Proving the Inequality of Upper Darboux Sums for Riemann Integrals

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SUMMARY

The discussion focuses on proving the inequality of Upper Darboux sums for Riemann integrals, specifically showing that if partition P is contained within partition Q, then U(Q,f) ≤ U(P,f). The proof utilizes the supremum of the function f over subintervals defined by the partitions. The participants highlight the need to generalize the proof to account for multiple refinements in partition Q, rather than just a single additional point.

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Homework Statement


Suppose that function f: [a,b] --> R is bounded, and P and Q be 2 partition of [a,b]. Prove that if P is in Q then U(Q,f)<= U(P,f)

Homework Equations


The Attempt at a Solution


P is in Q so suppose there's a c that is in Q but not in P such that c is in between x_i-1 and x_i
M_k = sup{f(x)| x in [x_k-1,x_k]

let's denote:
M_i = sup{f(x)| x in [x_i-1,x_i]
r_1 = sup{f(x)| x in [x_i-1, c]
r_2 = sup{f(x)| x in [c ,x_i]

therefore M_i = max{r_1,r_2}

by definition U(P,f) = [k=1]\Sigma[/n] M_k*(x_k - x_k-1)
= [k=1]\Sigma[/i-1]M_k*(x_k - x_k-1) + M_i(x_i -x_i-1) + [k=i+1]\Sigma[/n]M_k*(x_k - x_k-1)
>= [k=1]\Sigma[/i-1]M_k*(x_k - x_k-1) + r_1(c - x_i-1) + r_2(x_i - c) [k=i+1]\Sigma[/n]M_k*(x_k - x_k-1)
= U(Q,f)
as desired.

Do you see anything wrong with this proof? Thanks very much
 
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Your calculation works for the case where Q is a refinement of P by one single additional point c. In general Q refines P by adding some finite number m of points, so you need to observe that the argument can be repeated, say by letting P \subset P_1 \subset \cdots \subset P_{m-1} \subset Q be a sequence of partitions each of which refines the previous by adding a single point.
 
ystael said:
Your calculation works for the case where Q is a refinement of P by one single additional point c. In general Q refines P by adding some finite number m of points, so you need to observe that the argument can be repeated, say by letting P \subset P_1 \subset \cdots \subset P_{m-1} \subset Q be a sequence of partitions each of which refines the previous by adding a single point.

yes you're right. i totally forgot about the general case with Q refines P by any finite number of points more than 1. Thanks very much for your response.
 

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