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I know that certain functions are not integrable. Geometrically integration is finding the area under the curve of the graph of the given function. So, in another way why is it not possible to find the area under the curve in case of certain functions?

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    $\begingroup$ It's not easy to explain this without invoking some high-powered stuff. One can find a non-measureable subset of the interval $[0,1]$. Then the function that is 1 on that set and zero elsewhere won't be integrable. $\endgroup$ – Gerry Myerson Aug 27 '13 at 12:36
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    $\begingroup$ If you take integrable to imply finite as well, simply take for example $f(x)=1/x$ on $(0,1)$. Otherwise, see Gerry Myerson's comment $\endgroup$ – Clayton Aug 27 '13 at 12:37
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    $\begingroup$ A partly inaccurate translation of Gerry Myerson's comment into layman terms is that there are bounded subsets of the plane that are so irregular that we are unable to talk about their area. If you accept this, then it is not too much of a surprise that some functions are so irregular that the collection of points under their graph forms such a subset of the plane. $\endgroup$ – Jyrki Lahtonen Aug 27 '13 at 12:42
  • $\begingroup$ By the way the idea of integration as area under the curve of the graph is more the idea of Riemann's integration : dividing the input space into subdivisions. Lebesgues is more about dividing the space fo function's values. $\endgroup$ – Bertrand R Aug 27 '13 at 12:50
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    $\begingroup$ There are different definitions of integration too, and some functions integrable by one method may not be integrable by another - that is in part why new definitions were developed. The function on the real interval $[0,1]$ which is equal to $1$ at rational points and $0$ everywhere else is Lebesgue integrable, but does not have a Riemann integral. $\endgroup$ – Mark Bennet Aug 27 '13 at 13:33
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Well, well, well, this question actually can go really deep into the elementary ideas of integration, or, in general, finding the area under a curve.

1: The Old School Way - The Method of "Exhaustion"

In the ancient times, and especially during the period when mathematicians such as Eudoxus and Archimedes lived, the main way in which one could calculate the area between non-linear curves was the Method of Exhaustion. We will examine, in short, how would this work on calculating the area of a circle.

Let a circle $\mathcal{C}$ of radius $R$ and let two squares, $A_1A_2A_3A_4$ (red) and $B_1B_2B_3B_4$ (blue) such that $A_1A_2A_3A_4$ is inscribed in $\mathcal{C}$ and $B_1B_2B_3B_4$ is circumscribed to $\mathcal{C}$. Then, we have the following situation:

First step in "exhausting" the circle

Now, the next step is to enclose the circle in some "tighter" canonical polygons. An easy construction is to construct the octagons that are induced from the two squares $A_1A_2A_3A_4$ and $B_1B_2B_3B_4$, as it is depicted in the following shape:

Second step in "exhausting" the circle

Then one can move a step forward to create a deca-exagon (16 angles canonical polygon) etc. It is clear the the red polygon is always underestimating the area of the circle and the blue polygon is always oversetimating it.

Now, it is relatively easy, using the know formulas for the area of canonical polygons, to prove that both the red and the blue area do converge to the area of the circle. However, since back then calculus was not something "known" with the modern-day meaning (numbers were always treted as geometrical objects and not as abstract notions) they used the following form of Eudoxus' principle:

Eudoxus' Principle: Given two quantities $A,B$, if, from the first quantity $A$ we substract a quantity $C$ that is larger that the half of $A$ and, then, from the remnant ($A-C$ in modern notation) we substract a quantity that is larger than it half and we continue this way, then, after a finite number of steps, the remnant quantity will be less than $B$.

So, in that framework, it was sufficient to prove that each construction of a new $2^n-$sided polygon was "cutting" more than half of the error - which can be proved, but has a long and tiring proof and, hence, is out of the spectrum of this answer. After that, "convergence" (this is a modern term; it was not used back then) was ensured. What was missing was to show that these two areas (blue and red) were both "converging" to the area of the circle. To do so (since calculus and, hence, limits, were unknown) an argument - again tiring, in the general case - of double contradiction was used.

It is remarkable that, in the same way one could find the volume of the sphere and several other 3D shapes.

2: The Riemann-Darboux Way - Upper and Lower Sums

It is called Riemann Integral, however, in order to find the very solid geometric foundation of the notions of integral and integration we should investigate the so-called Darboux Upper and Lower Sums.

Let us consider a bounded function $f:[a,b]\to\mathbb{R}$ - boundedness is necessary for Rieman-integration; else, what would be the meaning of the area under that curve? Then, let a partition $\mathcal{P_n}=\{a=x_0<x_1<\dots<x_n=b\}$ of the interval $[a,b]$ and also let $m_k=\inf\limits_{x\in[x_k,x_{k+1}]}f(x)$ and $M_k=\sup_{x\in[x_k,x_{k+1}]}f(x)$, $k=0,1,\dots,n-1$. Then we define the lower, $L(f,\mathcal{P_n})$, and upper $U(f,\mathcal{P_n})$, Darboux sums for that partition and function as follows: $$\begin{align*} L(f,\mathcal{P}_n)&=\sum_{k=0}^{n-1}m_k(x_{k+1}-x_k)\\ U(f,\mathcal{P}_n)&=\sum_{k=0}^{n-1}M_k(x_{k+1}-x_k) \end{align*}$$

Also, we give the following definition:

Thinner partition: A partition $\mathcal{Q}$ is called thinner than $\mathcal{P}$ if: $$\mathcal{P}\subseteq \mathcal{Q}$$

Now, as we go forward to thinner partitions $\mathcal{P_k}$, with $k>n$, we say that $f$ is Riemann integrable if: $$\lim_{k\to\infty}U(f,\mathcal{P_k})=\lim_{k\to\infty}L(f,\mathcal{P_k})=I\in\mathbb{R}$$ for every sequence of "decreasing" partitions (every partition of the sequence is thinner than every previous one).

Note now that, in this way, Rieman integration is "just" a more generalised implementation of the Method of "Exhaustion" that ancient mathematicians used. Instead of canonical polygons, we do use the polygons induced from the partition of $f$'s domain we choose and these polygons are, in a way inscribed and circmuscribed to that curve. Also, thinner partitions are, in a sense, the same thing as going form a canonical $n-$polygon to another canonical $m-$polygon, with $m>n$. Moreover, what we seek, in order to be able to say that a function is Riemann integrable, is for the two limits of the sequences of the estimations given by the Darboux sums do converge to the same number, which was also the wanted in the Method of Exhaustion, in spite of not being proved with modern day calculus but with "old-fashioned" - however, interesting - geometrical arguments.

So, to conclude, a function is not Riemann integrable exactly when the Method of Exhaustion cannot be applied to the curve that represents it graph.

However, this does not apply to Lebesgue integrability... (I don't feel this was the case the OP requested, however, I can extend to Lebesgue integration, if requested).

3: Appendix - Integration as infinite non-countable summation and the Heuristic Method of Archimedes

In a "palimpseston" (Παλίμψηστον) - a script written on parchement that has been scratched and re-wrtiten in a direction vertical to the previous one - that was found in the early 20th century in an orthodox monastery close to Constantinople, after a small "adventurous" trip and, eventually, the proper scientific investigation, it was found that, the original script was a letter that Archimedes had writen to Eratosthenes in which, among other things, he was explaining the way though which he could find the results he should prove, however, he notes that he does not consider this method to be a method of prooving his results, just a way to "get a felling" of what the goal is.

In one of his examples he describes the way he found the area of a parabolic part - see this article in wikipedia: https://en.wikipedia.org/wiki/Archimedes_Palimpsest - and at some point he mentions that the reason he cannot consider this a proof is that of using a "summation of segments over an interval of real numbers" - this is in modern terms, not the ones used by Archimedes! So, with respect to the historical frame, Archimedes was very close to the notion of infinite, non-countable summation that we give to the Riemann integral, nowadays.

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I think this is more or less due to Banach-Tarski paradox, which shows a ball can be decomposed into a finite number of point sets and reassembled into two balls identical to the original one. This means we cannot define measurement (a conception rather related with integration) for every point set.

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    $\begingroup$ No, Banach-Tarski only comes up in higher dimensions, non-integrability already comes up in one dimension. $\endgroup$ – Gerry Myerson Aug 27 '13 at 12:40
  • $\begingroup$ You are right that the existence of non-measurable sets is why not every function is (Lesbegue) integrable, but the Banach-Tarski paradox is something that is completely unrelated. (It does involve non-measurable sets, though.) $\endgroup$ – Potato Aug 27 '13 at 12:41
  • $\begingroup$ I'm trying to explain why "area" (volume in this case) is not always well defined, and Banach-Tarski paradox is an example that we can construct point set that goes against our intuitive about area. Even we want measurement be additive in finite case there are non-measurable set exist. $\endgroup$ – Peter Wu Aug 27 '13 at 12:53
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    $\begingroup$ Maybe a better way of saying this would be that the Banach-Tarski paradox is due to this phenomenon. $\endgroup$ – Ryan Reich Dec 15 '13 at 18:52

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