Why Use Supremum Instead of Maximum in Riemann Sums?

AxiomOfChoice
Messages
531
Reaction score
1
Let f be a Riemann integrable function defined on an interval [a,b], and let P = \{a = x_0 < x_1 < \ldots < x_n = b\} be a partition of [a,b]. I don't understand why the definition of (say) the upper Riemann sum of f associated with P is always given as
<br /> U(f,P) = \sum_{i=1}^n M_i (x_i - x_{i-1}),<br />

where

<br /> M_i = \sup\limits_{x\in [x_{i-1},x_i]} f(x).<br />

Why does it need to be \sup? Why can't it just be \max? I can't think of an instance where \max wouldn't be suitable; that is, I can't think of a Riemann integrable function that doesn't actually attain its max somewhere on any given [x_{i-1},x_i]. Can someone provide an example for me?
 
Physics news on Phys.org
y=x when x\neq 1, y=0 when x=1 on interval [0,1]. Take interval [x_{n-1},x_n] from the partition.
 
Back
Top