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Note: I am asking two questions. Replace "area" by "volume" with appropriate adjustments to get the second question

Hello everyone. I am studying mathematics in college and I am trying to find the formal definition of the notion "area" in geometry. What is area? I can understand it when the sides involved are integers, like when the side is 3, we have nine 1x1 grids, so the area is 9 if we think of area as the number of "unit grids". Maybe we can also define area for squares with rational sides in the same vein, but this is not a good way of thinking because how to define the area of a square of side PI or an irrational number for example?

Please give me a formal definition of area. I am looking for the most precise definition. I am not finding any results on the internet. Thanks in advance

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    $\begingroup$ en.m.wikipedia.org/wiki/Lebesgue_measure $\endgroup$ – Praise Existence Aug 12 '16 at 22:00
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    $\begingroup$ @PraiseExistence I know! This is how I like to think about it. But how do people talk about areas and volumes in classical geometry? This is what puzzles me $\endgroup$ – user00000 Aug 12 '16 at 22:01
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    $\begingroup$ By classical geometry do you mean Euclidean geometry? $\endgroup$ – M10687 Aug 12 '16 at 22:02
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    $\begingroup$ I doubt Euclid ever defined area. In fact, as far as I know there was no truly precise definition until the late 19th century. $\endgroup$ – Eric Wofsey Aug 12 '16 at 22:03
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    $\begingroup$ In other words, how was area defined in Hilbert's axiomatization of Euclidean geometry? $\endgroup$ – littleO Aug 12 '16 at 22:16
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Never underestimate the value of primitive notions! That said:

Here's one way to define area for a wide range of shapes in $\mathbb{R}^2$ without resorting to measure theory (and is arguably closer to the intuitive definition of area - but see below). First, define by fiat that the area of an $a\times b$ rectangle is $ab$. Next, we say that two shapes have the same area if we can cut one shape into finitely many pieces by finitely many cuts, and rearrange those pieces through finitely many rigid motions to get the other shape. See page 2 of this paper by Hales.

(EDIT: the definition of area of a rectangle introduces nontrivial issues around well-definedness, see the comments by Eric Wofsey below. If we want to avoid this issue, we can just define the area of a shape to be the set of all shapes it is scissors-equivalent to.)

Now, this approach has a couple drawbacks. First of all, it utterly fails to measure the area of non-polygons. But arguably that's fine - maybe in the context of classical Euclidean geometry, we really only care about polygons, and indeed this approach does successfully compute the area of every polygon: any polygon can be cut and rearranged appropriately into a rectangle. The more serious problem comes when we try to generalize to higher dimensions: it turns out that even in dimension $3$, things break down! There are polyhedra with the same volume which cannot be cut and rearranged into each other. See Hilbert's third problem. This can be fixed by adding more complicated operations than just the scissor operations, but things rapidly get complicated. In particular, I'm not aware of any nicely-describable version of the scissor congruence which works for $n$-dimensional polyhedra; which, in my mind, is a good argument for the naturality of Lebesgue measure.

Note that Lebesgue measure is really two separate topics: first, outer measure ($\mu(A)$ is the inf over all covers by boxes of the sum of the measures of the boxes involved), and second, the algebra of measurable sets. But this second issue can be laid aside for "classical" geometry, and what we're left with is a formal notion that - in my opinion - does a very good job of capturing the informal concept of area/volume/etc. In particular, arguments by exhaustion for calculating the area of (say) the circle feel (to me) grounded in a view of area very close to this one. I go back and forth over which (scissors or measure) corresponds better to the "pre-theoretic" notion of area/volume/etc (and even over whether there is such a notion in the first place in any meaningful sense!). Right now, I fall very slightly on the scissors side, but that may change.

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  • $\begingroup$ Actually, you don't need anything more than scissors-congruence if you take the theory of real numbers as given. You can't prove directly when two polyhedra have the same value, but you can bound the volume of any polyhedron above and below (by sandwiching it between unions of tiny cubes, for instance) and thus uniquely determine volumes of all polyhedra. $\endgroup$ – Eric Wofsey Aug 12 '16 at 22:36
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    $\begingroup$ A difficulty with the scissors-congruence approach that you're glossing over is proving that it's well-defined. That is, how do you know you can't get two different areas for the same polygon by cutting it up in different ways? If you're treating geometry axiomatically and taking area as a primitive notion, you can just assume this as an axiom. But in a modern context where we model geometry within some broader foundation for mathematics, this is a nontrivial theorem to prove. $\endgroup$ – Eric Wofsey Aug 12 '16 at 23:13
  • $\begingroup$ @EricWofsey I mean, the right way to define area via scissors-congruence is to say that the area of a polygon is the set of all polygons it is scissors-congruent to. Now there's no problem of well-definedness; the fact that this corresponds to a map to $\mathbb{R}$, of course, is a nontrivial theorem. My point wasn't that these results are easy, just that the underlying concept is different from, and in certain ways more concrete (not simpler) than, that of outer measure. $\endgroup$ – Noah Schweber Aug 12 '16 at 23:16
  • $\begingroup$ Right--I just felt like this needed to be pointed out since the way you wrote it in the answer seemed to be starting with a map to $\mathbb{R}$ (for instance, since you define area of rectangles using multiplication). $\endgroup$ – Eric Wofsey Aug 12 '16 at 23:18
  • $\begingroup$ @EricWofsey Oh, yes, that was sloppy of me. I've added a note. $\endgroup$ – Noah Schweber Aug 12 '16 at 23:24
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So in Euclidean geometry we have a list of axioms, together with some undefined terms. These undefined terms include: "point," "line" and "area," among others. So "area" has no precise definition in Euclidean geometry. See here.

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    $\begingroup$ You are misreading the link you supplied. Your link confirms that Euclid's notion of area is well-defined. $\endgroup$ – Rob Arthan Aug 12 '16 at 23:52
  • $\begingroup$ Ok, I kind of messed up with that link. The real undefined term there is "region." I recalled from a geometry book I had a while ago that "area" was undefined, but in that book the term "region" had a definition if I'm remembering correctly. Good point. $\endgroup$ – M10687 Aug 13 '16 at 1:34
  • $\begingroup$ @RobArthan: Your link confirms that Euclid's notion of area is well-defined. No, the answer is exactly right. The link gives a list of axioms about area, but it treats area as a primitive, undefined term. That's how area is treated in any approach to Euclidean geometry that is analogous to Euclid's original approach. $\endgroup$ – Ben Crowell Jan 6 '17 at 16:10
  • $\begingroup$ @Ben Crowell: the link says that the area of a unit square is unity. That (together with the other axioms) constitutes an implicit definition. $\endgroup$ – Rob Arthan Jan 6 '17 at 21:06
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Jordan Measure defines a notion of area that works for "classical geometry". I am sure that Archimedes would have been quite happy with it once he'd understood our modern obsession for Cartesian coordinate systems.

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    $\begingroup$ I'm surprised this wasn't mentioned by the other answers. It also gives a rigorous proof for en.wikipedia.org/wiki/Cavalieri%27s_principle, which in turn is closest to the ancient Greek method of exhaustion. =) $\endgroup$ – user21820 Aug 13 '16 at 8:40
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The other answers so far have mentioned basically three kinds of unsigned area:

  1. Scissors-congruence to compare areas. Note that this approach does not define area itself as a magnitude, unless you further add the notion of real numbers. It also fails for non-polygons unless you take limits, in which case it becomes equivalent to the next kind of area.

  2. Jordan measure (equal to both the limit of the total area of grid squares contained in the figure as the grid size tends to zero and the limit of the total area of grid squares needed to contain the figure as the grid size tends to zero, if these limits are equal). Any bounded object in classical geometry has finite Jordan measure, and this is the simplest intuitive notion that the classical geometers would surely agree with, since the method of exhaustion is essentially based on the same principle. Jordan area defined is finitely-additive and hence classical geometry works as usual.

  3. Lebesgue measure, which assigns a countably-additive area to some sets of the plane, and is compatible with Jordan measure whenever the Jordan measure is defined.

All of the above generalize easily to higher dimensions.

However, there is also signed area, which can drastically reduce the number of cases in Euclidean plane geometry when it comes to polygons. We define (signed) area of a polygon using the shoelace algorithm, and then check that $area(XP...QYR...S) = area(XP...QY) + area(YR...SX)$ (the area of a polygon is the sum of the areas of its parts), and that area is preserved when the vertices are cycled but negated when they are reversed. This allows us to handle most geometry problems without having unnecessary cases, unlike if we use unsigned area.

Nevertheless, it is not so easy to extend signed area to non-polygonal regions. We can define the (signed) area of a region bounded by a (directed) rectifiable closed curve as the limit of the area of the approximating polygon. It turns out that the limit exists, but is not so easy to prove. (To see that the condition of rectifiability cannot be dropped, note that an Osgood curve is a continuous closed curve with positive Lebesgue measure, and hence its interior is not Jordan measurable, which implies that the limit does not exist.)

Also, signed volume is even more troublesome to define, so it is not very useful compared to the Lebesgue measure for general sets in $\mathbb{R}^3$.

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