[problem: Riemann Integrability Of A Piecewise Constant Function | 1] Let $f: [0, 3] \to \mathbb{R}$ be defined by \begin{align*} f: [0, 3] &\to \mathbb{R} \\ x &\mapsto \begin{cases} 2 & \text{if } x \in [0, 1), \\ -1 & \text{if } x \in [1, 2), \\ 3 & \text{if } x \in [2, 3]. \end{cases} \end{align*} Prove directly from the Darboux criterion that $f \in \mathcal{R}([0, 3])$ and compute $\int_0^3 f(x)\, dx$. [/problem]

[solution] **Step 1: Construct a partition isolating the discontinuities.** The function $f$ is discontinuous at $x = 1$ and $x = 2$ (jump discontinuities). For any $\varepsilon > 0$, define the partition $P_\varepsilon = \{0, 1 - \varepsilon/12, 1 + \varepsilon/12, 2 - \varepsilon/12, 2 + \varepsilon/12, 3\}$. This partition has six points, creating five subintervals. On each subinterval that avoids the discontinuities, $f$ is constant, so $M_i = m_i$ on that subinterval. **Step 2: Estimate the Darboux sum difference on each subinterval.** On $[0, 1 - \varepsilon/12]$, $f = 2$ identically, so $M_1 - m_1 = 0$. On $[1 + \varepsilon/12, 2 - \varepsilon/12]$, $f = -1$ identically, so $M_3 - m_3 = 0$. On $[2 + \varepsilon/12, 3]$, $f = 3$ identically, so $M_5 - m_5 = 0$. On the two transition intervals $[1 - \varepsilon/12, 1 + \varepsilon/12]$ and $[2 - \varepsilon/12, 2 + \varepsilon/12]$, the function takes two values, so the oscillation is at most $|2 - (-1)| = 3$ on the first and $|{-1} - 3| = 4$ on the second. **Step 3: Compute the total Darboux sum difference.** Each transition interval has length $\varepsilon/6$, so \begin{align*} U(f, P_\varepsilon) - L(f, P_\varepsilon) &= 0 + 3 \cdot \frac{\varepsilon}{6} + 0 + 4 \cdot \frac{\varepsilon}{6} + 0 = \frac{3\varepsilon + 4\varepsilon}{6} = \frac{7\varepsilon}{6}. \end{align*} For the original $\varepsilon$, this quantity is less than $2\varepsilon$. Since $\varepsilon > 0$ is arbitrary, the Darboux criterion is satisfied and $f \in \mathcal{R}([0, 3])$. **Step 4: Compute the integral using additivity.** By the additivity of the Riemann integral over intervals, \begin{align*} \int_0^3 f(x)\, dx &= \int_0^1 2\, dx + \int_1^2 (-1)\, dx + \int_2^3 3\, dx \\ &= 2(1 - 0) + (-1)(2 - 1) + 3(3 - 2) \\ &= 2 - 1 + 3 = 4. \end{align*} The value at the finitely many discontinuity points does not affect the integral (changing $f$ at finitely many points does not change any Darboux sum by more than a quantity that vanishes as the mesh tends to zero). [/solution]

[problem: Upper And Lower Sums For A Quadratic | 2] Let $f: [0, 1] \to \mathbb{R}$ be defined by $f(x) = x^2$. For the uniform partition $P_n$ of $[0, 1]$ into $n$ equal subintervals, compute $U(f, P_n)$ and $L(f, P_n)$ explicitly in terms of $n$, show that $U(f, P_n) - L(f, P_n) \to 0$ as $n \to \infty$, and verify that $\int_0^1 x^2 \, dx = 1/3$. [/problem]

[solution] **Step 1: Identify the partition and extrema.** The uniform partition $P_n = \{0, 1/n, 2/n, \ldots, (n-1)/n, 1\}$ has $n$ subintervals $[(i-1)/n, \, i/n]$ for $i = 1, \ldots, n$, each of width $\Delta x_i = 1/n$. Since $f(x) = x^2$ is strictly increasing on $[0, 1]$, the supremum and infimum on the $i$-th subinterval are \begin{align*} M_i = f(i/n) = \frac{i^2}{n^2}, \qquad m_i = f((i-1)/n) = \frac{(i-1)^2}{n^2}. \end{align*} **Step 2: Compute the upper and lower sums using the sum-of-squares identity.** The classical identity $\sum_{i=1}^n i^2 = n(n+1)(2n+1)/6$ gives \begin{align*} U(f, P_n) &= \sum_{i=1}^n \frac{i^2}{n^2} \cdot \frac{1}{n} = \frac{1}{n^3} \sum_{i=1}^n i^2 = \frac{1}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6n^2}. \end{align*} Similarly, using $\sum_{i=0}^{n-1} i^2 = \sum_{i=1}^{n-1} i^2 = (n-1)n(2n-1)/6$, \begin{align*} L(f, P_n) &= \sum_{i=1}^n \frac{(i-1)^2}{n^2} \cdot \frac{1}{n} = \frac{1}{n^3} \sum_{i=0}^{n-1} i^2 = \frac{1}{n^3} \cdot \frac{(n-1)n(2n-1)}{6} = \frac{(n-1)(2n-1)}{6n^2}. \end{align*} **Step 3: Compute the Darboux sum difference and verify convergence.** \begin{align*} U(f, P_n) - L(f, P_n) &= \frac{(n+1)(2n+1) - (n-1)(2n-1)}{6n^2}. \end{align*} Expanding the numerator: $(n+1)(2n+1) = 2n^2 + 3n + 1$ and $(n-1)(2n-1) = 2n^2 - 3n + 1$. The difference is $6n$, so \begin{align*} U(f, P_n) - L(f, P_n) = \frac{6n}{6n^2} = \frac{1}{n}. \end{align*} This confirms $U(f, P_n) - L(f, P_n) \to 0$ as $n \to \infty$. The Darboux criterion is satisfied, confirming $f \in \mathcal{R}([0, 1])$. **Step 4: Evaluate the integral.** Taking $n \to \infty$ in either sum: \begin{align*} \int_0^1 x^2 \, dx = \lim_{n \to \infty} U(f, P_n) = \lim_{n \to \infty} \frac{(n+1)(2n+1)}{6n^2} = \lim_{n \to \infty} \frac{2n^2 + 3n + 1}{6n^2} = \frac{2}{6} = \frac{1}{3}. \end{align*} This agrees with the Second Fundamental Theorem of Calculus, which gives $\int_0^1 x^2\, dx = [x^3/3]_0^1 = 1/3$. [/solution]

[problem: The Riemann Integral And Modification On Null Sets | 3] Let $f \in \mathcal{R}([a, b])$ and let $g: [a, b] \to \mathbb{R}$ be a bounded function that agrees with $f$ except on a set $E \subseteq [a, b]$ of Lebesgue measure zero. Prove that $g \in \mathcal{R}([a, b])$ and $\int_a^b g(x)\, dx = \int_a^b f(x)\, dx$. [/problem]

[solution] **Step 1: Establish that $h := g - f$ is zero except on a null set.** Define $h: [a, b] \to \mathbb{R}$ by $h(x) := g(x) - f(x)$. By hypothesis, $h(x) = 0$ for all $x \in [a, b] \setminus E$, where $E$ has Lebesgue measure zero. The function $h$ is bounded because both $f$ and $g$ are bounded; let $M := \sup_{[a,b]} |h|$. **Step 2: Show $h \in \mathcal{R}([a, b])$ with integral zero using the Darboux criterion.** Let $\varepsilon > 0$. Since $E$ has Lebesgue measure zero, there exists a countable collection of open intervals $\{(a_k, b_k)\}_{k=1}^\infty$ with $E \subseteq \bigcup_{k=1}^\infty (a_k, b_k)$ and $\sum_{k=1}^\infty (b_k - a_k) < \varepsilon / (2M)$. By compactness of $[a, b]$, finitely many of these intervals cover $E \cap [a, b]$, say $(a_1, b_1), \ldots, (a_N, b_N)$ with $\sum_{k=1}^N (b_k - a_k) < \varepsilon / (2M)$. **Step 3: Construct a partition subordinate to the cover.** Let $P$ be any partition of $[a, b]$ whose points include all endpoints $a_1, b_1, \ldots, a_N, b_N$ (after intersecting with $[a, b]$). Separate the subintervals of $P$ into two groups: those $I_j$ entirely contained in some $(a_k, b_k)$, and those $I_j$ that are disjoint from $E$. On the latter, $h = 0$ identically, so $M_j - m_j = 0$. On the former, $M_j - m_j \le 2M$. The total length of the former subintervals is at most $\sum_{k=1}^N (b_k - a_k) < \varepsilon / (2M)$. Therefore \begin{align*} U(h, P) - L(h, P) &= \sum_{\text{near } E} (M_j - m_j) \Delta x_j + \sum_{\text{away from } E} 0 \cdot \Delta x_j \le 2M \cdot \frac{\varepsilon}{2M} = \varepsilon. \end{align*} By the Darboux criterion, $h \in \mathcal{R}([a, b])$. Since $0 \le U(h, P)$ and $L(h, P) \le 0$ can be arranged (the function $h$ takes value $0$ on most of $[a, b]$), passing to the limit gives $\int_a^b h\, dx = 0$. **Step 4: Conclude by linearity.** Since $g = f + h$ and both $f$ and $h$ are in $\mathcal{R}([a, b])$, linearity of the Riemann integral gives \begin{align*} \int_a^b g(x)\, dx = \int_a^b f(x)\, dx + \int_a^b h(x)\, dx = \int_a^b f(x)\, dx + 0 = \int_a^b f(x)\, dx. \end{align*} This result shows that the Riemann integral "does not see" modifications on null sets — a property shared with, and ultimately formalised more completely by, the Lebesgue integral. [/solution]

[problem: Closure Failure Of The Riemann Integral Under Pointwise Limits | 4] Let $\{q_1, q_2, q_3, \ldots\}$ be an enumeration of $\mathbb{Q} \cap [0, 1]$. For each $n \in \mathbb{N}$, define $f_n := \mathbb{1}_{\{q_1, \ldots, q_n\}}$. (a) Prove that each $f_n \in \mathcal{R}([0, 1])$ and compute $\int_0^1 f_n\, dx$. (b) Show that $f_n \to \mathbb{1}_\mathbb{Q}$ pointwise on $[0, 1]$ and that $\mathbb{1}_\mathbb{Q} \notin \mathcal{R}([0, 1])$. (c) Explain why this failure does not arise in the Lebesgue theory. State precisely which theorem applies and what it concludes. [/problem]

[solution] **Step 1: Each $f_n$ is Riemann integrable with integral zero.** The function $f_n$ is zero everywhere except at the $n$ points $q_1, \ldots, q_n$, where it equals $1$. The discontinuity set of $f_n$ is contained in $\{q_1, \ldots, q_n\}$, which is finite and therefore has Lebesgue measure zero. By the Lebesgue criterion, $f_n \in \mathcal{R}([0, 1])$. To compute the integral directly from Darboux sums: for any $\varepsilon > 0$, choose a partition $P$ that encloses each $q_i$ in a subinterval of length less than $\varepsilon / n$. On the $n$ subintervals containing the $q_i$, $M_j = 1$ and $m_j = 0$ (since irrationals are dense), so $(M_j - m_j)\Delta x_j < \varepsilon / n$. On all other subintervals, $f_n = 0$ identically, so $M_j = m_j = 0$. Therefore \begin{align*} U(f_n, P) - L(f_n, P) \le n \cdot \frac{\varepsilon}{n} = \varepsilon, \end{align*} confirming integrability. Since $L(f_n, P) = 0$ for every partition (every subinterval contains an irrational where $f_n = 0$), the integral is $\int_0^1 f_n\, dx = 0$. **Step 2: Pointwise convergence to $\mathbb{1}_\mathbb{Q}$ and non-integrability of the limit.** If $x \in \mathbb{Q} \cap [0, 1]$, then $x = q_k$ for some $k$, so $f_n(x) = 1$ for all $n \ge k$. If $x \notin \mathbb{Q}$, then $f_n(x) = 0$ for all $n$. In both cases, $f_n(x) \to \mathbb{1}_\mathbb{Q}(x)$. The Dirichlet function $\mathbb{1}_\mathbb{Q}$ is discontinuous at every point of $[0, 1]$ (both $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ are dense, so every neighbourhood of any point contains points where $\mathbb{1}_\mathbb{Q}$ takes both values $0$ and $1$). The discontinuity set is $D = [0, 1]$, which has Lebesgue measure $1 \neq 0$. By the Lebesgue criterion, $\mathbb{1}_\mathbb{Q} \notin \mathcal{R}([0, 1])$. **Step 3: Resolution in the Lebesgue theory.** In the Lebesgue framework, $\mathbb{1}_\mathbb{Q}$ is a measurable function (as the indicator of a countable, hence measurable, set). The Lebesgue Dominated Convergence Theorem applies to the sequence $\{f_n\}$: each $f_n$ is measurable, $f_n \to \mathbb{1}_\mathbb{Q}$ pointwise, and $|f_n| \le 1 =: g$ where $g \in L^1([0, 1], \mathcal{L}^1)$. The theorem concludes \begin{align*} \lim_{n \to \infty} \int_{[0,1]} f_n \, d\mathcal{L}^1 = \int_{[0,1]} \mathbb{1}_\mathbb{Q} \, d\mathcal{L}^1 = \mathcal{L}^1(\mathbb{Q} \cap [0,1]) = 0. \end{align*} Both sides equal $0$, and the interchange $\lim \int = \int \lim$ holds. The failure in the Riemann theory is not that the integral values disagree — if the Riemann integral of the limit existed, it would have to be $0$ as well — but that the Riemann integral cannot even *formulate* the conclusion, because the limit function lies outside $\mathcal{R}([0,1])$. The Lebesgue integral resolves this by working in the strictly larger and limit-closed space $L^1$. [/solution]