Thread: Proof that Upper Riemann Integral > Lower Riemann Integral...?

heyyy guys....
im new at this level of maths so i would appreciate a lil bit of help here and there hehe

i want to show that
lower riemann integral upper riemann integral

i have worked this out so far but im not sure if its correct: please help me on my solution.

proof:
if P,Q are partitions of P[a,b] (i.e P,Q P[a,b] then
L(f,P) U(f,Q)
If we let R:=P U Q, be a common refinement of P & Q then we see that:

L(f,P) L(f,R) &
U(f,R) U(f,Q)

We know L(f,R) U(f,R) so now,
L(f,P) L(f,R) U(f,R) U(f,Q)

This deduces that L(f,P) U(f,Q).
Therefore,
sup L(f,P) inf U(f,P).... which is the same as

lower riemann integral < upper riemann integral..


Is this proof totally correct? or is there another solution to proving this? thanks a lot xxx

Originally Posted by joanne_q

heyyy guys....
im new at this level of maths so i would appreciate a lil bit of help here and there hehe

i want to show that
lower riemann integral upper riemann integral

i have worked this out so far but im not sure if its correct: please help me on my solution.

proof:
if P,Q are partitions of P[a,b] (i.e P,Q P[a,b] then
L(f,P) U(f,Q)
If we let R:=P U Q, be a common refinement of P & Q then we see that:

L(f,P) L(f,R) &
U(f,R) U(f,Q)

We know L(f,R) U(f,R) so now,
L(f,P) L(f,R) U(f,R) U(f,Q)

This deduces that L(f,P) U(f,Q).
Therefore,
sup L(f,P) inf U(f,P).... which is the same as

lower riemann integral < upper riemann integral..


Is this proof totally correct? or is there another solution to proving this? thanks a lot xxx

Remove the redundance, and don't claim things that you are going to
show before you show that they are true and correct the slight errors:

proof:

Let P,Q be partitions of [a,b] (i.e P,Q P[a,b]), and R be a common refinement of P & Q.

We know L(f,R) U(f,R) hence L(f,P) L(f,R) U(f,R) U(f,Q)

Which tells us that L(f,P) U(f,Q).

Therefore, sup L(f,P) inf U(f,P).... which is the same as

lower riemann integral upper riemann integral..

RonL

i would have said the same

thanks for the reply captain.

Originally Posted by CaptainBlack

Remove the redundance, and don't claim things that you are going to
show before you show that they are true and correct the slight errors:

> which parts were u referring to that should be made redundant? and where were the errors please?

Originally Posted by CaptainBlack

proof:
Which tells us that L(f,P) U(f,Q).

Therefore, sup L(f,P) inf U(f,P).... which is the same as

lower riemann integral upper riemann integral..

RonL

> here you showed that : (a) L(f,P) U(f,Q).
but how does this deduce to the fact that, (b) sup L(f,P) inf U(f,P)?

this is because the darboux sums of (b) involve only the partition P... whereas (a) involves both P and Q? ...

Originally Posted by joanne_q

thanks for the reply captain.



> which parts were u referring to that should be made redundant? and where were the errors please?

The proposed proof that I posted was what you had with the redundant stuff
deleted (as the quoting what you were going to demonstrate before you had done).

The only substantive error that I corrected was to replace < by <= in the
last line.

RonL

Originally Posted by joanne_q

tha

> here you showed that : (a) L(f,P) U(f,Q).
but how does this deduce to the fact that, (b) sup L(f,P) inf U(f,P)?

If L(f,P) U(f,Q) for all partitions P and Q of [a,b], then
in particular:

L(f,P) U(f,P)

hence:

sup L(f,P) inf U(f,P),

where the supremum and infimum are over all partitions P.

RonL

thank you captainblack, you were very helpful x
i will give you a thanks