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

Is this correct?