
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?

Basically is the fact that $\displaystyle \mathcal{U}(f,P) $ and $\displaystyle \mathcal{L}(f,P) $ are not Riemann sums they key idea here?