Upper Riemann Integral is greater than or equal to lower Riemann Integral

1.3k Views Asked by At

I need to prove that $$ \underline{\int_{a}^{b}} f dx \le \overline{\int_{a}^{b}} f dx$$ which is same as showing $$\sup_P L(f;P) \le \inf_P U(f;P)$$

Where $U(f;P)$ , $L(f;P)$ are Upper and Lower Riemann Sums, $P$ is Partition and $f$ is a function. I proceed as below $$L(f;P) \le U(f;P) \ \forall P$$ $$\sup_P L(f;P) \le U(f;P) \ \forall P$$ $$\sup_P L(f;P) \le \inf_P U(f;P)$$ Is this correct reasoning?

1

There are 1 best solutions below

3
On BEST ANSWER

It is not correct. The step $$L(f;P) \le U(f;P) \ \forall P \Rightarrow \sup_P L(f;P) \le U(f;P) \ \forall P$$ is not justified. Theoretically it could be possible that $L(f;P') > U(f;P)$ for distinct partitions $P, P'$. But fortunately we have $$L(f;P') \le U(f;P) \ \forall P', P \tag{1}$$ which leads to a correct argument.

To prove $(1)$, observe that if $Q$ refines $P$, then $$L(f;P) \le L(f;Q) \quad, \quad U(f;Q) \le U(f;P) .$$

For any two $P',P$ we can choose a common refinement $Q$ and get $$L(f;P') \le L(f;Q) \le U(f;Q) \le U(f;P) .$$