Learning to use the Lebesgue Integral

[ice109]

please don't hang me for this but is there anywhere I can find a text or a tutorial on how to actually do integrals using Lebesgue Integration? Everywhere I look all I see is theory and proof exercises. I know, I know that lots of functions that are Lebesgue Integrable are Riemann Integrable and the Lebesgue Integral is equal to the Riemann Integral but right now I want to learn how to perform a Lebesgue Integral and not simply perform the integral. Theory is good, it's great fun to decipher, but right now I just want to learn about it the same way I learned about the Riemann Integral in my calc book.

 

[mrandersdk]

You can't do the leasbegue integral (exept for some simple cases like on identity functions on a measureble set), but as you say some nice functions have the property that the leasbegue integral agree with the riemann integral.

So the riemann integral is practical to calculate, so why do we need the lesbegue integral?

It is used to do proof and theory because it have some nice properties, fx it is easy to determine if the following is allowed

$$\frac{\partial}{\partial x}\int f(x) dx = \int \frac{\partial}{\partial x} f(x) dx$$

which can be hard to determine for the riemann integral, so we need both, harsh said one to calculate and one to do proofs.

 

[HallsofIvy]

Well, you can "do" a Lebesque integral in a sense: If a function is NOT Riemann integrable, it can still be Lebesque integrable. Of course, in that case, there must exist a Riemann integrable function equal to the given function everywhere except on a set of measure 0. To find the Lebesque integral, you find the Riemann integral of such a function.

For example, f(x)= x² if x is irrational, 2x if x is rational, 0<= x<= 1, is not Riemann integrable because it is discontinuous everywhere. But it is equal to x² "almost everywhere" (everywhere except on the set of rational numbers which is a set of measure 0) so the Lebesque integral, from 0 to 1, of f(x) is the same as the Riemann integral of x^{2[sup, 1/3.}

 

[ice109]

well thank you for the advice

 

[morphism]

If a function is NOT Riemann integrable, it can still be Lebesque integrable. Of course, in that case, there must exist a Riemann integrable function equal to the given function everywhere except on a set of measure 0.

Hold on - this is not true. This would imply that every (bounded) Lebesgue integrable function is equal to a continuous function almost everywhere, which is not the case (e.g. look at the characteristic function of a fat Cantor set in [a,b]). What is true is that if a function f, which is defined on a set of finite measure, is Lebesgue integrable then f is equal to a continuous function except on a set of arbitrarily small measure (Lusin's theorem).

Anyway, Lebesgue integration isn't really about computing integrals. It was introduced as an attempt to remedy the shortcomings of the Riemann integral. Here are four advantages to using the Lebesgue integral instead of the Riemann integral:

(1) Riemann integrable functions are still Lebesgue integrable, so we're not losing anything by using the Lebesgue integral.

(2) We no longer have to restrict integration to intervals and the like. The Lebesgue integral allows us to integrate over more general sets (those which are measurable).

(3) We now have very powerful and useful convergence theorems, something which the Riemann integral lacks.

(4) Consider the vector space C[0,1] of continuous, real-valued functions on [0,1]. There is a natural norm on this space defined by $\|f\|_1 = \int_0^1 |f(x)| dx$ (this is the Riemann integral). However, C[0,1] is not a complete space under $\|\cdot\|_1$; its completion is actually the space of functions which are Lebesgue integrable on [0,1].

The importance of the Lebesgue integral really lies in its construction. The use of simple functions instead of step functions allows us to introduce the notion of integration to spaces where there are no "intervals." As a consequence we get some very deep and useful theorems, like the Riesz representation theorem.