Since any two Euclidean shapes have an infinite number of points inside of them, shapes with different area have the same infinite number of points in them (and any object has the same number of points in it as are inside the plane it is inside). So area isn't a measure of the amount of points in an object, right? And if this is true, what does the area of a shape actually represent? In other words beyond the abstract idea of the amount of space in an object, what does an area of, for example, $4$, mean?

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    $\begingroup$ Area is a measure, just like length is. On a line between two points, there are an infinite number of points. The length of the line is a measure of the line, it's not counting the number of points. $\endgroup$ – Glen O May 23 '13 at 0:38

Area (in $\mathbb R^2$) starts with agreeing that the area of a rectangle of sides $a$ and $b$ has area equal to $a\cdot b$. You may take that as an axiom. Now, the area of more complicated shapes is obtained by approximating the shape by rectangles. This can be done in different ways and there are plenty of things to prove here, as well as some surprises. For instance, one can approximate a given shape from the outside: cover it by countably many disjoint rectangles, and add their areas. Take the infimum over all such covers, and that will give you a measure of the area of the shape measured from the outside. Similarly, you can inscribe rectangles inside the shape, add their areas, and take the supremum over all such numbers. This will give you a measure of the area of the shape measured from the inside.

Your question is a prelude to measure theory, a very important part of modern analysis. One of the important surprises is that the concept of area (and even the concept of length) can't be extended sensibly to measure all subsets of the plane (line). A famous example is Vitali's Set. So, the resulting theory is quite subtle.

  • $\begingroup$ Okay. Going from the idea of the metric of a space to area as $ab$ felt strange, but if it's just a definition used to generally define area that makes sense. Thanks. $\endgroup$ – user71641 May 23 '13 at 0:54
  • $\begingroup$ "Now, the area of more complicated shapes is obtained by approximating the shape by rectangles." This could get better said by saying "can get obtained by... using the areas of rectangles" instead of "is obtained by" since using the areas of polygons will work also (yes rectangles are polygons, but polygons aren't rectangles). Actually, I think one can make a case that using the areas of polygons comes as much more elegant than using the areas of rectangles here. $\endgroup$ – Doug Spoonwood May 23 '13 at 1:03
  • $\begingroup$ @nsanger: If going to area as $ab$ feels strange, then take a step back and ask what it is you would expect the concept of area to do. Two not-unreasonable expectations are (1) congruent figures have equal areas, and (2) a (finitely) sub-divided figure has area equal to the sum of the areas of its sub-figures. Now, suppose we assign the symbol "$u$" to the "area" (whatever that means) of a $1\times 1$ square. Notice that an $a\times b$ rectangle (for integers $a$, $b$) can be sub-divided into $ab$ squares of "area" $u$; its "area" is $uab$. (continued) $\endgroup$ – Blue Sep 15 '13 at 5:59
  • $\begingroup$ (Part 2) A little thought reveals that the same formula holds for $a$, $b$ rational, and it's not difficult to extend this to include $a$, $b$ irrational. Every rectangle then has "area" $uab$. We (again not-unreasonably) simply agree to call $u$ ---the "area" of the $1\times 1$ square--- by the name "1 square unit". And we move on from there. (Interestingly, in non-Euclidean geometry, there are no rectangles or squares, so area is a trickier notion. It turns out that the expectations (1) & (2) lead to a definition of "area" that stem from the sums of the measures of angles in a triangle.) $\endgroup$ – Blue Sep 15 '13 at 6:06

It tells you that you could cover the region with 4 $1\times 1$ squares. If it's a weird shape, it would be more informative to say that if you cut it out of material and a $1\times 1$ square of that material weighs 1 ounce, then the shape you cut out would weigh 4 ounces.

  • $\begingroup$ What about non-measurable subsets? $\endgroup$ – Calvin Lin May 23 '13 at 0:41
  • $\begingroup$ You specified an area of 4 square units (like square inches), so I will measure a 1 inch by 1 inch square. You cannot see points, let alone count uncountably many of them, so drop that. @Calvin Lin, hush up! :) $\endgroup$ – Ted Shifrin May 23 '13 at 0:48
  • $\begingroup$ Three upvotes for @Ittay and a downvote for me for saying explicitly what he took as an axiom. I guess I should give up! It would help to know the mathematical level of our OP. $\endgroup$ – Ted Shifrin May 23 '13 at 0:53
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    $\begingroup$ I didn't vote initially, but gave you a +1 as the -1 didn't make sense in my opinion. I stayed out of answering this question as it gets too hairy. $\endgroup$ – Calvin Lin May 23 '13 at 0:56
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    $\begingroup$ Dear Ted, I just noticed this and I like your answer. Regards, $\endgroup$ – Matt E Sep 15 '13 at 5:40

Yes, area isn't a measure of the amount of points of an object. A positive measure P of an object J implies that J can get further subdivided into parts, since P consists of a positive real number and for all positive numbers, there exists a real number X such (X+X)=P. A point T consists of something which has no part, making subdivision into parts impossible and thus T has measure of zero.

The area of a shape represents how much space a 2-dimensional object would take up in a 2-dimensional plane.

  • $\begingroup$ not only points have measure $0$ and the amount of space a 2-dimensional object takes up is $0$. $\endgroup$ – Ittay Weiss May 23 '13 at 8:29

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