# Riemann Epsilon

• Mar 15th 2009, 11:14 PM
manjohn12
Riemann Epsilon
Lemma. Let $[a,b] \subseteq K \subseteq \mathbb{R}$, and let $f: \mathbb{K} \to \mathbb{R}$ be a function. Let $P$ be a partition of $[a,b]$, and let $\epsilon >0$ be a real number. Suppose that whenever $\mathcal{R}_{1}(f,P)$ and $\mathcal{R}_{2}(f,P)$ are Riemann sums of $f$ on $P$, then $|\mathcal{R}_{1}(f,P)- \mathcal{R}_{2}(f,P)| < \epsilon$. Then these statements follow:

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

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

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

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

Is this correct?

• Mar 17th 2009, 01:21 PM
manjohn12
Basically is the fact that $\mathcal{U}(f,P)$ and $\mathcal{L}(f,P)$ are not Riemann sums they key idea here?