In a calculus course the integral of \(f\) from \(a\) to \(b\) is computed as \(F(b)-F(a)\) , where \(F\) is an antiderivative of \(f\) . As a method this is excellent. As a definition it fails on three counts. It says nothing when no antiderivative is known, or when none exists. It does not explain why the number obtained should be the area under the graph. And it makes the Fundamental Theorem of Calculus true by decree, when it ought to be a theorem connecting two independently defined things.

So the integral has to be defined without mentioning derivatives. The idea is as old as Archimedes: trap the area between sums of rectangles that overestimate it and sums that underestimate it, and call the function integrable if the two kinds of estimate can be forced together. Then one must ask which functions are integrable. Some bounded functions are not. Continuous functions are, and the reason is uniform continuity. With the integral in hand, the Fundamental Theorem becomes a statement with hypotheses and a proof, and its hypotheses turn out to matter.

7.1 Upper and lower sums

Throughout, \(f\) is a bounded real function on a closed bounded interval \([a,b]\) with \(a<b\) . Boundedness is needed for the definitions to make sense.

Clearly \(L(f,P)\le U(f,P)\) . The lower sum is the total area of rectangles lying under the graph and the upper sum of rectangles containing it, at least when \(f\ge 0\) . A partition \(Q\) is a refinement of \(P\) if \(P\subseteq Q\) .

So every lower sum is a lower bound for the set of all upper sums, whatever partitions are used. This is exactly the situation of Proposition 1.15(d).

This formulation by upper and lower sums is due to Darboux. Riemann’s original definition used sums \(\sum f(t_i)\Delta x_i\) with arbitrary sample points; Section 7.4 shows the two agree.

Example 7.1: The integral of the squaring function from the definition

Let \(f(x)=x^2\) on \([0,1]\) and let \(P_n\) be the partition into \(n\) equal pieces, \(x_i=i/n\) . Since \(f\) is increasing, \(m_i=f(x_{i-1})\) and \(M_i=f(x_i)\) . Using \(\sum_{k=1}^nk^2=n(n+1)(2n+1)/6\) ,

\[ U(f,P_n)=\frac1n\sum_{i=1}^n\frac{i^2}{n^2}=\frac{(n+1)(2n+1)}{6n^2},\qquad L(f,P_n)=\frac1n\sum_{i=1}^{n}\frac{(i-1)^2}{n^2}=\frac{(n-1)(2n-1)}{6n^2}. \]

For \(n=4\) these are \(U=15/32=0.46875\) and \(L=7/32=0.21875\) . In general \(U(f,P_n)-L(f,P_n)=1/n\) , and both sums tend to \(1/3\) .

Now the definition does the rest. For every \(n\) ,

\[ L(f,P_n)\le L(f)\le U(f)\le U(f,P_n). \]

The outer terms converge to \(1/3\) , so \(L(f)=U(f)=1/3\) . Hence \(f\) is integrable and \(\int_0^1x^2\,dx=1/3\) , with no antiderivative in sight.

7.2 Which functions are integrable

To prove integrability one rarely computes \(L(f)\) and \(U(f)\) . One shows that upper and lower sums for the same partition can be made close.

The difference has a useful form:

where \(M_i-m_i\) is the oscillation of \(f\) on the \(i\) th subinterval. The criterion asks for the oscillation to be small on most of the interval, measured by length. Moreover, as in Example 7.1, if \((P_n)\) is a sequence of partitions with \(U(f,P_n)-L(f,P_n)\to 0\) , then \(f\) is integrable and both \(U(f,P_n)\) and \(L(f,P_n)\) converge to \(\int_a^bf\) .

Uniform continuity is exactly what the proof needs: one \(\delta\) that controls the oscillation on all the subintervals at once. With pointwise continuity alone, each subinterval would demand its own fineness and there would be no way to choose the partition.

A monotone function can have infinitely many discontinuities, though at most countably many (Exercise 5.21): the function on \([0,1]\) equal to \(0\) at \(0\) and to \(1/n\) on each interval \((1/(n+1),1/n]\) is increasing and jumps at every point \(1/n\) with \(n\ge 2\) . So integrability does not require continuity. How much discontinuity can be tolerated?

The proof isolates the bad points in intervals of small total length, where the oscillation may be large but the length is negligible, and controls the oscillation everywhere else. The same strategy handles a function with infinitely many discontinuities, provided they can be confined in this way.

7.3 Properties of the integral

It is convenient to set \(\int_a^af=0\) and \(\int_b^af=-\int_a^bf\) . With these conventions the additivity formula holds for any three points \(a,b,c\) of an interval on which \(f\) is integrable, in any order.

The converse of the statement about \(\lvert f\rvert\) is false: the function equal to \(1\) on the rationals and \(-1\) on the irrationals is not integrable, while its absolute value is constant.

7.4 Riemann sums

Since \(m_i\le f(t_i)\le M_i\) , every Riemann sum satisfies \(L(f,P)\le S(f,P,t)\le U(f,P)\) .

Consequently, for an integrable function, \(\int_a^bf\) is the limit of the Riemann sums for any sequence of tagged partitions whose mesh tends to \(0\) , for instance equal subdivisions with left endpoints, right endpoints, or midpoints as tags. Many limits of sums can be recognised as integrals in this way; Example 7.6 is one.

7.5 The Fundamental Theorem of Calculus

There are two theorems, one for each order in which differentiation and integration can be performed.

Combining the two: a continuous function \(f\) on \([a,b]\) has an antiderivative, namely \(F(x)=\int_a^xf\) ; any other antiderivative differs from it by a constant, by Corollary 6.8(a); and the integral is the change in any antiderivative. This is the bridge that Calculus uses computationally (math.calculus.fundamental-theorem-bridge). Here the hypotheses are visible. Theorem 7.14 needs continuity of \(f\) at the point. Theorem 7.15 needs \(f\) to be integrable, which does not follow from its being a derivative. The next example shows both hypotheses failing.

7.6 Improper integrals and the integral test

The Riemann integral is defined for bounded functions on bounded intervals. Other cases are handled by limits. If \(f\) is integrable on \([a,b]\) for every \(b>a\) , the improper integral \(\int_a^\infty f\) is defined as \(\lim_{b\to\infty}\int_a^bf\) when that limit exists, and is then said to converge. For example, by Theorem 7.15, \(\int_1^bx^{-2}\,dx=1-1/b\to 1\) , so \(\int_1^\infty x^{-2}\,dx=1\) , while \(\int_1^bx^{-1}\,dx=\log b\) is unbounded, so \(\int_1^\infty x^{-1}\,dx\) diverges.

Exercises

1. For \(f(x)=1/x\) on \([1,2]\) and the partition \(P=\{1,\tfrac43,\tfrac53,2\}\) , compute \(L(f,P)\) and \(U(f,P)\) as exact fractions, and state the bounds on \(\log 2\) that result.

   Answer

   Each piece has length \(\tfrac13\) , and \(1/x\) is decreasing, so suprema are at left endpoints and infima at right endpoints. \(U(f,P)=\tfrac13\bigl(1+\tfrac34+\tfrac35\bigr)=\tfrac{47}{60}\approx 0.7833\) and \(L(f,P)=\tfrac13\bigl(\tfrac34+\tfrac35+\tfrac12\bigr)=\tfrac{37}{60}\approx 0.6167\) . Hence \(\tfrac{37}{60}\le\log 2\le\tfrac{47}{60}\) .

2. For \(f(x)=x\) on \([0,1]\) and the partition \(P_n\) into \(n\) equal pieces, compute \(U(f,P_n)\) and \(L(f,P_n)\) , and prove from Definition 7.4 that \(f\) is integrable with \(\int_0^1x\,dx=\tfrac12\) .

   Solution

   With \(x_i=i/n\) and \(f\) increasing, \(M_i=i/n\) and \(m_i=(i-1)/n\) . Using \(1+2+\cdots+n=n(n+1)/2\) ,

   \[ U(f,P_n)=\frac1n\sum_{i=1}^n\frac in=\frac{n+1}{2n},\qquad L(f,P_n)=\frac1n\sum_{i=1}^n\frac{i-1}{n}=\frac{n-1}{2n}. \]

   By the definitions of the lower and upper integrals as a supremum and an infimum, and by Lemma 7.3, for every \(n\)

   \[ \frac{n-1}{2n}=L(f,P_n)\le L(f)\le U(f)\le U(f,P_n)=\frac{n+1}{2n}. \]

   Both outer terms converge to \(\tfrac12\) as \(n\to\infty\) . The numbers \(L(f)\) and \(U(f)\) do not depend on \(n\) , so by the order theorem for limits \(\tfrac12\le L(f)\le U(f)\le\tfrac12\) . Hence \(L(f)=U(f)=\tfrac12\) : the function is integrable and \(\int_0^1x\,dx=\tfrac12\) .

   The difference \(U(f,P_n)-L(f,P_n)=1/n\) is the area of the \(n\) small squares of side \(1/n\) along the diagonal.

3. Let \(f(x)=1\) at \(x=\tfrac12\) and \(f(x)=0\) at every other point of \([0,1]\) . Prove directly from Theorem 7.5 that \(f\) is integrable, and find its integral.

   Solution

   Lower sums. \(f\ge 0\) , and every subinterval of positive length contains points other than \(\tfrac12\) , where \(f=0\) . So \(m_i=0\) for every subinterval and \(L(f,P)=0\) for every partition.

   Upper sums. Let \(0<\varepsilon<1\) and take \(P=\{0,\ \tfrac12-\tfrac\varepsilon4,\ \tfrac12+\tfrac\varepsilon4,\ 1\}\) . On the middle subinterval, of length \(\varepsilon/2\) , the supremum of \(f\) is \(1\) ; on the other two it is \(0\) . So \(U(f,P)=\varepsilon/2\) .

   Hence \(U(f,P)-L(f,P)=\varepsilon/2<\varepsilon\) , and \(f\) is integrable by Theorem 7.5. Its integral equals the lower integral, which is \(0\) because all lower sums are \(0\) .

   A single point, however tall the spike above it, carries no area: it can be enclosed in an interval as short as we please. The same argument handles any finite number of exceptional points, which is Exercise 4.

4. Let \(f\in\mathcal R[a,b]\) and let \(g\) be a function on \([a,b]\) that differs from \(f\) at only finitely many points. Prove that \(g\) is integrable and \(\int_a^bg=\int_a^bf\) .

   Hint

   Let \(h=g-f\) , which vanishes except at finitely many points \(c_1,\dots,c_k\) and is bounded by some \(H\) . As in Exercise 3, enclose each \(c_j\) in a subinterval of length less than \(\varepsilon/(2kH)\) to get \(-\varepsilon<L(h,P)\le U(h,P)<\varepsilon\) . So \(h\) is integrable with integral \(0\) , and \(g=f+h\) is integrable with the same integral as \(f\) by Theorem 7.9.

5. Let \(f(x)=x\) for rational \(x\) and \(f(x)=0\) for irrational \(x\) , on \([0,1]\) . Compute the upper integral \(U(f)\) and the lower integral \(L(f)\) , and conclude that \(f\) is not integrable.

   Solution

   Lower integral. \(f\ge 0\) , and every subinterval of positive length contains an irrational number, where \(f=0\) . So \(m_i=0\) for every subinterval of every partition, all lower sums are \(0\) , and \(L(f)=0\) .

   Upper integral. On \([x_{i-1},x_i]\) the values of \(f\) are \(0\) and the rational numbers of the interval, all at most \(x_i\) ; and rationals of the interval come arbitrarily close to \(x_i\) by density. So \(M_i=x_i\) . Therefore

   \[ U(f,P)=\sum_{i=1}^nx_i\,\Delta x_i , \]

   which is exactly the upper sum \(U(g,P)\) of the function \(g(x)=x\) for the same partition. The two functions have the same upper sums for every partition, hence the same upper integral, and by Exercise 2, \(U(g)=\int_0^1x\,dx=\tfrac12\) . So \(U(f)=\tfrac12\) .

   Since \(L(f)=0<\tfrac12=U(f)\) , the function is not Riemann integrable. It is continuous only at \(0\) ; elsewhere the oscillation on any interval around a point \(x\) is at least \(x\) .

6. Let \(f\) be increasing on \([a,b]\) . Using the formula in the proof of Theorem 7.7, find the smallest number \(n\) of equal pieces that guarantees \(U(f,P_n)-L(f,P_n)<0.01\) for \(f(x)=e^x\) on \([0,1]\) .

   Answer

   \(U-L=(1-0)(e-1)/n\) . This is below \(0.01\) exactly when \(n>100(e-1)\approx 171.83\) . The smallest such \(n\) is \(172\) .

7. Let \(f\) be continuous on \([a,b]\) with \(f(x)\ge 0\) for all \(x\) and \(\int_a^bf=0\) . Prove that \(f(x)=0\) for every \(x\) . Show by example that the conclusion fails if continuity is dropped.

   Solution

   Suppose \(f(c)>0\) for some \(c\in[a,b]\) . By continuity at \(c\) (Exercise 5.3) there is \(\delta>0\) such that \(f(x)>f(c)/2\) for all \(x\in[a,b]\) with \(\lvert x-c\rvert<\delta\) . The set of such \(x\) contains a closed interval \(J=[u,v]\) with \(u<v\) (if \(c\) is an endpoint of \([a,b]\) , take \(J\) on one side of \(c\) ).

   By additivity (Theorem 7.10), \(\int_a^bf=\int_a^uf+\int_u^vf+\int_v^bf\) . The first and third integrals are nonnegative by the order property, because \(f\ge 0\) . On \(J\) we have \(f\ge f(c)/2\) , so \(\int_u^vf\ge(v-u)f(c)/2>0\) . Hence \(\int_a^bf>0\) , contradicting the hypothesis. Therefore \(f(x)=0\) for all \(x\) .

   Without continuity the conclusion fails: the function of Exercise 3 is nonnegative, has integral \(0\) , and is not identically zero. Continuity is what spreads positivity at one point to a whole interval, and only intervals have positive length.

8. Prove the mean value theorem for integrals: if \(f\) is continuous on \([a,b]\) , there is a point \(c\in[a,b]\) with \(\int_a^bf=f(c)\,(b-a)\) .

   Solution

   By the Extreme Value Theorem \(f\) attains a minimum \(m=f(x_{\min})\) and a maximum \(M=f(x_{\max})\) on \([a,b]\) . Since \(m\le f(x)\le M\) for all \(x\) , the order property gives

   \[ m\,(b-a)\le\int_a^bf\le M\,(b-a). \]

   So the number \(\mu=\dfrac{1}{b-a}\displaystyle\int_a^bf\) satisfies \(m\le\mu\le M\) . By Corollary 5.8 the image \(f([a,b])\) is the whole interval \([m,M]\) , so there is \(c\in[a,b]\) with \(f(c)=\mu\) . That is, \(\int_a^bf=f(c)(b-a)\) .

   The number \(\mu\) is the average value of \(f\) , and the theorem says that a continuous function takes its average value. For a discontinuous function it need not: the step function of Example 7.4 has average \(\tfrac12\) on \([0,2]\) and takes only the values \(0\) and \(1\) .

9. Give an example of a function \(f\) on \([0,1]\) such that \(\lvert f\rvert\) is integrable and \(f\) is not, and explain which step of the proof of Theorem 7.11 cannot be reversed.

   Hint

   Take \(f=1\) on the rationals and \(f=-1\) on the irrationals, so \(\lvert f\rvert=1\) . The proof of Theorem 7.11 bounds the oscillation of \(\varphi\circ f\) by a multiple of the oscillation of \(f\) ; there is no inequality in the other direction, and here the oscillation of \(\lvert f\rvert\) is \(0\) on every interval while that of \(f\) is \(2\) .

10. Prove the Cauchy–Schwarz inequality for integrals: for \(f,g\in\mathcal R[a,b]\) , \(\bigl(\int_a^bfg\bigr)^2\le\int_a^bf^2\cdot\int_a^bg^2\) . (Imitate the proof of Lemma 4.2.)

    Hint

    All the functions involved are integrable by Theorem 7.11. For real \(t\) , \(0\le\int_a^b(f-tg)^2=A-2tC+t^2B\) with \(A=\int f^2\) , \(B=\int g^2\) , \(C=\int fg\) . If \(B>0\) take \(t=C/B\) . If \(B=0\) , the inequality \(A-2tC\ge 0\) for all real \(t\) forces \(C=0\) .

11. Find the error: “By the Fundamental Theorem, \(\int_{-1}^{1}x^{-2}\,dx=\bigl[-1/x\bigr]_{-1}^{1}=-1-1=-2\) .” Explain how one can see at once that the answer must be wrong.

    Answer

    The integrand is positive wherever it is defined, so a negative integral is impossible. The function \(x^{-2}\) is unbounded near \(0\) and undefined at \(0\) , so it is not Riemann integrable on \([-1,1]\) ; and \(-1/x\) is not continuous on \([-1,1]\) , so it is not an antiderivative in the sense required by Theorem 7.15.

12. On \([0,2]\) let \(f(t)=t\) for \(0\le t\le 1\) and \(f(t)=2\) for \(1<t\le 2\) . Find \(F(x)=\int_0^xf\) explicitly, determine where \(F\) is differentiable, and compare with what Theorem 7.14 guarantees.

    Solution

    On \([0,1]\) . Here \(f(t)=t\) , which is continuous with antiderivative \(t^2/2\) , so by Theorem 7.15, \(F(x)=x^2/2\) for \(0\le x\le 1\) . In particular \(F(1)=\tfrac12\) .

    On \((1,2]\) . By additivity \(F(x)=F(1)+\int_1^xf\) . On \([1,x]\) the function \(f\) equals the constant \(2\) except at the single point \(1\) , and changing a function at one point does not change its integral (Exercise 4). So \(\int_1^xf=2(x-1)\) and

    \[ F(x)=\tfrac12+2(x-1)=2x-\tfrac32\qquad(1\le x\le 2). \]

    For instance \(F(2)=\tfrac52\) .

    Differentiability. For \(x<1\) , \(F'(x)=x=f(x)\) ; for \(x>1\) , \(F'(x)=2=f(x)\) . At \(x=1\) the left difference quotient is \((x^2/2-\tfrac12)/(x-1)=(x+1)/2\to 1\) and the right difference quotient is \(2\) . They differ, so \(F\) is not differentiable at \(1\) .

    Comparison. Theorem 7.14 guarantees that \(F\) is continuous everywhere, which it is (\(F(1)=\tfrac12\) from both formulas), and differentiable with \(F'=f\) wherever \(f\) is continuous, that is, at every point except \(1\) . At the jump of \(f\) the theorem promises nothing, and \(F\) has a corner there whose two slopes are the two one-sided limits of \(f\) .

13. Let \(f\) be continuous on \([0,\infty)\) . Find the derivative of \(G(x)=\int_0^{x^2}f(t)\,dt\) , justifying each step, and evaluate it for \(f(t)=\sqrt{1+t^3}\) .

    Answer

    Write \(G=\Phi\circ\varphi\) with \(\Phi(u)=\int_0^uf\) and \(\varphi(x)=x^2\) . Since \(f\) is continuous, \(\Phi'=f\) by Theorem 7.14, and the chain rule gives \(G'(x)=f(x^2)\cdot 2x\) . For \(f(t)=\sqrt{1+t^3}\) : \(G'(x)=2x\sqrt{1+x^6}\) .

14. Recognise each limit as an integral and evaluate it: (a) \(\lim_{n\to\infty}\dfrac1n\sum_{k=1}^n\sqrt{k/n}\) ; (b) \(\lim_{n\to\infty}\sum_{k=1}^n\dfrac{k}{n^2+k^2}\) .

    Answer

    (a) The sum is a Riemann sum of \(\sqrt x\) on \([0,1]\) with right endpoints, so the limit is \(\int_0^1\sqrt x\,dx=\tfrac23\) . (b) Write the term as \(\dfrac1n\cdot\dfrac{k/n}{1+(k/n)^2}\) ; the limit is \(\int_0^1\dfrac{x}{1+x^2}\,dx=\tfrac12\log 2\approx 0.3466\) .

15. Prove directly from Definition 7.17, by bounding the integrand, that \(1-\dfrac1x\le\log x\le x-1\) for every \(x>0\) .

    Solution

    For \(x\ge 1\) . On \([1,x]\) the integrand satisfies \(\dfrac1x\le\dfrac1t\le 1\) . By the order property of the integral,

    \[ \frac{x-1}{x}=\int_1^x\frac{dt}{x}\le\int_1^x\frac{dt}{t}\le\int_1^x1\,dt=x-1 , \]

    that is, \(1-\dfrac1x\le\log x\le x-1\) .

    For \(0<x<1\) . By the convention on reversed limits, \(\log x=-\int_x^1dt/t\) . On \([x,1]\) we have \(1\le\dfrac1t\le\dfrac1x\) , so

    \[ 1-x\le\int_x^1\frac{dt}{t}\le\frac{1-x}{x}. \]

    Multiplying by \(-1\) reverses the inequalities: \(-\dfrac{1-x}{x}\le\log x\le-(1-x)\) , which is again \(1-\dfrac1x\le\log x\le x-1\) .

    So the inequality holds for all \(x>0\) , with equality only at \(x=1\) . For example at \(x=2\) it reads \(0.5\le\log 2\le 1\) , and at \(x=\tfrac12\) it reads \(-1\le\log\tfrac12\le-0.5\) . Replacing \(x\) by \(1+x\) recovers Exercise 6.7, this time without the Mean Value Theorem.

16. Prove that \(\log\) is uniformly continuous on \([1,\infty)\) and is not uniformly continuous on \((0,1]\) .

    Hint

    On \([1,\infty)\) the derivative \(1/x\) is bounded by \(1\) , so \(\log\) is Lipschitz there by Corollary 6.8(c). On \((0,1]\) it is unbounded, since \(\log(2^{-n})=-n\log 2\) , whereas a uniformly continuous function on a bounded set is bounded (Exercise 5.18).

17. Let \(\gamma_n=1+\tfrac12+\cdots+\tfrac1n-\log n\) . Use the inequalities in the proof of Theorem 7.19 to show that \((\gamma_n)\) is decreasing and bounded below by \(0\) , so that it converges. (Its limit is Euler’s constant.)

    Solution

    Apply the inequality \(f(n+1)\le\int_n^{n+1}f\le f(n)\) from the proof of Theorem 7.19 to the decreasing function \(f(x)=1/x\) ; by Definition 7.17 and additivity, \(\int_n^{n+1}dt/t=\log(n+1)-\log n\) .

    Decreasing. From the definition of \(\gamma_n\) ,

    \[ \gamma_n-\gamma_{n+1}=\log(n+1)-\log n-\frac{1}{n+1}=\int_n^{n+1}\frac{dt}{t}-\frac{1}{n+1}\ge 0 . \]

    Bounded below. Summing \(\int_k^{k+1}dt/t\le 1/k\) over \(k=1,\dots,n-1\) gives \(\log n\le 1+\tfrac12+\cdots+\tfrac1{n-1}\) . Hence

    \[ \gamma_n=\Bigl(1+\tfrac12+\cdots+\tfrac1{n-1}-\log n\Bigr)+\frac1n\ge\frac1n>0 . \]

    A decreasing sequence that is bounded below converges, by the monotone convergence theorem. So \(\gamma=\lim\gamma_n\) exists and \(0\le\gamma\le\gamma_1=1\) .

    Numerically \(\gamma_1=1\) , \(\gamma_{10}\approx 0.6264\) , \(\gamma_{100}\approx 0.5822\) , \(\gamma_{10000}\approx 0.57727\) ; the limit is \(0.5772\) to four places. The harmonic series diverges, but it does so exactly like \(\log n\) , up to a constant.

18. Let \(T\) be Thomae’s function and \(F(x)=\int_0^xT\) . Find \(F\) , and use it to show that the function \(F\) of Theorem 7.14 can be differentiable at a point \(c\) where \(f\) is discontinuous, with \(F'(c)\ne f(c)\) .

    Answer

    The argument of Example 7.3 applies on every interval \([0,x]\) , so \(F(x)=0\) for all \(x\in[0,1]\) . Thus \(F'(c)=0\) at every point. At a rational \(c=p/q\) the function \(T\) is discontinuous and \(T(c)=1/q\ne 0=F'(c)\) . Continuity of \(f\) at \(c\) is sufficient for \(F'(c)=f(c)\) , and without it \(F'(c)\) may exist and differ from \(f(c)\) .

19. Let \(f\) be a bounded function on \([a,b]\) and suppose there is a number \(A\) such that for every \(\varepsilon>0\) there is \(\delta>0\) with \(\lvert S(f,P,t)-A\rvert<\varepsilon\) for all tagged partitions of mesh less than \(\delta\) . Prove that \(f\in\mathcal R[a,b]\) and \(\int_a^bf=A\) .

    Hint

    Given \(\varepsilon\) , take \(\delta\) as in the hypothesis and fix one partition \(P\) of mesh less than \(\delta\) . Choosing tags with \(f(t_i)>M_i-\varepsilon/(b-a)\) produces a Riemann sum exceeding \(U(f,P)-\varepsilon\) , so \(U(f,P)<A+2\varepsilon\) ; similarly \(L(f,P)>A-2\varepsilon\) . Conclude with Theorem 7.5 and the fact that the integral lies between \(L(f,P)\) and \(U(f,P)\) .

20. Let \(f\) be a bounded function on \([0,1]\) that is continuous at every point not in the Cantor set \(C\) of Example 4.3. Prove that \(f\) is Riemann integrable.

    Solution

    Let \(\lvert f\rvert\le M\) with \(M>0\) and let \(\varepsilon>0\) . We imitate the proof of Theorem 7.8, with the Cantor set playing the role of the finite set of discontinuities.

    Covering the bad set. By Example 4.3, \(C\subseteq C_n\) , where \(C_n\) is a union of \(2^n\) closed intervals of total length \((2/3)^n\) . Choose \(n\) with \((2/3)^n<\varepsilon/(8M)\) , and enlarge each of the \(2^n\) intervals \([u,v]\) to an open interval \((u-\eta,v+\eta)\) , with \(\eta>0\) so small that \(2^{n+1}\eta<\varepsilon/(8M)\) . We obtain open intervals \(V_1,\dots,V_r\) , \(r=2^n\) , that cover \(C\) and have total length less than \(\varepsilon/(4M)\) .

    The good set. The set \([0,1]\setminus(V_1\cup\dots\cup V_r)\) is a union of finitely many closed intervals \(J_1,\dots,J_s\) (possibly degenerate ones, which we discard). No point of any \(J_k\) lies in \(C\) , so \(f\) is continuous at every point of each \(J_k\) , and hence integrable on it by Theorem 7.6. Choose partitions \(Q_k\) of \(J_k\) with \(U(f,Q_k)-L(f,Q_k)<\varepsilon/(2s)\) .

    Assembling. Let \(P\) be the partition of \([0,1]\) formed by all points of the \(Q_k\) together with the endpoints of the intervals \(J_k\) . Its subintervals are of two kinds: those inside some \(J_k\) , which contribute less than \(s\cdot\varepsilon/(2s)=\varepsilon/2\) to \(U(f,P)-L(f,P)\) ; and the gaps between consecutive \(J_k\) , each contained in the closure of a union of sets \(V_j\) , with total length less than \(\varepsilon/(4M)\) and oscillation at most \(2M\) , contributing less than \(\varepsilon/2\) . Hence \(U(f,P)-L(f,P)<\varepsilon\) , and \(f\) is integrable by Theorem 7.5.

    Only two facts about \(C\) were used: it can be covered by finitely many intervals of arbitrarily small total length, and away from it \(f\) is continuous. The indicator function of \(C\) is an example: it is discontinuous at every point of the uncountable set \(C\) and is integrable, with integral \(0\) .

21. Evaluate, justifying the use of Theorem 7.16: (a) \(\int_0^1xe^x\,dx\) ; (b) \(\int_0^1\dfrac{2x}{1+x^2}\,dx\) .

    Answer

    (a) Parts with \(F(x)=x\) , \(G(x)=e^x\) : \(\int_0^1xe^x\,dx=\bigl[xe^x\bigr]_0^1-\int_0^1e^x\,dx=e-(e-1)=1\) . All four functions are continuous, so Theorem 7.16(a) applies. (b) Substitution with \(\varphi(t)=1+t^2\) , which has a continuous derivative, and \(f(x)=1/x\) , continuous on \([1,2]\) : the integral is \(\int_1^2dx/x=\log 2\) .

22. For which real \(p\) does \(\int_1^\infty x^{-p}\,dx\) converge? For which \(p\) does the limit \(\lim_{a\to 0^+}\int_a^1x^{-p}\,dx\) exist? Is there a \(p\) for which both hold?

    Answer

    \(\int_1^bx^{-p}\,dx=(b^{1-p}-1)/(1-p)\) for \(p\ne 1\) and \(\log b\) for \(p=1\) ; it has a limit as \(b\to\infty\) exactly when \(p>1\) . Similarly \(\int_a^1x^{-p}\,dx\) has a limit as \(a\to 0^+\) exactly when \(p<1\) . No \(p\) satisfies both: a power of \(x\) is never integrable at both ends of \((0,\infty)\) .

23. Give bounded functions \(f\) and \(g\) on \([0,1]\) for which the upper integrals satisfy \(U(f+g)<U(f)+U(g)\) , and explain why this does not contradict Theorem 7.9.

    Answer

    Let \(f=D\) be Dirichlet’s function and \(g=1-D\) . Then \(U(f)=U(g)=1\) , while \(f+g=1\) has \(U(f+g)=1<2\) . Theorem 7.9 assumes that \(f\) and \(g\) are integrable, and neither is; for integrable functions the upper integrals are the integrals and add. In general only \(U(f+g)\le U(f)+U(g)\) holds, the integral form of Example 1.5.