# Proof that Upper Riemann Integral > Lower Riemann Integral...?

• Aug 3rd 2007, 05:00 PM
joanne_q
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 $\int^b_a f(x)dx \leq$ upper riemann integral $\int^b_a f(x)dx$

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 $\epsilon$ P[a,b] then
L(f,P) $\leq$ U(f,Q)
If we let R:=P U Q, be a common refinement of P & Q then we see that:

L(f,P) $\leq$ L(f,R) &
U(f,R) $\leq$ U(f,Q)

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

This deduces that L(f,P) $\leq$ U(f,Q).
Therefore,
sup L(f,P) $\leq$ 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
• Aug 3rd 2007, 10:15 PM
CaptainBlack
Quote:

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 $\int^b_a f(x)dx \leq$ upper riemann integral $\int^b_a f(x)dx$

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 $\epsilon$ P[a,b] then
L(f,P) $\leq$ U(f,Q)
If we let R:=P U Q, be a common refinement of P & Q then we see that:

L(f,P) $\leq$ L(f,R) &
U(f,R) $\leq$ U(f,Q)

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

This deduces that L(f,P) $\leq$ U(f,Q).
Therefore,
sup L(f,P) $\leq$ 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 $\epsilon$ P[a,b]), and R be a common refinement of P & Q.

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

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

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

lower riemann integral $\le$ upper riemann integral..

RonL
• Aug 4th 2007, 07:56 AM
smoothman
i would have said the same
• Aug 4th 2007, 07:59 AM
joanne_q

Quote:

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?

Quote:

Originally Posted by CaptainBlack
proof:
Which tells us that L(f,P) $\leq$ U(f,Q).

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

lower riemann integral $\le$ upper riemann integral..

RonL

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

this is because the darboux sums of (b) involve only the partition P... whereas (a) involves both P and Q? ...
• Aug 4th 2007, 08:46 AM
CaptainBlack
Quote:

Originally Posted by joanne_q

> 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
• Aug 4th 2007, 09:02 AM
CaptainBlack
Quote:

Originally Posted by joanne_q
tha

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

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

L(f,P) $\leq$ U(f,P)

hence:

sup L(f,P) $\leq$ inf U(f,P),

where the supremum and infimum are over all partitions P.

RonL
• Aug 4th 2007, 10:48 AM
joanne_q
thank you captainblack, you were very helpful :) x
i will give you a thanks