Let's walk through **Section #2 (Basic Understanding and True-or-False Questions)** step by step in Tutor Mode, explaining each reasoning. --- ## #2. Basic Understanding and True-or-False Questions Let's analyze each statement one by one. --- ###a) If \( f \) is Riemann integrable on \([a, b]\), then \( f \) is continuous on \([a, b]\). **Step 1:** Recall the definition of Riemann integrable functions: A function \( f \) is Riemann integrable on \([a, b]\) if and only if it is **bounded** and the set of its discontinuities has measure zero (i.e., is a set of zero length). **Step 2:** Counterexample: The function \( f(x) = \begin{cases} 1, & x= \\ , & x \neq \end{cases} \) on \([, 1]\) is Riemann integrable (since the set of discontinuities is just \(\{\}\), which has measure zero), but it is not continuous. **Conclusion:** **False**. Riemann integrable functions can have discontinuities as long as they are not "too many" (set of measure zero). --- ### (b) If \( |f| \) is Riemann integrable on \([a, b]\), then \( f \) is Riemann integrable on \([a, b]\). **Step 1:** Recall: Riemann integrability is not always preserved under "taking the absolute value" in reverse. **Step 2:** Counterexample: Suppose \( f \) is the Dirichlet function on \([, 1]\) defined by \( f(x) = 1 \) if \( x \) rational, \( -1 \) if \( x \) irrational. Then \( |f(x)| = 1 \), which is Riemann integrable, but \( f \) itself is **not** Riemann integrable (since it is discontinuous everywhere). **Conclusion:** **False**. --- ### (c) If \( f \) is continuous on \([a, b]\) and on \([b, c]\), then \( f \) is Riemann integrable on \([a, c]\). **Step 1:** Continuity on both subintervals implies that \( f \) is continuous on all of \([a, c]\) except possibly at \( b \). **Step 2:** Even if \( f \) is discontinuous at \( b \), this is just a single point—set of measure zero. **Step 3:** The set of discontinuities is at most countable (in this case, one point), so \( f \) is Riemann integrable. **Conclusion:** **True**. --- ### (d) If \( f \) is locally integrable on \( (a, b) \), then \( |f| \) is locally integrable on \( (a, b) \). **Step 1:** By definition, "locally integrable" means for every compact subinterval \( [c, d] \subset (a, b) \), \( f \) is integrable on \( [c, d] \). **Step 2:** If \( f \) is integrable on \( [c, d] \), then \( |f| \) is also integrable on \( [c, d] \) (by properties of Riemann/Lebesgue integrals). **Conclusion:** **True**. --- ### (e) If \( f \) is improperly integrable on \( [, \infty) \), then \( |f| \) is also improperly integrable on \( [, \infty) \). **Step 1:** Improper integrability means \( \int_^\infty f(x) dx \) converges. **Step 2:** However, it is possible for \( f \) to have cancellation of positive and negative areas, causing \( \int_^\infty f(x) dx \) to converge while \( \int_^\infty |f(x)| dx \) diverges. **Example:** \( f(x) = \frac{\sin x}{x} \) on \( [1, \infty) \): - \( \int_1^\infty \frac{\sin x}{x} dx \) converges - \( \int_1^\infty \left|\frac{\sin x}{x}\right| dx \) diverges **Conclusion:** **False**. --- ### (f) If \( f \) is improperly integrable on \( [, \infty) \), then \( \lim_{x \to \infty} f(x) = \). **Step 1:** For the improper integral to converge, it is necessary (but not sufficient) that \( f(x) \to \) as \( x \to \infty \). **Step 2:** If \( f(x) \) does not tend to zero, then the integral over the tail would not converge. **Conclusion:** **True**. --- ## Summary Table | Statement | True/False | Justification | |-----------|------------|--------------| | (a) | False | Discontinuities of measure zero allowed | | (b) | False | Counterexample: Dirichlet function | | (c) | True | At most finite discontinuity at \( b \) | | (d) | True | Integrability of \( f \implies \) integrability of \( |f| \) locally | | (e) | False | Absolute integrability may diverge | | (f) | True | Must tend to zero for convergence | --- Would you like to proceed with another section or need details on any specific question?