How Does the Riemann Integral Affect Lp Space Completeness?

[dumbQuestion]

"All Lp spaces, (except where p=∞) fail to be complete under the Reimann integral"?

I am trying to learn about the Lebesgue integral and Lebesgue measurability. None of my textbooks really cover it from the basics, but I found this document online which seems to be pretty through in explaining the motivation behind developing the Lebesgue integral http://web.media.mit.edu/~lifton/snippets/measure_theory.pdf However, there is a statement I am having a hard time grasping, on the bottom of the first page:

"Third, all L_p spaces except for L_∞ fail to be complete under the Riemann
integral"

Here is what I understand: when saying "L_p spaces" I'm assuming this means metric spaces, right? I know from functional analysis that a complete metric space is one where there are no "gaps", or formally, where every Cauchy sequence has a limit that's also in the space. (That's why, for example the rational numbers with the st. Euclidean metric (L_p with p=2) is not complete, because we have gaps at all the irrational places) Here's what I don't understand: what does it mean to be complete under the Riemann integral? I don't understand what this means. I thought a metric space would be a set of numbers, with a metric defined on it, and it would be complete or incomplete just based on that information alone. Where does the Reimann integral come into play?

 

[micromass]

No, the $L^p$-spaces are a special kind of Banach space. They are defined as (for $1\leq p <+\infty$):

$$L^p([a,b])=\left\{f:[a,b]\rightarrow \mathbb{R} ~\text{measurable}~\left|~ \int_a^b |f|^p < +\infty \right.\right\}$$

The norm on this space is

$$\|f\|_p=\sqrt[p]{\int_a^b |f|^p}$$

The thing is that the above integral must be the Lebesgue integral. If we just focus on the Riemann-integral, then we find out there are not enough Riemann integrable functions and the space will be incomplete.

 

[dumbQuestion]

oh ok, thank you very much for clearing up my confusion

 

[Bacle2]

An example for L^1(ℝ) , which you can generalize:

Take an enumeration {a_n} of the Rationals, and Let X_S be

the characteristic function of the set S .

Define:

f_1=χ_a1

...
f_n=χ_a1+...x_an

...

So that your function f_n is 1 at each of the rationals ≤ a_n, and is

0 everywhere else. Then each of the f_n is R-integrable, but the

sequence converges to the characteristic function of the Rationals, which is not

R-integrable.

Notice the limit functions is Lebesgue integrable. As a nitpick, remember: Riemann, not Reimann; I don't mind so much, but your

prof. may cringe.

 

[blue_raver22]

I can provide a response to this content. The Riemann integral is a method for calculating the area under a curve by dividing it into small rectangles and taking the limit as the width of the rectangles approaches zero. It is commonly used in calculus and analysis. In contrast, the Lebesgue integral is a more general method for calculating the area under a curve, which was developed to address some of the limitations of the Riemann integral.

In the context of Lp spaces, the Riemann integral is used to define the norm (or distance) between two functions. A complete Lp space is one in which every Cauchy sequence (a sequence that gets closer and closer together) of functions has a limit that is also in the space. However, as mentioned in the statement, all Lp spaces (except for L∞) fail to be complete under the Riemann integral. This means that there are Cauchy sequences of functions in Lp spaces that do not have a limit in that same space when the Riemann integral is used.

This is where the Lebesgue integral comes into play. The Lebesgue integral is a more general method for calculating the area under a curve, and it allows for a wider range of functions to be integrated. In fact, the Lebesgue integral is used to define completeness in Lp spaces. So, while the Riemann integral may not be able to capture the full completeness of Lp spaces, the Lebesgue integral can.

In summary, the statement is saying that the Riemann integral is not sufficient to define completeness in Lp spaces, but the Lebesgue integral is. This highlights the importance of the Lebesgue integral in modern analysis and its role in defining completeness in Lp spaces. I hope this helps to clarify the concept for you.