# Upper and Lower Sums

Lemma. Let $\displaystyle [a,b] \subseteq K \subseteq \mathbb{R}$, and suppose that $\displaystyle f: K \to \mathbb{R}$ is bounded on $\displaystyle [a,b]$. Let $\displaystyle P$ be a partition of $\displaystyle [a,b]$, and let $\displaystyle \epsilon >0$. Then there exist Riemann sums $\displaystyle \mathcal{R}_{1}(f,P)$ and $\displaystyle \mathcal{R}_{2}(f,P)$ such that $\displaystyle \mathcal{U}(f,P) \leq \mathcal{R}_{1}(f,P) + \epsilon$ and $\displaystyle \mathcal{L}(f,P) \geq \mathcal{R}_{2}(f,P)- \epsilon$.
Proof. By definition, $\displaystyle \mathcal{U}(f,P) = \sum_{i=1}^{n} M_{i}(x_{i}-x_{i-1})$ and $\displaystyle \mathcal{L}(f,P) = \sum_{i=1}^{n} m_{i}(x_{i}-x_{i-1})$. Now $\displaystyle M_{i} = \sup \{f(x): x \in [x_{i-1}, x_{i}] \}$ and $\displaystyle m_i = \inf \{ f(x): x \in [x_{i-1}, x_i] \}$. Let $\displaystyle S = \{x: x \in [x_{i-1}, x_i] \}$. Then there exists $\displaystyle x_j \in S$ such that $\displaystyle |x_j-M_i| < \epsilon$.
Thus $\displaystyle M_i < x_j + \epsilon$. If we choose $\displaystyle x_j = x_{i}^{*}$ as the test points, then it follows that $\displaystyle \mathcal{U}(f,P) \leq \mathcal{R}_{1}(f,P)+ \epsilon$. The other case follows similarly. $\displaystyle \diamond$