# Thread: Lower and Upper Darboux Sums

1. ## Lower and Upper Darboux Sums

Let $\displaystyle f(x) : [a,b] -> R$ be a bounded function such that $\displaystyle U(f,P) = L(f,P)$ for any partition P of [a,b].
Prove that there exists a constant c such that $\displaystyle f(x) = c$ for any x in [a,b] (by contradiction) .
Appreciate if someone could help on this one. Thanks...

2. Originally Posted by seniorcalculus
Let $\displaystyle f(x) : [a,b] -> R$ be a bounded function such that $\displaystyle U(f,P) = L(f,P)$ for any partition P of [a,b].
Prove that there exists a constant c such that $\displaystyle f(x) = c$ for any x in [a,b] (by contradiction) .
Appreciate if someone could help on this one. Thanks...
it's really trivial! let $\displaystyle P=\{a,b\}.$ let $\displaystyle M=\sup_{x \in [a,b]}f(x), \ m=\inf_{x \in [a,b]}f(x).$ then: $\displaystyle 0=U(f,P)-L(f,P)=(M-m)(b-a).$ thus $\displaystyle M=m,$ which means $\displaystyle f$ is constant.

3. Okay, thanks. But that's a direct proof, right? I kinda need help for proof by contradiction though.

4. Originally Posted by seniorcalculus
Okay, thanks. But that's a direct proof, right? I kinda need help for proof by contradiction though.
it's the same proof! let $\displaystyle P,M, m$ be as above. suppose $\displaystyle f$ is not constant. then $\displaystyle M \neq m.$ but $\displaystyle 0=U(f,P)-L(f,P)=(M-m)(b-a).$ thus $\displaystyle b=a.$ contradiction!

5. Okay, I understand it now. Thanks, NonCommAlg
But, I realized that you have picked an instance of a partition in this case, which is P={a,b}. But in the proof we have to show that for any partition interval. Shouldn't we have to assume for any partition P?