# Upper and Lower Sums

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