Let's tackle each part step-by-step, with explanations after each step. --- ### **#1** #### **(a) If \( f \) is Riemann integrable on \([a,b]\), then \( f \) is continuous on \([a,b]\).** #### **Step 1:** Recall that a function can be Riemann integrable even if it is not continuous everywhere, as long as the set of discontinuities has measure zero (is a null set). **Explanation:** A classic counterexample is the characteristic function of the rationals, which is not Riemann integrable, **but** a function with a jump discontinuity at a single point (e.g., \( f(x) = 1 \) for \( x \neq c \), \( f(c) = 0 \)) is Riemann integrable. #### **Step 2:** Therefore, the statement is **false**. A Riemann integrable function need not be continuous everywhere on \([a,b]\). **Final Answer:** **False.** Counterexample: The function \( f(x) = \begin{cases} 1 & x \neq c \\ 0 & x = c \end{cases} \) on \([a,b]\) is Riemann integrable, but not continuous at \( x = c \). --- #### **(b) If \( |f| \) is Riemann integrable on \([a,b]\), then \( f \) is Riemann integrable on \([a,b]\).** #### **Step 1:** Consider the definition of Riemann integrability: a function is Riemann integrable if it is bounded and its set of discontinuities has measure zero. If \( |f| \) is Riemann integrable, \( f \) must be bounded, but \( f \) could still fail to be Riemann integrable if its discontinuities are too large. **Explanation:** Counterexample: Let \( f(x) = 1 \) if \( x \) is rational, \( f(x) = -1 \) if \( x \) is irrational. Then \( |f(x)| = 1 \), which is Riemann integrable, but \( f \) is not since the set of discontinuities is not measure zero. #### **Step 2:** So, the statement is **false**. **Final Answer:** **False.** Counterexample: The Dirichlet function as above. --- #### **(c) If \( f \) is continuous on \([a,b]\) and on \([b,c]\), then \( f \) is Riemann integrable on \([a,c]\).** #### **Step 1:** Continuity implies Riemann integrability on each interval, but at \( b \), \( f \) may have a jump (discontinuity only at a single point). **Explanation:** A function with a single jump discontinuity at \( b \) (e.g., different left and right limits at \( b \)) is still Riemann integrable on \([a, c]\) because the set of discontinuities is a single point. #### **Step 2:** Thus, the statement is **true**. **Final Answer:** **True.** At worst, \( f \) is discontinuous at a single point \( b \), which does not prevent Riemann integrability. --- #### **(d) If \( f \) is continuous on \([a,b]\) and on \([b,c]\) and is bounded on \([a,c]\), then \( f \) is Riemann integrable on \([a,c]\).** #### **Step 1:** Even if \( f \) is not continuous at \( b \), it's only one point of discontinuity. Boundedness ensures it doesn't blow up at any point. **Explanation:** A function which is continuous except at a single point, and bounded, is Riemann integrable (since the set of discontinuities is measure zero). #### **Step 2:** So, the statement is **true**. **Final Answer:** **True.** The function can have at most one discontinuity, which does not preclude Riemann integrability. --- ### **#2** #### **(a) If \( P = \{x_0, \ldots, x_n\} \) is any partition of \([a,b]\), prove that** \[ \sum_{j=1}^n (M_j(f) - m_j(f)) \Delta x_j \leq (f(b) - f(a))\|P\|, \] where \( M_j(f) \) and \( m_j(f) \) are the sup and inf of \( f \) on \([x_{j-1}, x_j]\), and \( \|P\| \) is the mesh size. #### **Step 1:** Since \( f \) is increasing, on each subinterval \([x_{j-1}, x_j]\), \( m_j(f) = f(x_{j-1}) \) and \( M_j(f) = f(x_j) \). So, \[ M_j(f) - m_j(f) = f(x_j) - f(x_{j-1}). \] **Explanation:** This follows from the monotonicity of \( f \). #### **Step 2:** So, \[ \sum_{j=1}^n (f(x_j) - f(x_{j-1})) \Delta x_j \leq \sum_{j=1}^n (f(x_j) - f(x_{j-1})) \|P\| = (f(b) - f(a)) \|P\|, \] because \( \Delta x_j \leq \|P\| \) for all \( j \). **Final Answer:** \[ \boxed{ \sum_{j=1}^n (M_j(f) - m_j(f)) \Delta x_j \leq (f(b) - f(a)) \|P\| } \] --- #### **(b) Prove that \( f \) is integrable on \([a,b]\).** #### **Step 1:** From part (a), we see that as \( \|P\| \to 0 \), the difference between upper and lower Riemann sums goes to zero. **Explanation:** This is the criterion for Riemann integrability: for every \( \epsilon > 0 \), there exists a partition \( P \) such that the difference between the upper and lower sums is less than \( \epsilon \). #### **Step 2:** Therefore, every increasing function on a closed interval is Riemann integrable. **Final Answer:** \[ \boxed{ \text{An increasing function on } [a, b] \text{ is Riemann integrable.} } \]

Let's analyze each part carefully, providing a step-by-step explanation. --- ## **#1** ### **(a) If \(f\) is Riemann integrable on \([a,b]\), then \(f\) is continuous on \([a,b]\).** **Step 1:** Recall the definition: Riemann integrability allows functions to have discontinuities, but only on a set of measure zero. Continuity everywhere is **not** necessary for Riemann integrability. **Explanation:** Counterexample: The characteristic function of rationals in \([a,b]\) (Dirichlet function) is not Riemann integrable. But consider a function with a single jump discontinuity, e.g., \[ f(x) = \begin{cases} 1 & x \neq c \\ 0 & x = c \end{cases} \] which is discontinuous only at \( c \), a single point. Such a function is Riemann integrable, but **not** continuous at \( c \). **Conclusion:** The statement is **false**. **Final answer:** **False** --- ### **(b) If \(|f|\) is Riemann integrable on \([a,b]\), then \(f\) is Riemann integrable on \([a,b]\).** **Step 1:** Riemann integrability depends on the set of discontinuities and boundedness. While \(|f|\) being Riemann integrable implies \(f\) is bounded, it does **not** guarantee \(f\) is Riemann integrable. **Explanation:** Counterexample: Take the Dirichlet function \(f(x) = \begin{cases} 1 & \text{if } x \in \mathbb{Q} \\ -1 & \text{if } x \notin \mathbb{Q} \end{cases}\). \(|f|\) is constantly 1, which is Riemann integrable, but \(f\) itself is not, because it is discontinuous everywhere. **Conclusion:** The statement is **false**. **Final answer:** **False** --- ### **(c) If \(f\) is continuous on \([a,b)\) and on \([b,c]\), then \(f\) is Riemann integrable on \([a,c]\).** **Step 1:** Since \(f\) is continuous on \([a,b)\) and \([b,c]\), it may have a discontinuity at \(b\). But a single point of discontinuity does **not** prevent Riemann integrability. **Explanation:** A function with finitely many discontinuities (or just one point) on a closed interval is Riemann integrable. **Conclusion:** The statement is **true**. **Final answer:** **True** --- ### **(d) If \(f\) is continuous on \([a,b)\) and on \([b,c]\), and bounded on \([a,c]\), then \(f\) is Riemann integrable on \([a,c]\).** **Step 1:** Since \(f\) is continuous everywhere except possibly at \(b\), and bounded, the only potential discontinuity is at the single point \(b\). **Explanation:** A function with at most finitely many discontinuities (here, one at \(b\)) on a closed interval is Riemann integrable. **Conclusion:** The statement is **true**. **Final answer:** **True** --- ## **#2** ### **(a) For an increasing function \(f\), prove that** \[ \sum_{j=1}^n (M_j(f) - m_j(f)) \Delta x_j \leq (f(b) - f(a)) \|P\| \] where \(M_j(f) = \sup_{x \in [x_{j-1}, x_j]} f(x)\), \(m_j(f) = \inf_{x \in [x_{j-1}, x_j]} f(x)\), and \(\|P\| = \max_j \Delta x_j\). --- **Step 1:** Since \(f\) is increasing, on each subinterval \([x_{j-1}, x_j]\), \[ M_j(f) = f(x_j), \quad m_j(f) = f(x_{j-1}). \] Thus, \[ M_j(f) - m_j(f) = f(x_j) - f(x_{j-1}). \] **Explanation:** This follows directly from the monotonicity: the supremum is achieved at the right endpoint, the infimum at the left. --- **Step 2:** Now, sum over \(j\): \[ \sum_{j=1}^n (f(x_j) - f(x_{j-1})) \Delta x_j. \] Note that \[ f(x_j) - f(x_{j-1}) \leq f(b) - f(a), \] and \[ \Delta x_j \leq \|P\| \quad \text{for all } j. \] Hence, \[ \sum_{j=1}^n (f(x_j) - f(x_{j-1})) \Delta x_j \leq (f(b) - f(a)) \|P\|. \] **Final answer:** \[ \boxed{ \sum_{j=1}^n (M_j(f) - m_j(f)) \Delta x_j \leq (f(b) - f(a)) \|P\| } \] --- ### **(b) Show that \(f\) is Riemann integrable on \([a, b]\).** **Step 1:** From part (a), as the mesh \(\|P\|\) becomes finer (\(\to 0\)), the difference between the upper sum (based on \(M_j(f)\)) and the lower sum (based on \(m_j(f)\)) goes to zero because: \[ \sup_P \left( \text{Upper Sum} - \text{Lower Sum} \right) \leq (f(b) - f(a)) \|P\| \to 0 \text{ as } \|P\| \to 0. \] **Explanation:** Since the difference between upper and lower sums can be made arbitrarily small, this implies \(f\) is Riemann integrable. **Conclusion:** Every increasing function on a closed interval is Riemann integrable. **Final answer:** \[ \boxed{ \text{An increasing function on } [a, b] \text{ is Riemann integrable}. } \] --- ## **Summary of final answers:** - **(a):** False - **(b):** False - **(c):** True - **(d):** True - **(a) in #2:** The sum inequality holds for increasing \(f\). - **(b) in #2:** Increasing functions are Riemann integrable. Let me know if you'd like further clarification!