Proof. Let \(c_1,\,\ldots,\,c_n,\,d_1,\,\ldots,\,d_m \in {\mathbb{R}}\) be such that \(\varphi(x) = c_i\) for \(x \in (x_{i-1},\,x_i)\) and \(i=1,\,\ldots,\,n\) as well as \(\varphi(y) = d_j\) for \(y \in (y_{j-1},\,y_j)\) and \(j=1,\,\ldots,\,m.\) We distinguish two cases:
Case 1: Every point \(x_i\) of \(Z_1\) is also in \(Z_2.\) Under this assumption there exist \(k_0 < k_1 < \ldots < k_n\) with \(k_0=0\) and \(k_n = m\) such that \(x_i = y_{k_i}\) for \(i=1,\,\ldots,\,n.\) In particular, for all \(i=1,\,\ldots,\,n\) we have \[\begin{gathered} x_{i-1} = y_{k_{i-1}} < y_{k_{i-1}+1} < \ldots < y_{k_i-1 } < y_{k_i} = x_i \quad \text{and} \quad d_j = c_i \\ \text{for } j = k_{i-1}+1,\,\ldots,\,k_i. \end{gathered}\] Then we obtain \[\begin{aligned} I_{Z_2}(\varphi) &= \sum_{j=1}^m d_j\cdot(y_j-y_{j-1}) \\ &= \sum_{i=1}^n \sum_{j=k_{i-1}+1}^{k_i} c_i\cdot(y_j - y_{j-1}) \\ &= \sum_{i=1}^n c_i\cdot(x_i - x_{i-1}) = I_{Z_1}(\varphi). \end{aligned}\]
Case 2: The partitions \(Z_1\) and \(Z_2\) are arbitrary. In this case we consider the refinement \(Z\) of \(Z_1\) and \(Z_2.\) Then by using the first case we get \[I_{Z_1}(\varphi) = I_Z(\varphi) = I_{Z_2}(\varphi).\]

Proof.

1. W. l. o. g. we can assume that \(\varphi_1\) and \(\varphi_2\) are staircase functions with respect to the same partition \(Z: a=x_0 < \ldots < x_n = b,\) otherwise we consider the refinement of the two partitions. Let \(c_1,\,\ldots,\,c_n,\,d_1,\,\ldots,\,d_n \in {\mathbb{R}}\) such that \[\varphi_1(x) = c_i \quad \text{and} \quad \varphi_2(x) = d_i \quad \text{for } x \in (x_{i-1},x_i), \text{ and } i=1,\,\ldots,\,n.\] Then it holds that \[\begin{aligned} I(\varphi_1+ \varphi_2) &= \sum_{i=1}^n(c_i+d_i)(x_i-x_{i-1}) \\ &= \sum_{i=1}^n c_i(x_i-x_{i-1}) + \sum_{i=1}^n d_i(x_i-x_{i-1}) = I(\varphi_1) + I(\varphi_2). \end{aligned}\] Analogously, one shows \(I(\lambda\varphi_1) = \lambda I(\varphi_1).\)

2. This statement can be shown analogously to i.

Proof. “\(\Longrightarrow\)”: Let \(f\) be integrable and \(\varepsilon > 0.\) Then by the characterization of the infimum and supremum (cf. Theorem 1.2) there exist staircase functions \(\psi,\,\varphi: [a,b] \to {\mathbb{R}}\) with \(\psi \le f \le \varphi\) as well as \[\overline{\int_a^b}f(x)\,\mathrm{d}x \ge I(\varphi) - \frac{\varepsilon}{2} \quad \text{and} \quad \underline{\int_a^b}f(x)\,\mathrm{d}x \le I(\psi) + \frac{\varepsilon}{2}.\] Because of the equality of the upper and lower integral this implies \[I(\varphi) - I(\psi) \le \overline{\int_a^b}f(x)\,\mathrm{d}x + \frac{\varepsilon}{2} - \underline{\int_a^b}f(x)\,\mathrm{d}x + \frac{\varepsilon}{2} = \varepsilon.\]
“\(\Longleftarrow\)”: Let \(\varepsilon > 0\) be arbitrary and let \(\psi,\,\varphi:[a,b] \to {\mathbb{R}}\) be staircase functions such that \(\psi \le f \le \varphi\) and \(I(\varphi) - I(\psi) \le \varepsilon.\) With Remark 7.2 we obtain \[0 \le \overline{\int_a^b}f(x)\,\mathrm{d}x - \underline{\int_a^b}f(x)\,\mathrm{d}x \le I(\varphi) - I(\psi) \le \varepsilon.\] Since \(\varepsilon > 0\) has been chosen arbitrarily, the integrability of \(f\) follows.

Proof.

1. Let \(\varepsilon > 0\) be arbitrary. Then there exist staircase functions \(\varphi_1,\,\psi_1,\,\varphi_2,\,\psi_2:[a,b] \to {\mathbb{R}}\) with \(\psi_1 \le f_1 \le \varphi_1\) and \(\psi_2 \le f_2 \le \varphi_2\) such that \[I(\varphi_1) - I(\psi_1) \le \frac{\varepsilon}{2} \quad \text{and} \quad I(\varphi_2) - I(\psi_2) \le \frac{\varepsilon}{2}.\] This directly implies \(\psi_1+\psi_2 \le f_1+f_2 \le \varphi_1+\varphi_2.\) Because of the linearity of \(I\) (cf. Theorem 7.2) we also obtain \[\begin{aligned} I(\varphi_1+\varphi_2) - I(\psi_1+\psi_2) &= I(\varphi_1) + I(\varphi_2) - I(\psi_1) - I(\psi_2) \\ &\le \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{aligned}\] Hence, by Theorem 7.3, \(f_1+f_2\) is integrable. The formula for the integral then follows from \[\begin{aligned} \overline{\int_a^b}(f_1+f_2)(x)\,\mathrm{d}x &\le \overline{\int_a^b}f_1(x)\,\mathrm{d}x + \overline{\int_a^b}f_2(x)\,\mathrm{d}x \quad \text{and} \\ \underline{\int_a^b}(f_1+f_2)(x)\,\mathrm{d}x &\ge \underline{\int_a^b}f_1(x)\,\mathrm{d}x + \underline{\int_a^b}f_2(x)\,\mathrm{d}x. \end{aligned}\] The latter follows from Definition 7.5 (exercise!). Analogously, one can show the second condition.

2. The proof of this statement is analogous to the proof of statement i.

Proof. By the extreme value theorem (Theorem 4.9), \(f\) is bounded and hence, we can apply Theorem 7.3. Let \(\varepsilon > 0\) be arbitrary. By the Heine-Cantor theorem (Theorem 4.13), \(f\) is uniformly continuous, i. e., for \(\widetilde{\varepsilon} := \frac{\varepsilon}{2(b-a)} > 0\) there exists a \(\delta > 0\) such that for all \(x,\,\widetilde{x} \in [a,b]\) it holds that \[\tag{7.2} \left| x-\widetilde{x}\right| < \delta \quad \Longrightarrow \quad \left|f(x)- f(\widetilde{x})\right| < \widetilde{\varepsilon}.\] Let \(n \in {\mathbb{N}}\) be such that \(\frac{b-a}{n} < \delta\) and consider the partition \(Z: a = x_0 < \ldots < x_n = b\) with \(x_k = a+k \frac{b-a}{n},\) \(k=0,\,\ldots,\,n\) as well as the staircase functions \(\varphi,\,\psi:[a,b] \to {\mathbb{R}}\) with \[\psi(x) := f(x_i)-\widetilde{\varepsilon} \quad \text{and} \quad \varphi(x) := f(x_i)+\widetilde{\varepsilon} \quad \text{for all } x \in (x_{i-1},x_i) \text{ and } i=1,\,\ldots,\,n,\] as well as \[\tag{7.3} \psi(x_i) = \varphi(x_i) = f(x_i) \quad \text{for } i=0,\,\ldots,\,n.\] Since \(|x-x_i| < \delta\) for all \(x \in (x_{i-1},x_i),\) \(i=1,\,\ldots,\,n\) and (7.2), we have \(|f(x)-f(x_i)| < \widetilde{\varepsilon}.\) With this we obtain \[\psi(x) \le f(x) \le \varphi(x)\quad \text{for all } x \in (x_{i-1},x_i) \text{ and } i=1,\,\ldots,\,n.\] Together with (7.3) we get \(\psi \le f \le \varphi.\) Moreover, with \(\varphi(x) - \psi(x) = 2\widetilde{\varepsilon}\) for all \(x \in [a,b] \setminus\{x_0,\,x_1,\,\ldots,\,x_n\}\) we obtain \[I(\varphi) - I(\psi) = I(\varphi-\psi) = 2\widetilde{\varepsilon}(b-a) = \varepsilon.\] Since \(\varepsilon > 0\) was chosen arbitrarily, the integrability of \(f\) follows with Theorem 7.3.

Proof. Exercise.

Proof.

1. Exercise!

2. The integrability of \(|f|\) follows from the fact that \(|f| = f^+ + f^- = \max\{f,0\} + \max\{-f,0\}\) and statement i. The first inequality in (7.4) is trivial, the second one follows from the monotonicity of the integral and the fact that \(f^- \ge 0.\) Then we get \[\begin{aligned} \left|\int_a^b f(x) \,\mathrm{d}x\right| &= \left|\int_a^b (f^+(x) - f^-(x)) \,\mathrm{d}x\right| \\ &\le \left|\int_a^b (f^+(x) + f^-(x)) \,\mathrm{d}x\right| \\ &= \left|\int_a^b |f(x)| \,\mathrm{d}x\right| = \int_a^b |f(x)| \,\mathrm{d}x. \end{aligned}\]

3. If \(f\) is bounded, then also \(|f|\) is bounded, i. e., there exists an \(S > 0\) such that \(|f| \le S.\) W. l. o. g. let \(S \le 1\) (otherwise we consider \(\frac{1}{S}\cdot|f|\) instead of \(|f|\)). Then \(0 \le |f| \le 1.\) Because of the integrability of \(|f|,\) Theorem 7.3 implies that for each \(\varepsilon > 0\) there exist staircase function \(\psi,\,\varphi:[a,b] \to {\mathbb{R}}\) with \(0 \le \psi \le |f| \le \varphi \le 1\) such that \[I(\varphi-\psi) = I(\varphi) - I(\psi) \le \frac{\varepsilon}{p}.\] Now consider the function \(h:[0,1] \to {\mathbb{R}},\) \(x \mapsto x^p.\) This function is differentiable with derivative \(h'(x) = px^{p-1}\) for all \(x \in [0,1].\) Theorem 6.11 (see also Remark 6.8) then delivers the estimate \[\left|h(x)-h(y)\right| \le \sup_{\xi \in [0,1]} |h'(\xi)| \cdot |x-y| = p\cdot|x-y|\] for all \(x,\,y \in [0,1].\) With this we obtain \[\begin{aligned} \varphi(x)^p - \psi(x)^p &= \left| h(\varphi(x)) - h(\psi(x)) \right| \\ &\le p \cdot|\varphi(x) - \psi(x)| = p\cdot(\varphi(x)-\psi(x)) \end{aligned}\] for all \(x \in [a,b],\) since \(0 \le \psi \le \varphi \le 1.\) Finally, with the monotonicity of the integral this implies \[I\left(\varphi^p\right) - I\left(\psi^p\right) = I\left(\varphi^p - \psi^p\right) \le p\cdot I(\varphi - \psi) \le \varepsilon.\] Since both \(\varphi^p\) and \(\psi^p\) are staircase functions with \(\psi^p \le |f|^p \le \varphi^p\) and since \(\varepsilon > 0\) was chosen arbitrarily, with Theorem 7.3 we conclude the integrability of \(|f|^p.\)

4. This statement follows from iii and the observation that \[f\cdot g = \frac{1}{4}\left( (f+g)^2 - (f-g)^2 \right) = \frac{1}{4}\left( |f+g|^2 - |f-g|^2 \right).\]