nice thing about this is that all the main properties you want (eg that areas behave correctly under linear maps and translations) come for free from the analogous properties of triangles, which are easy.

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    $\begingroup$ You can also allow for a countable triangulation, by taking the limit (you of course need to check that the limit is independant on the sequence of triangulation, but you already did the dirty work beforehand in proving the common subdivision lemma). Then you can measure any bounded element of the borelian tribe. It's easy to prove that it is the unique measure giving the usual formula for triangles, which satisfies the usual axiom of a measure. Personally I prefer using rectangles with sides parallel to a given system of cartesian coordinates. It's the same, but easier for students... $\endgroup$ Jan 26, 2013 at 20:33
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    $\begingroup$ @Loïc Teyssier : Countable triangulations are harder for two reasons. One, you have to worry about limits converging and what not. Two, it is much more technical to prove that countable triangulations have common subdivisions than finite triangulations, and then it is harder to prove that subdividing doesn't change the area since you can't express a subdivision as a finite sequence of "vertex additions". BTW, I agree that if you want to construct Lebesgue measure then rectangles are best -- for that, I recommend the treatment in Frank Jones's book "Lebesgue integration in Euclidean Space". $\endgroup$ Jan 26, 2013 at 20:46
  • $\begingroup$ (there are also issues with non-locally-finite triangulations that arise if you try to deal with sets that are not open) $\endgroup$ Jan 26, 2013 at 21:27
  • $\begingroup$ You keep adding the same answer :) $\endgroup$ Jan 26, 2013 at 21:37
  • $\begingroup$ @Anton Petrunin : My answer not quite the same as the others. For instance, it works correctly in higher dimensions. $\endgroup$ Jan 26, 2013 at 21:46

I don't know how much about areas you want to prove, and how developed a background the audience is supposed to have, but here is a definition of area of a bounded planar region.

Take such a region $D$ in $\mathbb{R}^2$, $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\ve}{{\varepsilon}}$ consider the $\varepsilon$-grid $(\varepsilon\bZ)^2$ and denote by $N_\ve(D)$ the number of $\ve$-pixels that touch $D$. (A pixel is one of the $\ve\times\ve$-tiny squarers of the grid.) We then declare $D$ to be measurable (i.e. to have area) if the limit $\newcommand{\eA}{\mathscr{A}}$

$$ \eA(D)= \lim_{\ve\to0}\ve^2 N_\ve(D) $$

exists. If this is the case, then we define the area to be the limit $\eA(D)$, and we set $\eA_\ve(D)=\ve^2N_\ve(D)$.

The first step is to prove $\newcommand{\bR}{\mathbb{R}}$ that if $L,U: [a, b]\to \bR$ are Riemann integrable functions and $D(U,L)$ is the region

$$D_f= \bigl\lbrace\; (x,y)\in [a,b]\times \bR;\;\;L(x) \leq y\leq U(x)\;\bigr\rbrace, $$

then $D(U,L)$ is measurable

$$\eA(D(U,L))=\int_a^b \bigl(\; U(x)-L(x)\;\bigr) dx. $$

The next thing to prove is a weak form of the inclusion-exclusion principle: if $D_1$, $D_2$ are measurable regions that intersect along the graph of a $C^1$-function, then $D_1\cup D_2$ is measurable and

$$\eA(D_1\cup D_2)=\eA(D_1)+\eA(D_2). $$

  • $\begingroup$ It is close to what I call definition with measuring grid. You do not need to use integrals at all, but you will need to sow that area of any unit square is the same. $\endgroup$ Jan 27, 2013 at 17:13
  • $\begingroup$ Those two facts I mentioned will prove this. Cut the square in question along the diagonal. $\endgroup$ Jan 27, 2013 at 19:35
  • $\begingroup$ right, but still, it is OK to define area which motivates a def of integral, but it seems to be wrong to use integral to define area. $\endgroup$ Jan 27, 2013 at 19:50

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 quasi-disjoint (intersect at most along a common edge) union of acceptable 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.


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