# Lemma of Lebesgue integration

Hello, I got a question about a lemma on Lebesgue integration (Riesz-Nagy approach).

Let f(x) be a Lebesgue integrable function on interval (a, b). Riesz and Nagy (pg. 50 of Lessons of Functional Analysis) say that if f(x) is not bounded, for all epsilon > 0 we can decompose f(x) into the sum of a BOUNDED INTEGRABLE FUNCTION g(x) and a function h(x) such that

(Integral from "a" to "b") |h(x)|dx < e.

From Apostol I knew this lema (10.19.b of Mathematical Analysis): if f(x) is the limit function of a increasing sequence of step functions (therefore a Lebesgue integrable function), for all epsilon > 0 we can decompose f(x) into the sum of a STEP FUNCTION s(x) and a function h(x) such that

(Integral from "a" to "b") |h(x)|dx < e.

I would thank if anyone can give me hints to pass from Apostol lemma to Riesz lemma.

May be someone can explain me how a not bounded function can be Lebesgue integrable.

I mean, only a referential explanation. I think I know more or less the basics of Lebesgue Integration, so I would need only that someone point out which lemma or concept I have forget that allows the existence of an unbounded but L-integrable function.

Thank you.

It seems nobody's interested in this. I will try again, nothing to lose.

Lebesgue integration can be studied through Fredrich Riesz and Bela Sz. Nagy approach (upper functions; Apostol book is an example). It is supposed to be more direct and economical that measure theory. I am trying to follow (pg. 50 of “Lessons of Functional Analysis”of Riesz-Nagy) the theorem which states that you need absolutely continuous functions to assure the Lebesgue version of the Fundamental Theorem of Calculus. In the proof Riesz-Nagy say:

Let f be a Lebesgue integrable function in closed interval (a, b). Suppose f is not bounded. As f is a L-integrable function, then, for all epsilon > 0, we can find two functions g and h that:

1. f(x) = g(x) + h(x).
2. g(x) is a BOUNDED integrable function
3. With h(x) we have (integral from a to b) |h(x)|dx < epsilon.

My question is: How to deduce the existence of the BOUNDED function g?

I know two lemmas that appear in Apostol book: if the function f is L-integrable and epsilon is > 0, then

1. There are upper functions u and v such that f = u – v, v is not negative and integral v(x) dx < epsilon.
2. There is a step function s and a L-integrable function t such that f = s + t and integral |t(x)|dx < epsilon.

But I can’t deduce from these lemmas the one mentioned by Riesz.

matt grime
Homework Helper
You appear to have overlooked more obvious results in Riemann integration that will help you understand what is going on (indeed, this result is equally true in ordinary Riemann integration). Just consider x^{-1/2}. Its definite integral from 0 to X exists, and it is unbounded, and you can see how to work out your lemma: just look at the function near the singularities and away from the singularities.

Matt:

Thank for your answer. From Improper Riemann integration I remember this (apologies, I cant use Latex in this PC, some technical problem):

Let f(x) = (integral from x to b) t^{-1/2}dt.

Then we define:

(integral from 0 to b) t^{-1/2}dt = lim f(x) (when x -> 0+).

So ( taking x < a < b) we have this equation "1":

(integral from 0 to b) t^{-1/2}dt = (integral from a to b) t^{-1/2}dt +
(integral from 0 to a) t^{-1/2}dt. (1)

Here
- The function t^{-1/2} is bounded in [a, b]
- we can make "(integral from 0 to a) t^{-1/2}dt" as small as we please.

But in (1) we do not have three functions, it is the same function in three different intervals of integration. Instead, in Riesz lemma we have three functions (f, g, h) with the same limits of integration.

Am i misguided? May be you can tell me something more.

I have read a paper according to which this Riesz Lemma is justified because "bounded functions are dense in Lebesgue integrable functions". Can someone tell me what does this mean and how to prove it?

mathwonk
Homework Helper
2020 Award
well dense in what sense?" from one point of view a lebesgue integrable function is rpecisely the limit, in the L1 sense of a step function, which of cousreis mbounded. hence lebesgue integrable functions are DEFINED as functions in which step fucntions, which are bounded, are dense.

i am an amateur here, but this is my memory of the topic.l

I guess if you have a question of this nature, it must be because you are using a different definition of lebesgue integrable, in which this definition becomes a theorem.

so what is your definition of lebesgue integrable. or if you wish me to, tomorrow i will look it up in riesz nagy.

The lemma is this (Riesz-Nagy, pg. 50). Let f be an unbounded Lebesgue integrable function in an interval ]0, b]. Then, for all epsilon > 0, we can find two functions g and h that:

1. f(x) = g(x) + h(x).
2. g(x) is a BOUNDED integrable function
3. With h(x) we have (integral from a to b) |h(x)|dx < epsilon

Matt Grime hints help me to develop what follows, valid for Riemann Integration (not Lebesgue).

Let h: ]0, b] -> R, defined this way:
If t belongs to ]0, c[, h(t) = 0
If t belongs to [c, b], h(t) = t^{-1/2}.
Then h is bounded and Riemann integrable.
(integral from 0 to b) h(t)dt = 2b^{1/2} - 2c^{1/2}.

Let g: ]0, b] -> R, define this way:
If t belongs to ]0, c[, h(t) = t^{-1/2}.
If t belongs to [c, b], h(t) = 0.
This function is unbounded and (improper) Riemann integrable.
(integral from 0 to b) g(t)dt = 2c^{1/2}.

Let f: ]0, b] -> R / f(t) = h(t) + g(t) = t^{-1/2}.
This function f is unbounded and (improper) Riemann integrable.
(integral from 0 to b) f(t)dt = 2b^{1/2}.

If we take c sufficiently near to 0 we can make the integral of function g (from 0 to b) as small as we please.

So here, thanks to Matt Grime hint, I have what I look for, but for Riemann integrals. How to translate this to Lebesgue integrals?

Mathwonk: Riesz (pg 50) mentioned the lemma to justify a step in his proof that if F(x) is an indefinite Lebesgue integral, then it is absolutely continuous. He make use of a bounded function that not necesarily is a step function. So step functions won't suffice.