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

  • #1
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Summary:

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 ...

Main Question or Discussion Point

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
 

Answers and Replies

  • #2
PeroK
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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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