# Riemann Epsilon

• Mar 15th 2009, 10:14 PM
manjohn12
Riemann Epsilon
Lemma. Let $\displaystyle [a,b] \subseteq K \subseteq \mathbb{R}$, and let $\displaystyle f: \mathbb{K} \to \mathbb{R}$ be a function. Let $\displaystyle P$ be a partition of $\displaystyle [a,b]$, and let $\displaystyle \epsilon >0$ be a real number. Suppose that whenever $\displaystyle \mathcal{R}_{1}(f,P)$ and $\displaystyle \mathcal{R}_{2}(f,P)$ are Riemann sums of $\displaystyle f$ on $\displaystyle P$, then $\displaystyle |\mathcal{R}_{1}(f,P)- \mathcal{R}_{2}(f,P)| < \epsilon$. Then these statements follow:

1. $\displaystyle f$ is bounded on $\displaystyle [a,b]$.

2. $\displaystyle \mathcal{U}(f,P)- \mathcal{L}(f,P) \leq \epsilon$

3. If $\displaystyle Q$ is any refinement of $\displaystyle P$, and $\displaystyle \mathcal{R}(f,P)$ and $\displaystyle \mathcal{R}(f,Q)$ are Riemann sums, then $\displaystyle |\mathcal{R}(f,P)- \mathcal{R}(f,Q)| \leq \epsilon$

Proof. (1) Suppose $\displaystyle f$ is unbounded. Then there is a sequence $\displaystyle (y_n)$ in $\displaystyle [a,b]$ that converges to $\displaystyle y \in [a,b]$ such that for every $\displaystyle n \in \mathbb{N}$, $\displaystyle |f(y_n)| > n$. Then the set of Riemann sums of $\displaystyle f$ corresponding to $\displaystyle P$ is an unbounded set of real numbers. (2) $\displaystyle \mathcal{U}(f,P)$ and $\displaystyle \mathcal{L}(f,P)$ don't have to be Riemann sums. And since we are taking a "maximum" difference, we can get to $\displaystyle \epsilon$. (3) I think that $\displaystyle \mathcal{R}(f,Q) \leq \mathcal{R}(f,P)$. And so the inequaliy follows. $\displaystyle \diamond$

Is this correct?

• Mar 17th 2009, 12:21 PM
manjohn12
Basically is the fact that $\displaystyle \mathcal{U}(f,P)$ and $\displaystyle \mathcal{L}(f,P)$ are not Riemann sums they key idea here?