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

Gold Member

## 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: 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

Related Topology and Analysis News on Phys.org
PeroK
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.