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I am working on extended essay for my International Baccalaureate Programme Full Diploma, and I chose mathematics as my topic. My research questions is: “What are the shortcomings of Riemann integration, why do they exist, and how do other forms of integration overcome those limitations?”

The first alternative I will discuss is Lebesgue integration. By starting with a graphical analysis of a bijective function $f\left( x\right)$ whose limit will be evaluated from $x=a$ to $x=b$, I have attempted to write the process as a unified equation,

$$L=\small\lim_{n\to\infty}\sum_{i=1}^{n}\left(\left[ f^{-1}\left( a+i\left[\dfrac{f\left( b\right) -f\left( a\right)}{n}\right]\right)-f^{-1}\left( a+\left[ i-1\right]\left[\dfrac{f\left( b\right) -f\left( a\right)}{n}\right]\right)\right]\left( a+i\left[\dfrac{f\left( b\right) -f\left( a\right)}{n}\right]\right)\right)$$

By comparison, my formula for Riemann integration follows the form

$$R=\lim_{n\to\infty}\sum_{i=1}^{n}f\left( a+\frac{i\left( b-a\right)}{n}\right)\left(\frac{ b-a}{n}\right)$$

I am having extreme difficulty proving that $L=R$, since I do not know how to algebraically evaluate the limit of a series. However, I did try to test my formulae for $f\left( x\right)=x^2+1$, $a=0$ and $b=a$. While the formula $R$ has successfully yielded the correct answer under every trial, I have only gotten $L$ to converge to $R$ when integrating symmetrical trigonometric functions; in this case, however, $L$ ended up appearing to be undefined for all $x>0$.

Does anyone have any suggestions? Have I made a key mathematical error? an arithmetical one? I pulled my information from Arturo Magidin's answer under “Lebesgue Integration Basics.”

Thank you for any help you can give.

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gen-ℤ ready to perish
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  • I think this is a hugely ambitious project at your level. The two integrals are equal when they are both defined. However each can integrate things the other will not. Lebesgue is generalisable to multiple dimensions and has good results like the dominated convergence theorem. But the advantages are subtle and there are good reasons that a mathematician as good as Riemann thought his integral did the job. You will get some good answers on the mathematics, I am sure, but you should also mention the history - the Riemann Integral avoids some pathologies of naive area under the curve approaches. – Mark Bennet Jan 06 '17 at 20:53
  • The Lebesgue integral is strictly more general than the Riemann integral. The improper Riemann integral can integrate things the Lebesgue integral can't, but the Riemann integral is only defined for bounded functions on a bounded interval. – Chappers Jan 06 '17 at 21:02
  • I would second Mark's comments and add that the essential ingredient here is measure theory. Trying to prove the formula above is a fairly thankless task. – copper.hat Jan 06 '17 at 21:16

1 Answers1

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I don't know where you're getting that formula for Lebesgue integration. Here's how I would show that the Lebesgue integral agrees with the Riemann integral for Riemann integrable functions.

  1. First look at step functions. Show it's true there.
  2. If $f$ is Riemann integrable on $[a,b]$, then there are sequences of step functions $g_n$ and $h_n$ such that $g_n \le f \le h_n$ and $\int_a^b h_n(x)\; dx$ and $\int_a^b g_n(x)\; dx$ both converge to the Riemann integral $\int_a^b f(x)\; dx$ as $n \to \infty$.
  3. Use this to show that $f$ is Lebesgue integrable and that this same limit is its Lebesgue integral.
Robert Israel
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