[problem:Thomae Function Is Riemann Integrable | 2] Prove that the Thomae function $f: [0, 1] \to \mathbb{R}$ defined by $f(x) = 1/q$ if $x = p/q$ in lowest terms and $f(x) = 0$ if $x$ is irrational, is Riemann integrable with $\int_0^1 f = 0$. [/problem]

[solution] **Step 1: Lower integral.** Since $f(x) \geq 0$ for all $x$ and $f(x) = 0$ at every irrational, every subinterval $[x_{j-1}, x_j]$ contains irrationals, so $m_j = \inf f([x_{j-1}, x_j]) = 0$. Therefore $s(f, \mathcal{D}) = 0$ for every partition $\mathcal{D}$, giving $I_*(f) = 0$. **Step 2: Upper integral via the finitely-many-bad-points technique.** Fix $\varepsilon > 0$. Choose $N \in \mathbb{N}$ with $1/N < \varepsilon/2$. The set $B = \{x \in [0, 1] : f(x) \geq 1/N\}$ consists of rationals $p/q$ with $q \leq N$ — at most $N^2$ points. **Step 3: Construct the partition.** Place each point of $B$ inside an interval of length at most $\varepsilon/(2N^2)$ (this requires at most $2|B|$ extra partition points). On these "bad" intervals, $M_j \leq 1$; their total contribution to the upper sum is at most $|B| \cdot 1 \cdot \varepsilon/(2N^2) \leq \varepsilon/2$. On all remaining intervals, $f(x) < 1/N < \varepsilon/2$, so $M_j < \varepsilon/2$; their total contribution is at most $(\varepsilon/2) \cdot 1 = \varepsilon/2$. **Step 4: Conclude.** $S(f, \mathcal{D}) \leq \varepsilon/2 + \varepsilon/2 = \varepsilon$. Since $\varepsilon > 0$ was arbitrary, $I^*(f) = 0 = I_*(f)$, so $f$ is Riemann integrable with $\int_0^1 f = 0$. The technique — isolating finitely many large-value points and confining them to tiny intervals — is the same one used to prove continuity of the Thomae function at irrationals. This connection is not a coincidence: the Riemann integrability of a bounded function on $[a, b]$ is equivalent to the set of discontinuities having Lebesgue measure zero (Lebesgue's criterion), and the Thomae function is continuous except on $\mathbb{Q} \cap [0, 1]$, which is countable and hence has measure zero. [/solution]

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[problem:Uniform Limit Of Continuous Functions | 2] Let $f_n: [a, b] \to \mathbb{R}$ be continuous for each $n$, and suppose $f_n \to f$ uniformly on $[a, b]$. Prove that $f$ is continuous on $[a, b]$. [/problem]

[solution] **Step 1: Set up the $\varepsilon$-$\delta$ argument.** Fix $c \in [a, b]$ and $\varepsilon > 0$. We need $\delta > 0$ with $|x - c| < \delta \implies |f(x) - f(c)| < \varepsilon$. **Step 2: Use the triangle inequality to decompose.** For any $n$: \begin{align*} |f(x) - f(c)| \leq |f(x) - f_n(x)| + |f_n(x) - f_n(c)| + |f_n(c) - f(c)|. \end{align*} The first and third terms are controlled by uniform convergence; the middle term by continuity of $f_n$. **Step 3: Choose $n$ from uniform convergence.** Since $f_n \to f$ uniformly, there exists $N$ such that $\|f_N - f\|_\infty < \varepsilon/3$. This makes the first and third terms each less than $\varepsilon/3$, simultaneously for all $x$. **Step 4: Choose $\delta$ from continuity of $f_N$.** Since $f_N$ is continuous at $c$, there exists $\delta > 0$ with $|x - c| < \delta \implies |f_N(x) - f_N(c)| < \varepsilon/3$. **Step 5: Combine.** For $|x - c| < \delta$: $|f(x) - f(c)| < \varepsilon/3 + \varepsilon/3 + \varepsilon/3 = \varepsilon$. $\checkmark$ The "$\varepsilon/3$ trick" — splitting a bound into three pieces using the triangle inequality, then controlling each piece with a different tool — is one of the most common techniques in analysis. The key insight is that *uniform* convergence is essential: if the convergence were merely pointwise, we could not choose $N$ independently of $x$, and the argument would collapse. The sequence $f_n(x) = x^n$ on $[0, 1]$ converges pointwise to a discontinuous limit ($0$ for $x \in [0, 1)$ and $1$ for $x = 1$), showing that pointwise limits of continuous functions need not be continuous. [/solution]

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[problem:Square Root Is Uniformly Continuous On Half-Line | 2] Prove that $f(x) = \sqrt{x}$ is uniformly continuous on $[0, \infty)$, even though the domain is not compact. [/problem]

[solution] **Step 1: Algebraic identity.** For $x, y \geq 0$ with $x \neq y$: \begin{align*} |\sqrt{x} - \sqrt{y}| = \frac{|x - y|}{\sqrt{x} + \sqrt{y}}. \end{align*} **Step 2: Two-regime estimate.** Fix $\varepsilon > 0$. *When $\max(x, y) \geq \varepsilon^2$:* Then $\sqrt{x} + \sqrt{y} \geq \sqrt{\max(x, y)} \geq \varepsilon$, so $|\sqrt{x} - \sqrt{y}| \leq |x - y|/\varepsilon$. This is less than $\varepsilon$ whenever $|x - y| < \varepsilon^2$. *When $\max(x, y) < \varepsilon^2$:* Then $\sqrt{x} < \varepsilon$ and $\sqrt{y} < \varepsilon$, so $|\sqrt{x} - \sqrt{y}| \leq \sqrt{x} + \sqrt{y} < 2\varepsilon$. **Step 3: Choose $\delta$.** Set $\delta = \varepsilon^2$. If $|x - y| < \delta$: in the first regime, $|\sqrt{x} - \sqrt{y}| < \varepsilon$; in the second regime, $|\sqrt{x} - \sqrt{y}| < 2\varepsilon$. Replacing $\varepsilon$ with $\varepsilon/2$ throughout gives $\delta = \varepsilon^2/4$ working for all $x, y$. **Step 4: Observe the Lipschitz failure.** Despite uniform continuity, $f$ is *not* Lipschitz: $|\sqrt{x} - 0|/|x - 0| = 1/\sqrt{x} \to \infty$ as $x \to 0^+$. The modulus of continuity $\omega(\delta) = \sqrt{\delta}$ tends to zero, but $\omega(\delta)/\delta = 1/\sqrt{\delta} \to \infty$ — the function is uniformly continuous but not Lipschitz. This problem illustrates that Heine-Cantor is sufficient but not necessary for uniform continuity: the unbounded domain $[0, \infty)$ prevents the application of Heine-Cantor, yet the concavity of $\sqrt{x}$ provides uniform continuity through a direct argument. $\checkmark$ [/solution]

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[problem:Continuous Injection On An Interval Is Strictly Monotone | 3] Let $f: [a, b] \to \mathbb{R}$ be continuous and injective. Prove that $f$ is strictly monotone (either strictly increasing or strictly decreasing). [/problem]

[solution] **Step 1: Establish the sign of $f(b) - f(a)$.** Since $f$ is injective and $a \neq b$, we have $f(a) \neq f(b)$. Without loss of generality, assume $f(a) < f(b)$; we will show $f$ is strictly increasing. (The case $f(a) > f(b)$ gives strictly decreasing by applying the argument to $-f$.) **Step 2: Suppose for contradiction that $f$ is not strictly increasing.** Then there exist $c, d \in [a, b]$ with $c < d$ but $f(c) \geq f(d)$. Since $f$ is injective, $f(c) > f(d)$. **Step 3: Apply the IVT to derive a contradiction.** Consider the value $\gamma = f(d)$. We have $f(a) < f(b)$ and $f(c) > f(d)$, and we need to find two distinct points mapping to the same value. *Case 1: $f(d) \geq f(a)$.* Then $f(a) \leq f(d) < f(c)$. By the [Intermediate Value Theorem](/theorems/180) applied to $f$ on $[a, c]$, there exists $x_1 \in [a, c]$ with $f(x_1) = f(d)$. Since $x_1 \in [a, c]$ and $d > c$, we have $x_1 \neq d$ but $f(x_1) = f(d)$, contradicting injectivity. *Case 2: $f(d) < f(a)$.* Then $f(d) < f(a) < f(b)$. By the IVT applied to $f$ on $[d, b]$, there exists $x_2 \in [d, b]$ with $f(x_2) = f(a)$. Since $x_2 \in [d, b]$ and $a < d$, we have $x_2 \neq a$ but $f(x_2) = f(a)$, contradicting injectivity. **Step 4: Conclude.** In both cases, the assumption that $f$ is not strictly increasing leads to a contradiction with injectivity. Therefore $f$ is strictly increasing (given $f(a) < f(b)$). This result shows that continuous injections on intervals are automatically monotone — a purely topological constraint that fails for discontinuous injections (e.g., $f(x) = \mathbf{1}_\mathbb{Q}(x) + x$ is injective on $[0, 1]$ but not monotone). It also explains why continuous bijections from $[a, b]$ to $[f(a), f(b)]$ are automatically homeomorphisms: the inverse is continuous because it is the inverse of a strictly monotone continuous function. $\checkmark$ [/solution]