The fundamental theorem of calculus has two halves. The first says that differentiating an integral returns the integrand: if \(f\) is continuous and \(F(x)=\int_a^xf(t)\,dt\) , then \(F'(x)=f(x)\) . The second says that integrating a derivative returns the function: if \(F\) is continuously differentiable, then \(\int_a^bF'(t)\,dt=F(b)-F(a)\) . Both are statements about the Riemann integral and nicely behaved functions. What becomes of them for the Lebesgue integral?

The first half survives almost intact. For every integrable \(f\) , the derivative of \(\int_a^xf\) exists and equals \(f(x)\) at almost every \(x\) . This is the Lebesgue differentiation theorem. It is harder than it looks: the statement involves uncountably many intervals around each of uncountably many points, and the exceptional null set has to be controlled for all of them at once. The tool that does this, the maximal function, is one of the most useful inventions of twentieth-century analysis.

The second half does not survive without a new hypothesis. There is a continuous nondecreasing function on \([0,1]\) that rises from \(0\) to \(1\) and whose derivative exists and is \(0\) almost everywhere. The exact class of functions that are integrals of their derivatives is identified with the help of the Radon–Nikodym theorem.

The chapter ends by stepping back. Ten chapters have produced about a dozen theorems that license an interchange or an identification, each with its own hypotheses. The last section organizes them by the question they answer, because in practice the difficulty is rarely in applying a theorem and usually in knowing which one applies.

10.1 Averages and the maximal function

A function on \(\mathbb R\) is locally integrable if it is integrable on every bounded interval. For such \(f\) , \(x\in\mathbb R\) , and \(r>0\) , write

for the average of \(f\) over the interval of radius \(r\) about \(x\) . If \(f\) is continuous at \(x\) , then \(A_rf(x)\to f(x)\) as \(r\to0\) , because \(f\) is within \(\varepsilon\) of \(f(x)\) on the whole interval once \(r\) is small. The aim is the same conclusion for almost every \(x\) with no continuity assumed.

The maximal function records the worst average of \(\lvert f\rvert\) around \(x\) , at any scale. For each \(t\) the set \(\{Mf>t\}\) is open: if the average of \(\lvert f\rvert\) over \((x-r,x+r)\) exceeds \(t\) , then for \(y\) close to \(x\) the slightly larger interval of radius \(r+\lvert x-y\rvert\) about \(y\) contains \((x-r,x+r)\) and its average still exceeds \(t\) . In particular \(Mf\) is Borel measurable.

10.2 The Lebesgue differentiation theorem

A point \(x\) at which the displayed limit holds is a Lebesgue point of \(f\) . The second statement follows from the first because \(\lvert A_rf(x)-f(x)\rvert\le\frac1{2r}\int_{x-r}^{x+r}\lvert f(y)-f(x)\rvert\,dy\) .

The architecture is that of every theorem on almost-everywhere convergence. The result is easy for a dense class, here the continuous functions. A maximal inequality bounds the size of the set where things can go wrong, in terms of the norm of the function. The norm of the difference between a general \(f\) and a member of the dense class can be made arbitrarily small. Without the maximal inequality, density alone would be useless: an error that is small in \(L^1\) could still have bad averages on a large set.

So a measurable set has no fuzzy boundary in the sense of measure: almost every point is either a point where the set fills up nearly all of each small interval, or one where it fills up nearly none. A set that occupied exactly half of every interval would contradict this, and no such measurable set exists.

Example 10.2: Averages at a jump

Let \(f=\mathbf 1_{[0,1]}\) . At a point \(x\in(0,1)\) the average \(A_rf(x)\) equals \(1\) as soon as \(r<\min(x,1-x)\) , and at a point outside \([0,1]\) it is eventually \(0\) . These are Lebesgue points. At \(x=0\) , for \(r<1\) ,

\[ A_rf(0)=\frac1{2r}\int_0^r1\,dy=\frac12 . \]

The averages converge, but not to \(f(0)=1\) . Could a different value at the origin rescue the point? Changing \(f(0)\) to \(c\) does not change the equivalence class of \(f\) , and

\[ \frac1{2r}\int_{-r}^r\lvert f(y)-c\rvert\,dy=\frac{\lvert0-c\rvert+\lvert1-c\rvert}2\ge\frac12\qquad\text{for every } c . \]

So \(0\) is not a Lebesgue point of any function equal to \(f\) a.e., and likewise \(1\) . The exceptional set \(\{0,1\}\) is null, as the theorem requires. Figure 10.1 shows the averages for three radii together with the maximal function of Example 10.1.

10.3 Differentiating a measure, and functions that are integrals of their derivatives

By Theorem 2.14 a finite Borel measure \(\nu\) on \(\mathbb R\) is the Lebesgue–Stieltjes measure of a bounded nondecreasing right-continuous function \(F\) , and by Theorem 9.10 it splits as \(\nu=\nu_a+\nu_s\) with \(\nu_a\) having a density \(f\) and \(\nu_s\perp\lambda\) . The ratio \(\nu((x-r,x+r))/(2r)\) is a symmetric difference quotient of \(F\) . Its limit picks out the density and ignores the singular part.

On the line, then, the Radon–Nikodym derivative of the absolutely continuous part of a measure really is a derivative: the limit of mass per unit length over shrinking intervals. Which functions \(F\) correspond to measures with no singular part?

With a single interval this is uniform continuity. The force of the definition is in allowing many small intervals at once. Lipschitz functions are absolutely continuous, with \(\delta=\varepsilon/L\) .

The theorem identifies three things: absolutely continuous functions, indefinite integrals of integrable functions, and, for nondecreasing functions, Stieltjes measures that are absolutely continuous with respect to length. The two uses of the phrase “absolutely continuous”, for functions and for measures, name the same property.

10.4 Choosing the theorem

The theorems of this book are tools for a small number of recurring tasks, and the first step in any problem is to recognize the task.

Reading across a row gives the discipline that the subject demands: name the task, name the theorem, write down each hypothesis and the function or set that verifies it, and know the example that shows the hypothesis is needed. In practice a short list of questions sorts most problems. What two operations are being interchanged? Are the functions nonnegative, in which case the monotone convergence theorem or Tonelli’s theorem applies with nothing to check? If not, what is the dominator, or what is the integral of the absolute value? Are the measures \(\sigma\) -finite? And is the conclusion claimed everywhere, or only almost everywhere?

Exercises

1. Compute the averages \(A_rf(0)\) for \(f(x)=\lvert x\rvert\) and for \(f=\mathbf 1_{[0,\infty)}\) , find their limits as \(r\to0\) , and say in each case whether \(0\) is a Lebesgue point.

   Answer

   For \(f(x)=\lvert x\rvert\) : \(A_rf(0)=\frac1{2r}\int_{-r}^r\lvert y\rvert\,dy=\frac r2\to0=f(0)\) , and \(0\) is a Lebesgue point, as at any point of continuity. For \(f=\mathbf 1_{[0,\infty)}\) : \(A_rf(0)=\tfrac12\) for every \(r\) , which is not \(f(0)=1\) ; and \(\frac1{2r}\int_{-r}^r\lvert f(y)-c\rvert\,dy=\tfrac12(\lvert c\rvert+\lvert1-c\rvert)\ge\tfrac12\) for every \(c\) , so \(0\) is not a Lebesgue point of any function equal to \(f\) a.e.

2. Prove that every point of continuity of a locally integrable function is a Lebesgue point, and give a function with a Lebesgue point at which it is not continuous.

   Hint

   If \(f\) is continuous at \(x\) , then \(\lvert f(y)-f(x)\rvert<\varepsilon\) for \(\lvert y-x\rvert<\delta\) , so the average of \(\lvert f(y)-f(x)\rvert\) over \((x-r,x+r)\) is below \(\varepsilon\) for \(r<\delta\) . For the example take \(f=\mathbf 1_{\mathbb Q}\) : it is discontinuous everywhere, and every irrational \(x\) is a Lebesgue point, because \(f=0=f(x)\) almost everywhere.

3. For \(h>0\) let \(f_h=\tfrac1h\mathbf 1_{[0,h]}\) . Compute \(Mf_h\) and \(\lambda(\{Mf_h>t\})\) for \(0<t<\tfrac1{2h}\) , and deduce that the constant \(3\) in Theorem 10.3 cannot be replaced by any number smaller than \(1\) .

   Solution

   Since \(f_h\ge0\) , \(Mf_h(x)\) is the supremum over \(r\) of \(\frac1{2r}\cdot\frac1h\lambda\bigl([0,h]\cap(x-r,x+r)\bigr)\) .

   Inside. For \(0<x<h\) , small intervals about \(x\) lie in \([0,h]\) and give the average \(\tfrac1h\) , which is the largest possible value. So \(Mf_h=\tfrac1h\) on \((0,h)\) .

   To the right. Let \(x\ge h\) . The average is \(0\) for \(r\le x-h\) , increases for \(x-h<r\le x\) , where the interval captures more and more of \([0,h]\) , and equals \(\frac1{2r}\) for \(r\ge x\) , when it captures all of it. The maximum is at \(r=x\) : \(Mf_h(x)=\dfrac1{2x}\) . By symmetry about \(\tfrac h2\) , \(Mf_h(x)=\dfrac1{2(h-x)}\) for \(x\le0\) .

   The level sets. Let \(0<t<\tfrac1{2h}\) . On the right, \(\tfrac1{2x}>t\) means \(x<\tfrac1{2t}\) , and \(\tfrac1{2t}>h\) . On the left, \(x>h-\tfrac1{2t}\) . In between \(Mf_h\ge\tfrac1{2h}>t\) . So

   \[ \{Mf_h>t\}=\Bigl(h-\frac1{2t},\ \frac1{2t}\Bigr),\qquad\lambda(\{Mf_h>t\})=\frac1t-h . \]

   The constant. Every \(f_h\) has \(\lVert f_h\rVert_1=1\) . If the maximal inequality held with a constant \(C\) in place of \(3\) , then \(\tfrac1t-h\le\tfrac Ct\) for all small \(h\) , and letting \(h\to0\) gives \(C\ge1\) . For \(h=1\) the formula is the one in Example 10.1. As \(h\to0\) the functions \(f_h\) concentrate to a unit point mass, and the limit of their maximal functions, \(\tfrac1{2\lvert x\rvert}\) , lies just outside \(L^1\) in the precise sense of Exercise 7.17.

4. Apply the selection procedure in the proof of Lemma 10.2 to the intervals \((0,4)\) , \((3,5)\) , \((4.5,9)\) , \((8,10)\) , \((1,2)\) . List the chosen intervals in order and verify the inequality of the lemma.

   Solution

   The lengths are \(4\) , \(2\) , \(4.5\) , \(2\) , and \(1\) .

   First choice. The longest interval is \((4.5,9)\) . It meets \((3,5)\) , since both contain \(4.8\) , and \((8,10)\) ; it does not meet \((0,4)\) or \((1,2)\) . Discard \((4.5,9)\) , \((3,5)\) , and \((8,10)\) .

   Second choice. Of the remaining intervals \((0,4)\) and \((1,2)\) , the longer is \((0,4)\) . It meets \((1,2)\) , which is discarded with it. Nothing is left.

   The chosen intervals are \((4.5,9)\) and \((0,4)\) , in that order. They are disjoint, with total length \(4.5+4=8.5\) . The union of all five intervals is \((0,5)\cup(4.5,10)=(0,10)\) , of length \(10\) . The lemma asserts \(10\le3\times8.5=25.5\) , which holds with room to spare.

   To see the mechanism, triple the chosen intervals about their centres: \((4.5,9)\) becomes \((0,13.5)\) and \((0,4)\) becomes \((-4,8)\) . Each discarded interval lies in the tripled copy of the interval that eliminated it: \((3,5)\) and \((8,10)\) in \((0,13.5)\) , and \((1,2)\) in \((-4,8)\) . The factor \(3\) is wasteful here, but it is what the worst case requires: a discarded interval as long as the chosen one and touching it at one end.

5. Prove in detail that \(\{Mf>t\}\) is an open set for every locally integrable \(f\) and every \(t\) , and deduce that \(Mf\) is Borel measurable.

   Hint

   Suppose \(Mf(x)>t\) and choose \(r\) with \(\frac1{2r}\int_{x-r}^{x+r}\lvert f\rvert\,d\lambda>t\) . For \(\lvert y-x\rvert<\eta\) the interval of radius \(r+\eta\) about \(y\) contains \((x-r,x+r)\) , so the average of \(\lvert f\rvert\) over it is at least \(\frac{r}{r+\eta}\) times the average over \((x-r,x+r)\) , which exceeds \(t\) for small \(\eta\) . Open sets are Borel, so \(Mf\) is Borel measurable by Corollary 3.3.

6. Let \(f(x)=\sin(1/x)\) for \(x\ne0\) and \(f(0)=c\) . Show that \(A_rf(0)\to0\) as \(r\to0\) , and determine whether \(0\) is a Lebesgue point for some choice of \(c\) .

   Hint

   The function is odd, so \(\int_{-r}^rf\,d\lambda=0\) and \(A_rf(0)=0\) for every \(r\) . For the Lebesgue point condition substitute \(u=1/y\) : \(\frac1{2r}\int_{-r}^r\lvert\sin(1/y)-c\rvert\,dy\) becomes an average of \(\lvert\sin u-c\rvert\) and \(\lvert\sin u+c\rvert\) against the weight \(u^{-2}\) on \((1/r,\infty)\) , which tends to the mean of these functions over a period. That mean is positive for every \(c\) . So \(0\) is not a Lebesgue point for any \(c\) , although the averages of \(f\) itself converge.

7. Prove that there is no Lebesgue measurable set \(E\subseteq\mathbb R\) with \(\lambda(E\cap I)=\tfrac12\lambda(I)\) for every bounded interval \(I\) .

   Solution

   Suppose such a set \(E\) existed. Then for every \(x\in\mathbb R\) and every \(r>0\) ,

   \[ \frac{\lambda(E\cap(x-r,x+r))}{2r}=\frac12 . \]

   So the limit as \(r\to0\) exists at every point and equals \(\tfrac12\) . By Corollary 10.5 the limit equals \(1\) at almost every point of \(E\) . Since it is \(\tfrac12\) at every point, the set of points of \(E\) at which the corollary’s conclusion holds is empty, and therefore \(\lambda(E)=0\) . By the second half of the corollary the limit equals \(0\) at almost every point of the complement, and by the same reasoning \(\lambda(\mathbb R\setminus E)=0\) . Then \(\lambda(\mathbb R)=\lambda(E)+\lambda(\mathbb R\setminus E)=0\) , which is absurd.

   A measurable set cannot be spread evenly at all scales. Sets that meet every interval in positive measure without filling any interval do exist; what is impossible is a constant proportion. The proof used measurability in an essential way, through the differentiation theorem, and for non-measurable sets the outer measure can indeed behave more uniformly.

8. A student claims: “\(F\) is continuous and nondecreasing on \([0,1]\) and \(F'=0\) almost everywhere, so \(F\) is constant.” Give a counterexample, name the missing hypothesis, and state the correct theorem.

   Answer

   The Cantor function of Example 10.4 is continuous, nondecreasing, has derivative \(0\) almost everywhere, and is not constant. The missing hypothesis is absolute continuity. The correct statement is Theorem 10.10: an absolutely continuous function is the integral of its derivative, so if in addition \(F'=0\) a.e. then \(F\) is constant.

9. Prove that a Lipschitz function on \([a,b]\) is absolutely continuous, and that an absolutely continuous function whose derivative vanishes almost everywhere is constant.

   Solution

   Lipschitz implies absolutely continuous. Suppose \(\lvert F(x)-F(y)\rvert\le L\lvert x-y\rvert\) on \([a,b]\) , with \(L>0\) . Given \(\varepsilon>0\) put \(\delta=\varepsilon/L\) . For disjoint intervals \((a_k,b_k)\) with \(\sum_k(b_k-a_k)<\delta\) ,

   \[ \sum_k\lvert F(b_k)-F(a_k)\rvert\le L\sum_k(b_k-a_k)<L\delta=\varepsilon . \]

   The disjointness of the intervals was not even needed.

   Derivative zero implies constant. Let \(F\) be absolutely continuous on \([a,b]\) with \(F'=0\) almost everywhere. By Theorem 10.10 there is an integrable \(f\) with \(F(x)=F(a)+\int_a^xf\,d\lambda\) for all \(x\) , and \(F'=f\) a.e. Hence \(f=0\) a.e., and by Proposition 4.7(d), applied to \(f^+\) and \(f^-\) , \(\int_a^xf\,d\lambda=0\) for every \(x\) . So \(F(x)=F(a)\) on \([a,b]\) .

   The Cantor function shows that the conclusion fails for functions that are merely continuous and of bounded variation. The argument here relied on the implication from (a) to (b) in Theorem 10.10, whose proof in the manual is complete for nondecreasing \(F\) and quotes one fact, the absolute continuity of the total variation function, for the general case.

10. Compute the values of the Cantor function at \(\tfrac14\) , \(\tfrac13\) , and \(\tfrac79\) from the ternary expansions of these points.

    Solution

    The rule is: write the point in base \(3\) using only the digits \(0\) and \(2\) if possible, halve each digit, and read the result in base \(2\) . If the point lies in a removed interval, use the endpoint of that interval.

    The point \(\tfrac14\) . By Exercise 2.6, \(\tfrac14=0.020202\ldots\) in base \(3\) . Halving the digits gives \(0.010101\ldots\) in base \(2\) , which is \(\sum_{k\ge1}4^{-k}=\tfrac13\) . So \(c(\tfrac14)=\tfrac13\) .

    The point \(\tfrac13\) . In base \(3\) , \(\tfrac13=0.1=0.0222\ldots\) , and the second expansion uses only \(0\) and \(2\) . Halving gives \(0.0111\ldots\) in base \(2\) , which is \(\tfrac12\) . So \(c(\tfrac13)=\tfrac12\) , the value on the first removed interval, of which \(\tfrac13\) is the left endpoint.

    The point \(\tfrac79\) . In base \(3\) , \(\tfrac79=0.21=0.20222\ldots\) . Halving gives \(0.10111\ldots\) in base \(2\) , which is \(\tfrac12+\tfrac14=\tfrac34\) . So \(c(\tfrac79)=\tfrac34\) , the value on the removed interval \((\tfrac79,\tfrac89)\) .

    The three values \(\tfrac13,\tfrac12,\tfrac34\) are increasing with the points \(\tfrac14<\tfrac13<\tfrac79\) , as they must be. The first is not a dyadic rational, which reflects the fact that \(\tfrac14\) is a point of the Cantor set that is not an endpoint of any removed interval: on the removed intervals \(c\) takes only dyadic values.

11. Using the symmetry \(c(1-x)=1-c(x)\) of the Cantor function, compute \(\int_0^1c\,d\lambda\) , and compute \(\int x\,d\mu_c(x)\) for the Cantor measure \(\mu_c\) .

    Solution

    The integral of \(c\) . The substitution \(x\mapsto1-x\) preserves Lebesgue measure on \([0,1]\) (Proposition 2.9 and Exercise 4.13), so

    \[ \int_0^1c(x)\,dx=\int_0^1c(1-x)\,dx=\int_0^1\bigl(1-c(x)\bigr)\,dx=1-\int_0^1c(x)\,dx, \]

    whence \(\int_0^1c\,d\lambda=\tfrac12\) .

    The mean of the Cantor measure. Let \(\rho\) be the image of \(\mu_c\) under \(x\mapsto1-x\) , that is, \(\rho(B)=\mu_c(1-B)\) . For \(0\le a<b\le1\) , using that \(\mu_c\) has no atoms,

    \[ \rho((a,b])=\mu_c([1-b,1-a))=c(1-a)-c(1-b)=\bigl(1-c(a)\bigr)-\bigl(1-c(b)\bigr)=c(b)-c(a)=\mu_c((a,b]). \]

    The two measures agree on half-open intervals, so \(\rho=\mu_c\) by Theorem 1.17. Therefore \(\int x\,d\mu_c=\int(1-x)\,d\mu_c=1-\int x\,d\mu_c\) , and \(\int x\,d\mu_c(x)=\tfrac12\) .

    The two answers are the same number for a reason: for a continuous nondecreasing \(F\) on \([0,1]\) with \(F(0)=0\) and \(F(1)=1\) , Exercise 6.15 gives \(\int x\,d\mu_F+\int F\,d\lambda=1\) . Either symmetry argument therefore yields the other result. Notice that \(\int_0^1c\,d\lambda=\tfrac12\) is an honest Lebesgue integral of a continuous function, while \(\int_0^1c'\,d\lambda=0\) ; nothing is wrong with integrating \(c\) , only with recovering it from its derivative.

12. Let \(\nu\) be the measure on \([0,1]\) with \(\nu(E)=\int_E2x\,dx+\delta_{1/2}(E)\) . Determine \(\lim_{r\to0}\nu((x-r,x+r))/(2r)\) at every \(x\in(0,1)\) and relate the answer to Theorem 10.8.

    Answer

    For \(x\ne\tfrac12\) and \(r<\lvert x-\tfrac12\rvert\) the point mass is not in the interval, and the ratio is \(\frac1{2r}\int_{x-r}^{x+r}2y\,dy=2x\) . At \(x=\tfrac12\) the ratio is \(1+\tfrac1{2r}\to\infty\) . So the limit is the density \(2x\) of the absolutely continuous part except at the single point carrying the singular part, a \(\lambda\) -null set, as Theorem 10.8 states.

13. Let \(F\) and \(G\) be absolutely continuous on \([a,b]\) . Prove that \(FG\) is absolutely continuous and deduce the integration by parts formula \(\int_a^bF'G\,d\lambda+\int_a^bFG'\,d\lambda=F(b)G(b)-F(a)G(a)\) .

    Hint

    Both functions are bounded, say by \(C\) . From \(F(y)G(y)-F(x)G(x)=F(y)(G(y)-G(x))+G(x)(F(y)-F(x))\) , the sum of the increments of \(FG\) over disjoint intervals is at most \(C\) times the sum of the increments of \(F\) plus those of \(G\) . Then apply Theorem 10.10 to \(FG\) , whose derivative is \(F'G+FG'\) at every point where both are differentiable.

14. Evaluate \(\displaystyle\lim_{n\to\infty}\int_0^1\frac{nx^n}{1+x}\,dx\) . Explain why dominated convergence cannot be applied directly, and integrate by parts to reach a form in which it can.

    Solution

    Why not directly. The integrand \(\frac{nx^n}{1+x}\) tends to \(0\) for every \(x\in[0,1)\) . If a dominator existed, the integrals would tend to \(0\) . But \(nx^n\) is the sequence of Example 5.7, whose mass \(\frac n{n+1}\) concentrates at \(x=1\) , and near \(x=1\) the factor \(\frac1{1+x}\) is close to \(\tfrac12\) ; so one expects the limit \(\tfrac12\) , not \(0\) . Hence there is no dominator.

    Integration by parts. Both factors are continuously differentiable on \([0,1]\) , so with the antiderivative \(\frac{n}{n+1}x^{n+1}\) of \(nx^n\) ,

    \[ \int_0^1\frac{nx^n}{1+x}\,dx=\frac{n}{n+1}\Bigl[\frac{x^{n+1}}{1+x}\Bigr]_0^1+\frac{n}{n+1}\int_0^1\frac{x^{n+1}}{(1+x)^2}\,dx=\frac{n}{2(n+1)}+\frac{n}{n+1}\int_0^1\frac{x^{n+1}}{(1+x)^2}\,dx . \]

    The remaining integral. Now \(0\le\frac{x^{n+1}}{(1+x)^2}\le1\) on \([0,1]\) , the constant \(1\) is integrable there, and the integrand tends to \(0\) for \(x<1\) . By dominated convergence the last integral tends to \(0\) . Therefore

    \[ \lim_{n\to\infty}\int_0^1\frac{nx^n}{1+x}\,dx=\frac12 . \]

    Numerically the integrals are about \(0.3069\) , \(0.4751\) , and \(0.4975\) for \(n=1,10,100\) . The integration by parts moved the factor \(n\) onto a boundary term, where it does no harm, and left an integrand without it. When a sequence concentrates at a point, the limit of the integrals is a value of the other factor at that point, and the way to prove it is to remove the concentration before taking the limit.

15. Evaluate \(\displaystyle\int_0^1\sum_{n=1}^\infty\frac{(-1)^{n+1}x^n}{n}\,dx\) by integrating term by term, stating which theorem justifies the interchange and verifying its hypothesis.

    Answer

    The sum is \(\ln(1+x)\) , so the integral is \(\int_0^1\ln(1+x)\,dx=2\ln2-1\approx0.3863\) . Term by term: \(\sum_n\frac{(-1)^{n+1}}{n(n+1)}\) , the same number. The interchange is justified by Theorem 5.5, since \(\sum_n\int_0^1\frac{x^n}n\,dx=\sum_n\frac1{n(n+1)}=1<\infty\) . Theorem 4.5 does not apply because the terms alternate in sign.

16. Determine for which real \(a\) the function \((x+y)^{-a}\) is integrable over the unit square \((0,1)^2\) , naming the theorem that reduces the question to one-variable integrals.

    Answer

    The function is nonnegative and continuous on the open square, so by Tonelli’s theorem its integral is the iterated integral, finite or not. For \(a\ne1\) , \(\int_0^1(x+y)^{-a}\,dy=\frac{(x+1)^{1-a}-x^{1-a}}{1-a}\) , which for \(a>1\) behaves like \(\frac{x^{1-a}}{a-1}\) as \(x\to0\) and is integrable in \(x\) exactly when \(a-1<1\) . So the function is integrable exactly for \(a<2\) .

17. Let \(f\in L^1(\mathbb R)\) and \(F(x)=\int_{-\infty}^xf\,d\lambda\) . Prove that \(F\) is absolutely continuous on every bounded interval with \(F'=f\) a.e., and give an \(f\) for which \(F'(0)\) does not exist.

    Hint

    On \([a,b]\) , \(F(x)=F(a)+\int_a^xf\,d\lambda\) , so Theorem 10.10 applies. For the example take \(f=\mathbf 1_{[0,1]}\) : then \(F(x)=0\) for \(x\le0\) and \(F(x)=x\) for \(0\le x\le1\) , with left derivative \(0\) and right derivative \(1\) at the origin.