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I always thought that areas are defined by integrals, until I read Michael Spivak's Calculus p.289:

The desire to define area was the motivation, both in this book and historically, for the definition of the integral, but the integral does not really provide the best method of defining areas, although it is frequently the proper tool for computing them.

So, what is the best way to define areas? What is the weakness of using integrals to define areas?

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2 Answers 2

It is, at least partially, a chicken-and-egg problem. We'd like to define the Lebesgue integral which has nicer convergence properties when we integrate sequences of functions, and take limits of the functions and integrals -- but the definition of the Lebesgue integral requires that we already have a well-behaved measure (i.e., "area") concept to build it with.

The area measure we can get from ordinary freshman-calculus Riemann integrals is not defined for "wild" enough subsets of $\mathbb R^n$ that it can be used to bootstrap the Lebesgue integral into its full generality; in order to do that we need an integral-free definition of the measure to start with. Afterwards we can Lebesgue integrate many weird sets to get out the area measure we started with.

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I don't really understand (FYI, I'm still a freshman). The only integral that I know is the Riemann integral, and I was asking about areas in $\mathbb{R}^2$. So, is there any integral-free definition of areas? –  greg Oct 17 '11 at 2:54
    
We can list desired properties of area as axioms, including such items as the fact that the area of a $1 \times 1$ square is what it is, that area of a countable union of disjoint sets is the sum of the areas, that area is invariant under translation, rotation, reflection, a few more. –  André Nicolas Oct 17 '11 at 3:00
    
Hmm, I sort of assumed that the book you quote from would segue into defining the Lebesgue measure so I wouldn't have to describe it here. Now I've found a table-of-contents for it online, which makes that look rather unlikely. Gotta run now, but I'll check back later and see if someone else has answered this. –  Henning Makholm Oct 17 '11 at 3:00
    
Returning as promised, but I don't think I really have much to add to André's comment. The usual definition of the Lebesgue measure is to list axioms one wants it to satisfy -- Wikipedia's list looks fairly accessible. The one proves that there is a unique function that satisfies those properties, after which one can promptly forget how that proof worked. The properties are the important thing. –  Henning Makholm Oct 17 '11 at 23:10
    
Thanks. I kind of understand it now. –  greg Oct 18 '11 at 2:12

Consider the function $f$ that takes 1 on the rationals in $[0, 1]$ and 0 elsewhere. What's the area under $f$? You can't answer this question with the Riemann integral since the upper and lower limits of the Riemann sums don't agree. With the power of Lebesgue integration, we can say that the area under $f$ is in fact zero, because the rationals should in a sense have "zero length".

Your question is one of the motivations for the development of measure theory and Lebesgue integration. Probability theory is also based on measures, and the question of how to assign probabilities to events and areas to subsets are one and the same.

To even define an integral in the first place, you need to decide on a way to assign lengths to subsets of the real numbers, since this is what you're integrating against. Start with the real line for example: how does one measure the length of a subset of the line? For intervals $(a, b)$ we see that it's length should be $b-a$. But what should be the length of sets like the rational numbers, the irrationals, or the Cantor set? (Interestingly, not all sets can be assigned such a length, see Vitali sets for example). Measure theory tells us how to answer such questions.

The axioms of a measure lay out the properties that we'd like a suitable area/length/volume function to satisfy, and in particular, Lebesgue measure is the measure function you want when thinking of area in the normal way that assigns each interval its length. Measures can actually be thought of as integrals in many cases, which is essentially the content of the Radon-Nikodym theorem.

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Thanks. Are you saying that Lebesgue integration is a better way to compute areas than Riemann integration? –  greg Oct 18 '11 at 2:14
    
@greg: Lebesgue integration allows you compute the areas of more kinds of regions than Riemann integration does. –  Arturo Magidin Oct 18 '11 at 4:37
    
It's worth to add to Arturo's comment that if a set has an area which can be computed using the Riemann integral, then this area can also be computed using the Lebesgue integral and these two values are equal. –  Damian Sobota Oct 18 '11 at 22:37
    
Ahh I see. So Lebesgue integration is a generalization of Riemman integration, right? Or is it not so accurate to say that because they have completely different methods? –  greg Oct 19 '11 at 3:46
    
It is a generalization. And in a sense they are similar, because we can define the Riemann integral by means of the Jordan measure. The Lebesgue measure generalizes the Jordan measure. –  Damian Sobota Oct 19 '11 at 14:44

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