Exercises · 10. Integration

Easy

1Easy

Define $f$ on $[0,1]$ by $f(1/2) = 1$ and $f(x) = 0$ for $x \ne 1/2$ . Prove directly from upper and lower sums that $f$ is Riemann integrable with $\int_0^1 f = 0$ .

Upper and Lower Sums Close In

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The function $f$ is bounded ( $0 \le f \le 1$ ). For $0 < \delta < 1/2$ take the partition $P_\delta = \{0,\ 1/2 - \delta,\ 1/2 + \delta,\ 1\}$ . On the two outer intervals $f \equiv 0$ , so both the infimum and supremum are $0$ ; on the middle interval $[1/2-\delta,\ 1/2+\delta]$ the infimum is $0$ (points other than $1/2$ exist) and the supremum is $1$ (since $f(1/2) = 1$ ).
Hence the lower sum is $L(P_\delta, f) = 0$ and the upper sum is $U(P_\delta, f) = 1 \cdot 2\delta = 2\delta$ . Since $U - L = 2\delta \to 0$ as $\delta \to 0^+$ , the Riemann criterion gives integrability. The integral satisfies $0 = L(P_\delta, f) \le \int_0^1 f \le U(P_\delta, f) = 2\delta$ for every $\delta > 0$ , so it equals $0$ — the value at a single point does not affect the Riemann integral.

Medium

2Medium

Let $F(x) = x^2\sin(1/x^2)$ for $x \ne 0$ and $F(0) = 0$ . Show $F$ is differentiable at every point of $[-1,1]$ , but that $F'$ is unbounded on every neighborhood of $0$ . Conclude $F'$ is not Riemann integrable on $[-1,1]$ , and explain why this shows the hypothesis "the derivative is Riemann integrable" in FTC-II cannot be dropped.

F'(0) = \lim_{h \to 0} h\sin(1/h^2) = 0

F'(x) = 2x\sin(1/x^2) - \frac{2}{x}\cos(1/x^2) \quad (x \ne 0)

x_k = \frac{1}{\sqrt{2\pi k}}:\quad F'(x_k) = -2\sqrt{2\pi k} \to -\infty

The Area Function's Derivative Is the Function

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Get $F'(0)$ from the definition: $\dfrac{F(h) - F(0)}{h} = h\sin(1/h^2)$ , and $\lvert h\sin(1/h^2) \rvert \le \lvert h \rvert$ , so the squeeze theorem gives the limit $0$ . For $x \ne 0$ , the product and chain rules give $F'(x) = 2x\sin(1/x^2) + x^2\cos(1/x^2)\cdot(-2/x^3) = 2x\sin(1/x^2) - \dfrac{2}{x}\cos(1/x^2)$ . Hence $F$ is differentiable at every point of $[-1,1]$ .
But $F'$ is unbounded near $0$ . Put $x_k = \dfrac{1}{\sqrt{2\pi k}} \to 0^+$ ( $k = 1, 2, \dots$ ): then $\sin(1/x_k^2) = \sin(2\pi k) = 0$ and $\cos(1/x_k^2) = \cos(2\pi k) = 1$ , so $F'(x_k) = -\dfrac{2}{x_k} = -2\sqrt{2\pi k} \to -\infty$ . Every neighborhood of $0$ contains $x_k$ for large $k$ , so $F'$ is unbounded there.
A Riemann integrable function must be bounded (needed for finite upper sums), so $F'$ is not Riemann integrable on $[-1,1]$ . FTC-II asserts " $\int_a^b F' = F(b) - F(a)$ " but requires $F'$ to be Riemann integrable as a hypothesis — this example shows that without it the integral $\int_{-1}^{1} F'$ does not even exist, so the theorem cannot be applied. Differentiability alone is not enough.

3Medium

Let $f$ and $g$ be of bounded variation on $[a,b]$ . First show a BV function is bounded (hint: $\lvert f(x) - f(a) \rvert \le V_a^b(f)$ ), then prove the product $fg$ is also of bounded variation on $[a,b]$ . Find an upper bound for its total variation.

V_a^b(fg) \le M_f\,V_a^b(g) + M_g\,V_a^b(f), \qquad M_f = \sup_{[a,b]}\lvert f \rvert,\ M_g = \sup_{[a,b]}\lvert g \rvert

Functions of Bounded Variation

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For $x \in [a,b]$ , the partition $\{a, x, b\}$ gives $\lvert f(x) - f(a) \rvert \le V_a^b(f)$ , so by the triangle inequality $\lvert f(x) \rvert \le \lvert f(a) \rvert + V_a^b(f) < \infty$ . Hence $f$ is bounded with $M_f = \sup_{[a,b]}\lvert f \rvert < \infty$ , and likewise for $g$ .
For any partition $P = \{x_0, \dots, x_n\}$ , use the identity $f(x_i)g(x_i) - f(x_{i-1})g(x_{i-1}) = f(x_i)\bigl(g(x_i) - g(x_{i-1})\bigr) + g(x_{i-1})\bigl(f(x_i) - f(x_{i-1})\bigr)$ . Taking absolute values and summing, \[\sum_{i=1}^{n}\lvert (fg)(x_i) - (fg)(x_{i-1}) \rvert \le M_f \sum_{i=1}^{n}\lvert g(x_i) - g(x_{i-1}) \rvert + M_g \sum_{i=1}^{n}\lvert f(x_i) - f(x_{i-1}) \rvert \le M_f V_a^b(g) + M_g V_a^b(f).\] The right side is independent of $P$ , so taking the supremum gives $V_a^b(fg) \le M_f V_a^b(g) + M_g V_a^b(f) < \infty$ : $fg$ is of bounded variation.

Hard

4Hard

The uniform limit of continuous functions is continuous (the Weierstrass scene). Show this fails for bounded variation: with $g(x) = x^2\sin(1/x^2)$ for $x > 0$ and $g(0) = 0$ , put \[g_n(x) = \begin{cases} g(1/n) & 0 \le x \le 1/n \\ g(x) & 1/n \le x \le 1 \end{cases}.\] (a) Show $g$ is NOT of bounded variation on $[0,1]$ , using the oscillation at the points $x_k = \sqrt{2/((2k+1)\pi)}$ . (b) Show each $g_n$ is of bounded variation on $[0,1]$ and $g_n \to g$ uniformly.

x_k = \frac{1}{\sqrt{\pi/2 + k\pi}},\qquad g(x_k) = (-1)^k\,\frac{2}{(2k+1)\pi}

\sum_{k=1}^{N}\lvert g(x_{k-1}) - g(x_k) \rvert = \frac{2}{\pi}\sum_{k=1}^{N}\left(\frac{1}{2k-1} + \frac{1}{2k+1}\right) \to \infty

Functions of Bounded Variation

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(a) Put $x_k = \dfrac{1}{\sqrt{\pi/2 + k\pi}}$ ; then $x_k \downarrow 0$ and $\sin(1/x_k^2) = \sin(\pi/2 + k\pi) = (-1)^k$ , so $g(x_k) = (-1)^k x_k^2 = (-1)^k\,\dfrac{2}{(2k+1)\pi}$ . Also $x_0 = \sqrt{2/\pi} \approx 0.798 < 1$ . On the partition $0 < x_N < x_{N-1} < \cdots < x_1 < x_0 < 1$ , consecutive values have opposite signs, so \[\lvert g(x_{k-1}) - g(x_k) \rvert = \frac{2}{\pi}\left(\frac{1}{2k-1} + \frac{1}{2k+1}\right) \qquad (k = 1, \dots, N).\] Hence the variation sum over this partition is $\dfrac{2}{\pi}\sum_{k=1}^{N}\left(\dfrac{1}{2k-1} + \dfrac{1}{2k+1}\right) \ge \dfrac{2}{\pi}\sum_{k=1}^{N}\dfrac{1}{2k+1} \ge \dfrac{1}{\pi}\sum_{k=1}^{N}\dfrac{1}{k+1} \to \infty$ as $N \to \infty$ (harmonic series). The total variation is infinite, so $g$ is not of bounded variation.
(b) $g_n$ is continuous at $x = 1/n$ (both sides equal $g(1/n)$ ). It is constant, hence BV, on $[0, 1/n]$ , and $g$ is $C^1$ , hence BV, on $[1/n, 1]$ ( $V_{1/n}^1(g) \le \int_{1/n}^1 \lvert g' \rvert < \infty$ since $g'$ is continuous on a compact interval). So each $g_n$ is of bounded variation on $[0,1]$ . For uniform convergence: on $[1/n, 1]$ , $g_n = g$ , so the difference is $0$ ; on $[0, 1/n]$ , $\lvert g_n(x) - g(x) \rvert = \lvert g(1/n) - g(x) \rvert \le \lvert g(1/n) \rvert + \lvert g(x) \rvert \le \dfrac{1}{n^2} + \dfrac{1}{n^2} = \dfrac{2}{n^2} \to 0$ (using $\lvert g(t) \rvert \le t^2$ ). Thus $\lVert g_n - g \rVert_\infty \to 0$ .
Conclusion: each $g_n$ is of bounded variation and $g_n \to g$ uniformly, but the limit $g$ is not of bounded variation. Uniform limits preserve continuity but not bounded variation — variation can "leak out" in the limit.