# Lebesgue Integration of Simple Functions .... Lindstrom, Lemma 7.4.6 .... ....

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In summary, the post discusses Lemma 7.4.6 and its proof from Tom L. Lindstrom's book "Spaces: An Introduction to Real Analysis". The proof shows that \lim_{n\to\infty}\int_Bf_n\,d\mu \geq b\mu(B) and the author seeks assistance in demonstrating this formally and rigorously. The reader is also provided access to the relevant text from Lindstrom's book for further understanding. There is also a discussion about the notation f_n (x) \uparrow b, which is deemed to be incorrect. However, the information given in the statement of the lemma is sufficient for the rest of the proof to work.
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I am reading Tom L. Lindstrom's book: Spaces: An Introduction to Real Analysis ... and I am focused on Chapter 7: Measure and Integration ...

I need help with the proof of Lemma 7.4.6 ...

Lemma 7.4.6 and its proof read as follows:

In the above proof by Lindstrom we read the following:

" ... ... Since this holds for any number $$\displaystyle a$$ less than $$\displaystyle b$$ and any number $$\displaystyle m$$ less than $$\displaystyle \mu (B)$$, we must have $$\displaystyle \lim_{ n \to \infty } \int_B f_n d \mu \geq b \mu (B)$$ . ... ... "I need help in order to show, formally and rigorously, that $$\displaystyle \lim_{ n \to \infty } \int_B f_n d \mu \geq b \mu (B)$$ ... ...My thoughts are that we could assume that $$\displaystyle \lim_{ n \to \infty } \int_B f_n d \mu \lt b \mu (B)$$ ... ... and proceed to demonstrate a contradiction ... but I'm not sure how to formally proceed ... ...

Help will be much appreciated ...

Peter

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Readers of the above post may be assisted by access to Lindstrom's introduction to the integration of simple functions ... so I am providing access to the relevant text ... as follows:

Hope that helps ...

Peter

Peter said:
In the above proof by Lindstrom we read the following:

" ... ... Since this holds for any number $$\displaystyle a$$ less than $$\displaystyle b$$ and any number $$\displaystyle m$$ less than $$\displaystyle \mu (B)$$, we must have $$\displaystyle \lim_{ n \to \infty } \int_B f_n d \mu \geq b \mu (B)$$ . ... ... "I need help in order to show, formally and rigorously, that $$\displaystyle \lim_{ n \to \infty } \int_B f_n d \mu \geq b \mu (B)$$ ... ...
You want formal and rigorous, here it is!

First, Lindstrom appears to assume that $\lim_{n\to\infty}\int_Bf_n\,d\mu$ exists. Since $\left\{\int_Bf_n\,d\mu\right\}$ is an increasing sequence, it will either converge to a finite limit or tend to infinity. I don't know whether Lindstrom allows infinity as a possible limit. Even if he does, we do not need to worry about that case, because if $\int_Bf_n\,d\mu$ does go to infinity it will certainly eventually be larger than $b\mu(B)$.

To prove the inequality, given $\varepsilon>0$, choose $\delta$ such that $0<\delta<\dfrac{\varepsilon}{b+\mu(B)}$. Now choose $a<b$ and $m<\mu(B)$ with $a>b-\delta$ and $m>\mu(B)-\delta$. Then $$b\mu(B) - am = b(\mu(B) - m) + m(b-a) < \delta(b + m) < \delta(b + \mu(B)) = \varepsilon.$$ So $am > b\mu(B) - \varepsilon$. With $N$ as in Lindstrom's proof it follows that$$\displaystyle \int_Bf_n\,d\mu \geqslant am > b\mu(B) - \varepsilon$$ whenever $n\geqslant N$. Since that holds for all $\varepsilon>0$, $$\displaystyle \int_Bf_n\,d\mu \geqslant b\mu(B)$$.

[You will recognise that the above argument is just a variant of the proof that the limit of a product is the product of the two limits.]

Thanks for a most helpful post Opalg ...

Working carefully through your proof now ...

BUT ... I have another question ...

In the above proof by Lindstrom we read the following:

" ... ... Since $$\displaystyle f_n (x) \uparrow b$$ for all $$\displaystyle x \in B$$ ... ... "Unless I am misunderstanding the notation, $$\displaystyle f_n (x) \uparrow b$$ means $$\displaystyle f_n$$ tends to $$\displaystyle b$$ from below ... but ... all we are given is that $$\displaystyle \lim_{n \to \infty } f_n (x) \geq b$$ which surely is not the same ...

Can someone please clarify this issue ...

Peter

Last edited:
Peter said:
In the above proof by Lindstrom we read the following:

" ... ... Since $$\displaystyle f_n (x) \uparrow b$$ for all $$\displaystyle x \in B$$ ... ... "Unless I am misunderstanding the notation, $$\displaystyle f_n (x) \uparrow b$$ means $$\displaystyle f_n$$ tends to $$\displaystyle b$$ from below ... but ... all we are given is that $$\displaystyle \lim_{n \to \infty } f_n (x) \geq b$$ which surely is not the same ...
I completely agree with you. The statement of the lemma says that $\{f_n\}$ is an increasing sequence and that $\{f_n(x)\}$ has a limit that is greater than or equal to $b$. The statement in the proof of the lemma, that $f_n(x)\uparrow b$, is careless and wrong (because the limit could be greater than $b$). However, the information given in the statement of the lemma is sufficient to ensure that the sequence $\{A_n\}$ is increasing and that $$\displaystyle B = \bigcup_{n=1}^\infty A_n$$, which is what is needed for the rest of the proof to work.

Oh ... thanks Opalg ...

Peter

## 1. What is Lebesgue Integration?

Lebesgue Integration is a mathematical concept that extends the traditional Riemann Integral to a wider range of functions. It was developed by French mathematician Henri Lebesgue in the early 20th century and is used to calculate the area under a curve or the volume of a solid in higher dimensions.

## 2. What are Simple Functions?

Simple Functions are a type of function that takes on only a finite number of values over a given interval. They are often used as building blocks for more complex functions and are particularly useful in Lebesgue Integration as they can be easily approximated by step functions.

## 3. What is Lindstrom's Lemma 7.4.6?

Lindstrom's Lemma 7.4.6 is a theorem in Lebesgue Integration that states that if a sequence of simple functions converges pointwise to a measurable function, then the limit function is also measurable. This lemma is an important tool in proving the convergence of integrals of more complex functions.

## 4. How is Lebesgue Integration different from Riemann Integration?

Lebesgue Integration differs from Riemann Integration in several key ways. One major difference is that Lebesgue Integration allows for the integration of a wider range of functions, including non-continuous and unbounded functions. It also uses a different approach to defining the integral, based on measure theory rather than partitions of intervals.

## 5. What are some applications of Lebesgue Integration?

Lebesgue Integration has many applications in mathematics, physics, and engineering. It is used to calculate areas and volumes in higher dimensions, to solve differential equations, and to analyze the behavior of complex systems. It is also an important tool in probability theory and statistics, where it is used to calculate probabilities and expected values.

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