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.