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. $\displaystyle \diamond$

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