Undergrad Riemann Integration ... Existence Result .... Browder, Theorem 5.12 ....

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The discussion centers on understanding the proof of Theorem 5.12 in Andrew Browder's "Mathematical Analysis: An Introduction," specifically the inclusion of "+1" in the expression for choosing a positive integer n. Participants suggest that the "+1" is used to ensure n is always greater than the expression, preventing potential equality that could lead to insufficient values for n. This guarantees the existence of n necessary for the proof's conditions. The clarification emphasizes the importance of this adjustment in maintaining the integrity of the theorem's proof. Overall, the inclusion of "+1" serves a critical role in ensuring the validity of the argument presented in the theorem.
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I need help with an aspect of Andrew Browder's proof of the fact that if f is a monotone function on [a, b] then the Riemann integral of f exists ...
I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am currently reading Chapter 5: The Riemann Integral and am currently focused on Section 5.2 Existence Results ... ...

I need some help in understanding the proof of Theorem 5.12 ...Theorem 5.12 and its proof read as follows:
Browder ... Theorem 5.12 .png

In the above proof by Andrew Browder we read the following:

" ... ... [For instance, one can choose a positive integer ##n## such that ##n \gt [f(b) - f(a) + 1](b - a) / \epsilon## ... ... "My question is as follows:

Why does Browder have ##+1## in the expression ##[f(b) - f(a) + 1](b - a) / \epsilon## ... ... ?Surely ##[f(b) - f(a)](b - a) / \epsilon## will do fine ... since ...

##\mu ( \pi ) = (b - a)/ n##

and so

##\mu ( \pi ) [f(b) - f(a)] = [f(b) - f(a)] (b - a)/ n \lt \epsilon## ...

... so we only need ...

##n \gt [f(b) - f(a)](b - a) / \epsilon##

Hope someone can help ...

Peter
 
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Math Amateur said:
Why does Browder have ##+1## in the expression ##[f(b) - f(a) + 1](b - a) / \epsilon## ... ... ?

Surely ##[f(b) - f(a)](b - a) / \epsilon## will do fine ... since ...

##\mu ( \pi ) = (b - a)/ n##

I think you are right. I suspect he put in the ##1## to avoid equality, but he didn't need it.
 
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Likes bhobba, Math Amateur and member 587159


Hi Peter,

I believe the reason for the "+1" in the expression is to ensure that the chosen integer n is always greater than the given expression. In other words, if we just have [f(b) - f(a)](b - a) / epsilon, there is a possibility that the integer n might not be large enough to satisfy the condition. However, by adding the "+1", we guarantee that n will always be greater than the given expression, ensuring the existence of n.

I hope this helps clarify the proof for you.

 

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