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I'm trying to understand Lebesgue integral and integration but I'm completely mired by many author's motivation of this subject as only a tool for pathological functions such as the Dirichlet pathological function. I entirely appreciate the functionality of Lebesgue integration for this purpose but there just seems to be a huge gap between the classical calculus that I've taught since high school and this concept.

Can someone show how Lebesgue integration becomes Riemann integration for ordinary/smooth functions such as a constant or a polynomial?

Thank you

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    $\begingroup$ The Dirichlet function really is merely a pathological example (one can refine the definition of Riemann to capture these guys). What is not just pathological are the unbounded (continuous) functions over a precompact interval like $f(x)\equiv\frac{1}{\sqrt{x}}$ over $I=(0,1)$. There Riemann really fails! $\endgroup$ Aug 13 '14 at 21:22
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When we want to do Lebesgue integration to a Riemann integrable function like the ones you have described, it is helpful to recall Riemann sums.

For simplicity let $f$ be nonnegative, and let us integrate over a finite, closed interval $[a,b] \subset \mathbb R$.

We get the Lebesgue integral as the supremum of the integral of simple functions less than $f$.

Let us look at lower Riemann sums; for a partition $P = \{x_i | a < x_0, x_i < x_{i+1}, x_n < b \}$, the lower Riemann sum is the sum of rectangles "just barely" below the function on these intervals. This is the integral of a simple function, with value $\min\{f(x),x \in [x_i,x_{i+1}]\}$ on the interval $[x_i,x_{i+1}]$- so it is included in the supremum for the Lebesgue integral.

So for a riemann integrable function, as all the lower Riemann sums are integrals of simple functions, we can see that the Lebesgue Integral (the supremum of integrals of simple functions less than $f$) is greater than the Riemann integral (the supremum of lower riemann sums). From here it is not hard to show they are both bounded above by the upper Riemann integral.

For this simple case, Lebesgue is just a generalization of Riemann.

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Riemann Lebesgue Integration

There is a prove showing that $R(f,K)=L(f,K)$ if $R$ is the Riemann Integral and $L$ is the Lebesgue and $K$ is compact. So there is no "becoming" these are just two ways of integration that in this case yield the same result.
It is similiar to the statement that $R(f,[a,b]) = F(b)-F(a)$.

Simple example

The basis for lebesgue integration in $\mathbb{R}$ is just:

$$ \varphi(t) = \left\{\begin{matrix}1 & t\in [a,b] \\ 0 & t\not\in[a,b]\end{matrix}\right. \quad\quad \Rightarrow\quad\quad L(\varphi,\mathbb{R})=\lambda([a,b])=b-a$$

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To build on some of the previous answers:

  • for most well-behaved functions, Riemann integrable = Lebesgue integrable, as long as you're on a compact etc.

  • you have stronger theorem on Lebesgue integrable functions than on Riemann integrable functions for convergence, derivability, etc. For instance this is true for measurable functions, and is amazingly useful.

  • Lebesgue is at the basis of all the modern probability theory, and this is probably why it so important

  • as you say, some pathological function like the characteristic functions are measurable but you would not be able to compute a Riemann integral.

Last point: they are not exactly the same. Some functions are Lebesgue integrable but not Riemann $\chi_{\mathbb{Q}}$ for instance, and others are Riemann integrable but not Lebesgue (see this thread).

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  • $\begingroup$ No function is Riemann integrable but not Lebesgue. $\endgroup$
    – JLA
    Aug 10 '14 at 14:30
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    $\begingroup$ @JLA you should check the link I've put. $\endgroup$
    – Matt B.
    Aug 10 '14 at 15:29
  • $\begingroup$ but it's well known that Riemann integrable implies Lebesgue integrable. $\endgroup$
    – JLA
    Aug 10 '14 at 17:21
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    $\begingroup$ only for proper integrals. It is not the case for improper integrals. $\endgroup$
    – Matt B.
    Aug 10 '14 at 18:59
  • $\begingroup$ But an improper Riemann integral isn't a Riemann integral, so it doesn't contradict what I said. $\endgroup$
    – JLA
    Aug 10 '14 at 19:02
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Don't know if you find pictures convincing, but there is one at the Wikipedia page about Lebesgue integration: Intuitive interpretation . <quote> Folland[1] summarizes the difference between the Riemann and Lebesgue approaches thus: "to compute the Riemann integral of $f$, one partitions the domain $[a,b]$ into subintervals", while in the Lebesgue integral, "one is in effect partitioning the range of $f$".</quote> .

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  • $\begingroup$ Meh not really true though. $\endgroup$
    – JLA
    Aug 10 '14 at 14:28

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