What is the length of a set of real numbers? For an interval the answer is the difference of its endpoints, and for a finite union of disjoint intervals it is the sum of their lengths. The trouble begins when one asks for a length \(m(E)\) of every subset \(E\) of the line, subject to three requirements that seem beyond dispute: an interval \([a,b]\) has length \(b-a\) ; translating a set does not change its length; and the length of a union of countably many pairwise disjoint sets is the sum of their lengths. No such function exists. Section 2.5 constructs a subset of \([0,1)\) whose countably many disjoint translates, taken modulo one, fill \([0,1)\) exactly, so that its length can be neither zero nor positive.

Something has to give. Weakening the third requirement to finite unions destroys the limit theorems that are the whole reason for wanting a better integral. Strengthening it to uncountable unions is absurd, since an interval is an uncountable disjoint union of points of length zero. The resolution adopted a century ago, and never seriously challenged since, is to keep all three requirements and give up the demand that every set be measured. Length is defined on a family of sets rich enough to contain everything analysis produces and closed under the operations analysis performs.

This chapter answers three questions in the abstract. Which families of sets can serve as the domain of a measure? What follows from countable additivity alone? And when do the values of a measure on a few simple sets determine it everywhere? The construction of length itself is the business of Chapter 2.

1.1 Families closed under countable operations

The smallest \(\sigma\) -algebra on \(X\) is \(\{\emptyset,X\}\) and the largest is the power set \(\mathcal P(X)\) . Between them lie the interesting ones. If \(X\) is uncountable, the family of sets that are countable or have countable complement is a \(\sigma\) -algebra (Exercise 1.5), and it is strictly smaller than the power set.

The definition asks for little, and the rest follows from De Morgan’s laws.

Two limiting operations on sets are used so often that they have names.

Reading the quantifiers, \(x\in\limsup_n A_n\) exactly when for every \(n\) there is a \(k\ge n\) with \(x\in A_k\) , that is, when \(x\) lies in infinitely many of the sets. And \(x\in\liminf_n A_n\) exactly when \(x\) lies in all but finitely many of them. By Proposition 1.2 both sets are measurable when the \(A_n\) are. Always \(\liminf_n A_n\subseteq\limsup_n A_n\) .

Not every natural family passes the test. The open subsets of \(\mathbb R\) are closed under arbitrary unions but not under complements. The subsets of \(\mathbb N\) that are finite or have finite complement form an algebra, but not a \(\sigma\) -algebra: each singleton \(\{2k\}\) belongs to the family, while their union, the set of even numbers, is neither finite nor cofinite. The difference between an algebra and a \(\sigma\) -algebra is exactly the difference between finitely many steps and a limit, and Example 1.4 shows that it is not a technicality.

1.2 Generated sigma-algebras and the Borel sets

On an infinite set there is no hope of listing atoms. The \(\sigma\) -algebra generated by a family is defined from above, as an intersection.

The collection being intersected is nonempty, since \(\mathcal P(X)\) contains \(\mathcal E\) . So \(\sigma(\mathcal E)\) exists, contains \(\mathcal E\) , and is contained in every \(\sigma\) -algebra that contains \(\mathcal E\) . That last property is the only handle the definition provides, and it leads to the most common proof pattern in the subject. To show that every set in \(\sigma(\mathcal E)\) has some property, let \(\mathcal G\) be the family of all sets having the property, check that \(\mathcal G\) is a \(\sigma\) -algebra, and check that \(\mathcal E\subseteq\mathcal G\) . Then \(\sigma(\mathcal E)\subseteq\mathcal G\) . One never works with a typical member of \(\sigma(\mathcal E)\) , because there is no usable description of one. In particular, if \(\mathcal E\subseteq\sigma(\mathcal F)\) and \(\mathcal F\subseteq\sigma(\mathcal E)\) then \(\sigma(\mathcal E)=\sigma(\mathcal F)\) .

Closed sets are Borel, being complements of open sets. So are countable intersections of open sets and countable unions of closed sets. The rationals are a countable union of singletons, hence Borel, and therefore so are the irrationals. Almost any set one can describe is Borel, and for that reason it is useful to have small generating families.

1.3 Measures

The displayed property is countable additivity. Taking \(A_n=\emptyset\) from some index on shows that a measure is also finitely additive. Three examples are worth having at hand before any theory.

Counting measure. On any set \(X\) with \(\mathcal A=\mathcal P(X)\) , let \(\#(A)\) be the number of elements of \(A\) if \(A\) is finite and \(\infty\) otherwise. It is \(\sigma\) -finite exactly when \(X\) is countable.

Point mass. Fix \(x_0\in X\) and let \(\delta_{x_0}(A)=1\) if \(x_0\in A\) and \(0\) otherwise. In a disjoint union at most one set contains \(x_0\) , so both sides of the additivity equation are \(1\) or both are \(0\) .

Weighted sums. Let \(X\) be countable and let \(w:X\to[0,\infty]\) be any function. Put \(\mu_w(A)=\sum_{x\in A}w(x)\) . Countable additivity is the statement that a sum of nonnegative terms indexed by a disjoint union may be computed block by block. That is the rearrangement property of series with nonnegative terms, a result of Real Analysis: such a series has the same sum, finite or infinite, under every reordering and regrouping of its terms. Counting measure is the case of weight \(1\) everywhere, and a point mass is the case of a single nonzero weight.

Lebesgue measure, which assigns to each Borel set its length, is not on this list because proving that it exists takes all of Chapter 2.

1.4 Continuity along monotone sequences

Countable additivity is a statement about limits, and its most usable form says that a measure is continuous along monotone sequences of sets.

Part (a) is called continuity from below and part (b) continuity from above. In (b) it is enough that \(\mu(A_m)<\infty\) for some \(m\) , since the sequence may be started there. The left panel of Figure 1.1 shows the rings \(B_n\) of the proof for the intervals \(A_n=[0,1-2^{-n}]\) .

1.5 Null sets and completion

By countable subadditivity a countable union of null sets is null. An uncountable union need not be: once Lebesgue measure is available, each point of \([0,1]\) will be a null set while \([0,1]\) is not. The phrase “almost everywhere” is always relative to a measure, and when more than one measure is in play the measure is named, as in “\(\mu\) -a.e.”.

In Example 1.2 the null set \(\{4,5,6\}\) has non-measurable subsets. This is an inconvenience, not a paradox, and it can always be removed.

The measure space \((X,\overline{\mathcal A},\overline\mu)\) is the completion of \((X,\mathcal A,\mu)\) . In Example 1.2 the completion makes every subset of \(\{4,5,6\}\) measurable with measure zero, and \(\overline{\mathcal A}\) is the whole power set of \(X\) .

1.6 When two measures agree

A measure on \(\mathcal B(\mathbb R)\) should be determined by the masses it gives to intervals. That is true, but not for the reason one might first guess. Agreement on a generating family is not enough.

Exercises

1. List every \(\sigma\) -algebra on the set \(X=\{1,2,3\}\) , state how many there are, and explain why each one is the family of unions of blocks of a partition of \(X\) .

   Answer

   There are five: \(\{\emptyset,X\}\) ; \(\{\emptyset,\{1\},\{2,3\},X\}\) ; \(\{\emptyset,\{2\},\{1,3\},X\}\) ; \(\{\emptyset,\{3\},\{1,2\},X\}\) ; and \(\mathcal P(X)\) with \(8\) members. They correspond to the five partitions of a three-point set. In a finite \(\sigma\) -algebra the atom of a point \(x\) is the intersection of all members containing \(x\) ; the atoms partition \(X\) , and every member is the union of the atoms of its points.

2. Let \(\mathcal A\) be a \(\sigma\) -algebra on \(X\) . (a) For \(Y\subseteq X\) , prove that \(\{A\cap Y:A\in\mathcal A\}\) is a \(\sigma\) -algebra on \(Y\) . (b) For a map \(f:Z\to X\) , prove that \(\{f^{-1}(A):A\in\mathcal A\}\) is a \(\sigma\) -algebra on \(Z\) .

   Solution

   (a) Write \(\mathcal A_Y=\{A\cap Y:A\in\mathcal A\}\) . It contains \(Y=X\cap Y\) . The complement of \(A\cap Y\) inside \(Y\) is \(Y\setminus(A\cap Y)=A^c\cap Y\) , which is in \(\mathcal A_Y\) because \(A^c\in\mathcal A\) . For a sequence, \(\bigcup_n(A_n\cap Y)=\bigl(\bigcup_nA_n\bigr)\cap Y\) and \(\bigcup_nA_n\in\mathcal A\) . So all three defining conditions hold on the set \(Y\) .

   (b) Write \(\mathcal A_f=\{f^{-1}(A):A\in\mathcal A\}\) . Preimages commute with every set operation: \(f^{-1}(X)=Z\) , \(f^{-1}(A^c)=Z\setminus f^{-1}(A)\) , and \(f^{-1}\bigl(\bigcup_nA_n\bigr)=\bigcup_nf^{-1}(A_n)\) . Each identity is checked by chasing a point: for instance \(z\in f^{-1}(A^c)\) means \(f(z)\notin A\) , which means \(z\notin f^{-1}(A)\) . Hence \(Z\in\mathcal A_f\) , complements of members are members, and countable unions of members are members. Part (a) is the special case in which \(f\) is the inclusion map of \(Y\) into \(X\) .

3. On \(X=\{1,2,3,4,5\}\) find the atoms of the \(\sigma\) -algebra generated by \(\{1,2,3\}\) and \(\{3,4\}\) , and count its members without listing them.

   Answer

   The atoms are \(\{1,2\}\) , \(\{3\}\) , \(\{4\}\) , \(\{5\}\) : the points \(1\) and \(2\) are never separated, \(\{3\}\) is the intersection of the generators, \(\{4\}=\{3,4\}\setminus\{1,2,3\}\) , and \(\{5\}\) is the complement of their union. With four atoms the \(\sigma\) -algebra has \(2^4=16\) members.

4. Prove that an infinite \(\sigma\) -algebra is uncountable, by first showing that it contains an infinite sequence of pairwise disjoint nonempty sets.

   Hint

   Let \(\mathcal A\) be an infinite \(\sigma\) -algebra and pick \(A\in\mathcal A\) other than \(\emptyset\) and \(X\) . A member \(B\) is determined by the pair \((B\cap A,\ B\cap A^c)\) , so one of the two trace \(\sigma\) -algebras of Exercise 1.2(a), on \(A\) or on \(A^c\) , is infinite. Keep the other piece as \(E_1\) and repeat inside the infinite one. This produces pairwise disjoint nonempty sets \(E_1,E_2,\dots\) in \(\mathcal A\) . The unions \(\bigcup_{n\in S}E_n\) for \(S\subseteq\mathbb N\) are distinct members of \(\mathcal A\) , and there are uncountably many subsets \(S\) .

5. Let \(X\) be uncountable and let \(\mathcal C\) be the family of subsets that are countable or have countable complement. Prove that \(\mathcal C\) is a \(\sigma\) -algebra and that \(\mu(A)=0\) for countable \(A\) , \(\mu(A)=1\) for co-countable \(A\) , defines a measure on \(\mathcal C\) .

   Solution

   \(\mathcal C\) is a \(\sigma\) -algebra. The complement of \(X\) is empty, hence countable, so \(X\in\mathcal C\) . The definition is symmetric in \(A\) and \(A^c\) , so \(\mathcal C\) is closed under complements. Let \(A_1,A_2,\dots\in\mathcal C\) . If every \(A_n\) is countable, the union is a countable union of countable sets and is countable. Otherwise some \(A_m\) has countable complement, and \(\bigl(\bigcup_nA_n\bigr)^c\subseteq A_m^c\) is countable. Either way the union is in \(\mathcal C\) .

   \(\mu\) is well defined. No set is both countable and co-countable, for then \(X=A\cup A^c\) would be countable.

   \(\mu\) is countably additive. Let \(A_1,A_2,\dots\in\mathcal C\) be pairwise disjoint. If all are countable, so is the union, and both sides of the additivity equation are \(0\) . Suppose \(A_m\) is co-countable. Every other \(A_n\) is disjoint from \(A_m\) , so \(A_n\subseteq A_m^c\) is countable. Thus exactly one set in the sequence is co-countable, the union is co-countable, and the equation reads \(1=1+0+0+\cdots\) . Finally \(\mu(\emptyset)=0\) because the empty set is countable.

6. Prove that the closed intervals \([a,b]\) with rational endpoints generate \(\mathcal B(\mathbb R)\) , and deduce that every countable subset of \(\mathbb R\) is a Borel set.

   Hint

   By Proposition 1.7 it is enough to build each open interval with rational endpoints from closed ones: \((a,b)=\bigcup_n[a+\tfrac1n,b-\tfrac1n]\) . Conversely each closed interval is a closed set, hence Borel. A singleton \(\{x\}\) is the intersection of the intervals \([p,q]\) with rational \(p\le x\le q\) , and a countable set is a countable union of singletons.

7. For the intervals \(A_n=[0,1+(-1)^n/n]\) in \(\mathbb R\) , compute \(\liminf_nA_n\) and \(\limsup_nA_n\) , and say for which points \(x\) the sequence \(\mathbf 1_{A_n}(x)\) fails to converge.

   Answer

   \(\liminf_nA_n=[0,1)\) and \(\limsup_nA_n=[0,1]\) . A point \(x<1\) lies in \(A_n\) for all large \(n\) ; a point \(x>1\) lies in only finitely many; the point \(x=1\) lies in \(A_n\) for every even \(n\) and no odd \(n\) . The indicators fail to converge only at \(x=1\) , where they alternate between \(0\) and \(1\) .

8. Let \(A_1,A_2,\dots\) be measurable sets in a measure space. (a) Prove \(\mu(\liminf_nA_n)\le\liminf_n\mu(A_n)\) . (b) Prove \(\limsup_n\mu(A_n)\le\mu(\limsup_nA_n)\) when \(\mu(\bigcup_nA_n)<\infty\) . (c) Show by example that (b) can fail without the finiteness assumption.

   Solution

   (a) Put \(B_n=\bigcap_{k\ge n}A_k\) . Then \(B_n\uparrow\liminf_nA_n\) by Definition 1.3. For every \(k\ge n\) we have \(B_n\subseteq A_k\) , so \(\mu(B_n)\le\mu(A_k)\) , and therefore \(\mu(B_n)\le\inf_{k\ge n}\mu(A_k)\) . Continuity from below gives

   \[ \mu\bigl(\liminf_nA_n\bigr)=\lim_n\mu(B_n)\le\lim_n\inf_{k\ge n}\mu(A_k)=\liminf_n\mu(A_n). \]

   (b) Put \(C_n=\bigcup_{k\ge n}A_k\) . Then \(C_n\downarrow\limsup_nA_n\) and \(\mu(C_1)<\infty\) by hypothesis, so continuity from above applies. Since \(A_k\subseteq C_n\) for \(k\ge n\) , \(\mu(C_n)\ge\sup_{k\ge n}\mu(A_k)\) , and

   \[ \mu\bigl(\limsup_nA_n\bigr)=\lim_n\mu(C_n)\ge\lim_n\sup_{k\ge n}\mu(A_k)=\limsup_n\mu(A_n). \]

   (c) On \(\mathbb N\) with counting measure let \(A_n=\{n\}\) . Each point lies in exactly one \(A_n\) , so \(\limsup_nA_n=\emptyset\) has measure \(0\) , while \(\mu(A_n)=1\) for every \(n\) . Here \(\bigcup_nA_n=\mathbb N\) has infinite measure, and the inequality in (b) would read \(1\le 0\) .

9. Prove that \(\mu(A\cup B\cup C)=\mu(A)+\mu(B)+\mu(C)-\mu(A\cap B)-\mu(A\cap C)-\mu(B\cap C)+\mu(A\cap B\cap C)\) for sets of finite measure, and use it to count the integers from \(1\) to \(60\) divisible by at least one of \(2\) , \(3\) , \(5\) .

   Solution

   Apply Proposition 1.9(b) to the two sets \(A\cup B\) and \(C\) . All measures are finite, so terms may be moved across the equation:

   \[ \mu(A\cup B\cup C)=\mu(A\cup B)+\mu(C)-\mu\bigl((A\cup B)\cap C\bigr). \]

   Now \((A\cup B)\cap C=(A\cap C)\cup(B\cap C)\) , and the intersection of these two sets is \(A\cap B\cap C\) . Two more applications of Proposition 1.9(b) give \(\mu(A\cup B)=\mu(A)+\mu(B)-\mu(A\cap B)\) and \(\mu\bigl((A\cap C)\cup(B\cap C)\bigr)=\mu(A\cap C)+\mu(B\cap C)-\mu(A\cap B\cap C)\) . Substituting both yields the stated formula.

   For the count, use counting measure on \(\{1,\dots,60\}\) with \(A\) , \(B\) , \(C\) the multiples of \(2\) , \(3\) , \(5\) . The intersections are the multiples of \(6\) , \(10\) , \(15\) , and \(30\) . Hence the number of integers divisible by at least one of the three is \(30+20+12-10-6-4+2=44\) .

10. A student argues: “The intervals \((0,1/n)\) decrease to the empty set, so \(\mu((0,1/n))\to 0\) for every measure \(\mu\) on the Borel sets of \((0,1)\) .” Exhibit a measure for which the conclusion is false and identify the hypothesis that was not checked.

    Solution

    Let \(\mu(A)\) be the number of integers \(k\ge 2\) with \(1/k\in A\) , for Borel sets \(A\subseteq(0,1)\) . This is the weighted-sum measure carried by the countable set \(\{1/2,1/3,1/4,\dots\}\) , so it is countably additive. The interval \((0,1/n)\) contains \(1/k\) for every \(k>n\) , so \(\mu((0,1/n))=\infty\) for every \(n\) , although the intervals decrease to the empty set, which has measure \(0\) .

    The student applied continuity from above, Theorem 1.10(b), without checking that some set in the sequence has finite measure. For Lebesgue measure the check succeeds, since the first interval has length \(1\) ; for this \(\mu\) every interval in the sequence has infinite measure and the theorem says nothing. The decreasing half of continuity of measure is the half with a hypothesis.

11. Let \(\mu\) be a finitely additive function with values in \([0,\infty)\) on a \(\sigma\) -algebra \(\mathcal A\) , with \(\mu(\emptyset)=0\) . Prove that \(\mu\) is countably additive if and only if \(\mu(A_n)\to 0\) whenever \(A_n\downarrow\emptyset\) in \(\mathcal A\) .

    Solution

    Suppose \(\mu\) is countably additive. Then it is a finite measure, and if \(A_n\downarrow\emptyset\) , continuity from above (Theorem 1.10(b), applicable since \(\mu(A_1)<\infty\) ) gives \(\mu(A_n)\to\mu(\emptyset)=0\) .

    Conversely, suppose \(\mu(A_n)\to0\) whenever \(A_n\downarrow\emptyset\) . Let \(E_1,E_2,\dots\in\mathcal A\) be pairwise disjoint with union \(E\) , and put \(R_n=E\setminus(E_1\cup\dots\cup E_n)\) . The sets \(R_n\) are in \(\mathcal A\) , they decrease, and their intersection is empty because every point of \(E\) lies in some \(E_k\) . Finite additivity applied to the disjoint sets \(E_1,\dots,E_n,R_n\) gives

    \[ \mu(E)=\sum_{k=1}^n\mu(E_k)+\mu(R_n). \]

    All terms are finite. Letting \(n\to\infty\) and using \(\mu(R_n)\to0\) gives \(\mu(E)=\sum_{k=1}^\infty\mu(E_k)\) . Finiteness was used twice: to make the forward direction available, and implicitly to keep the remainder \(\mu(R_n)\) a real number that can tend to zero.

12. On \(\mathbb N\) let \(\mu\) be the measure with weights \(w(k)=1/(k(k+1))\) . Compute \(\mu(\mathbb N)\) , the tail masses \(\mu(\{n,n+1,\dots\})\) , and the mass of the set of even numbers in closed form.

    Solution

    Since \(w(k)=\frac1k-\frac1{k+1}\) , the tail sums telescope:

    \[ \mu(\{n,n+1,\dots\})=\sum_{k\ge n}\Bigl(\frac1k-\frac1{k+1}\Bigr)=\lim_{N\to\infty}\Bigl(\frac1n-\frac1{N+1}\Bigr)=\frac1n . \]

    In particular \(\mu(\mathbb N)=1\) , the case \(n=1\) . The tails decrease to the empty set and their masses \(1/n\) tend to \(0\) , as continuity from above requires for a finite measure.

    For the even numbers,

    \[ \mu(\{2,4,6,\dots\})=\sum_{j\ge1}\frac1{2j(2j+1)}=\sum_{j\ge1}\Bigl(\frac1{2j}-\frac1{2j+1}\Bigr)=\frac12-\frac13+\frac14-\frac15+\cdots . \]

    The alternating harmonic series \(1-\frac12+\frac13-\frac14+\cdots\) equals \(\ln 2\) , so the series above equals \(1-\ln 2\approx 0.3069\) . The odd numbers therefore have mass \(\ln 2\approx 0.6931\) , more than half, because the weight \(w(1)=\tfrac12\) alone is half the total.

13. Let \(X=\{1,2,3\}\) , \(\mathcal A=\{\emptyset,\{1\},\{2,3\},X\}\) , and let \(\mu\) be the measure with \(\mu(\{1\})=1\) and \(\mu(\{2,3\})=0\) . Describe the completion \((\overline{\mathcal A},\overline\mu)\) explicitly.

    Answer

    The only null sets are \(\emptyset\) and \(\{2,3\}\) , so the sets added are \(A\cup F\) with \(F\subseteq\{2,3\}\) . Every subset of \(X\) arises: \(\overline{\mathcal A}=\mathcal P(X)\) , and \(\overline\mu(E)=1\) if \(1\in E\) and \(0\) otherwise. The completion is the point mass \(\delta_1\) on the power set.

14. Let \(\mu\) be \(\sigma\) -finite. Prove that \(X\) is a countable union of pairwise disjoint sets of finite measure, and that any family of pairwise disjoint measurable sets of positive measure is countable.

    Hint

    For the first claim disjointify a sequence of finite-measure sets covering \(X\) , as in the proof of Proposition 1.9(c). For the second, let \(X=\bigcup_nX_n\) with \(\mu(X_n)<\infty\) and let \(\mathcal F\) be the family. For fixed \(n\) and \(m\) , at most \(m\,\mu(X_n)\) members \(E\) of \(\mathcal F\) can satisfy \(\mu(E\cap X_n)>1/m\) , by finite additivity. Every member of \(\mathcal F\) satisfies this for some pair \((n,m)\) , by continuity from below.

15. On \(X=\{1,2,3,4\}\) let \(\mathcal D\) be the family of subsets with an even number of elements. (a) Verify that \(\mathcal D\) is a \(\lambda\) -system but neither a \(\pi\) -system nor a \(\sigma\) -algebra. (b) Prove that two finite measures on \(\mathcal P(X)\) that agree on \(\mathcal D\) are nevertheless equal, and explain why this does not contradict Example 1.5.

    Solution

    (a) The whole set has \(4\) elements, so \(X\in\mathcal D\) . If \(A\subseteq B\) both have an even number of elements, then \(B\setminus A\) has \(\lvert B\rvert-\lvert A\rvert\) elements, an even number. An increasing sequence of subsets of a finite set is eventually constant, so its union is one of its terms and lies in \(\mathcal D\) . Hence \(\mathcal D\) is a \(\lambda\) -system. It is not a \(\pi\) -system, because \(\{1,2\}\cap\{2,3\}=\{2\}\) has one element, and so it is not a \(\sigma\) -algebra either.

    (b) Let \(\mu\) and \(\nu\) be finite measures on \(\mathcal P(X)\) agreeing on \(\mathcal D\) , and let \(d_i=\mu(\{i\})-\nu(\{i\})\) . Agreement on the two-point sets says \(d_i+d_j=0\) for all \(i\ne j\) . Then

    \[ 2d_1=(d_1+d_2)+(d_1+d_3)-(d_2+d_3)=0, \]

    and by symmetry every \(d_i\) vanishes: \(d_1=d_2=d_3=d_4=0\) . So the point masses agree and \(\mu=\nu\) .

    There is no contradiction with Example 1.5. Theorem 1.17 says that agreement on a generating \(\pi\) -system is sufficient for equality; it does not say that agreement on a family that is not a \(\pi\) -system must fail to force equality. The family in Example 1.5 contains only two of the six two-point sets, too few equations to pin down four point masses, whereas \(\mathcal D\) contains all six.

16. Prove that a finite measure \(\mu\) on \(\mathcal B(\mathbb R)\) is determined by the function \(F(x)=\mu((-\infty,x])\) , and explain which \(\pi\) -system and which part of Theorem 1.17 you used.

    Hint

    The rays \((-\infty,x]\) form a \(\pi\) -system, since the intersection of two rays is the shorter one, and they generate \(\mathcal B(\mathbb R)\) by Proposition 1.7. If two finite measures have the same \(F\) , they agree on the rays, and their total masses agree because \(F(n)\to\mu(\mathbb R)\) by continuity from below. Now apply Theorem 1.17(a).

17. Show that every measure on \((\mathbb N,\mathcal P(\mathbb N))\) has the form \(\mu_w\) for a unique weight function \(w:\mathbb N\to[0,\infty]\) , and point out the step of your proof that fails for the finitely additive set function \(\mu_0\) of Example 1.4.

    Answer

    Put \(w(k)=\mu(\{k\})\) . Any \(A\subseteq\mathbb N\) is the countable disjoint union of its singletons, so countable additivity gives \(\mu(A)=\sum_{k\in A}w(k)\) ; uniqueness holds because \(w(k)\) must be \(\mu(\{k\})\) . For \(\mu_0\) the step “\(\mu(A)\) is the sum over the singletons of \(A\) ” is exactly the countable additivity that \(\mu_0\) lacks: all its singleton weights are \(0\) and yet \(\mu_0(\mathbb N)=1\) .

18. Let \(A_1,A_2,\dots\) be measurable with \(\mu(A_n)\le 2^{-n}\) . (a) Show that almost every point lies in only finitely many \(A_n\) . (b) Give sets \(A_n\subseteq\mathbb N\) with \(\#(A_n)=1\) for all \(n\) and \(\limsup_nA_n=\mathbb N\) , so that the conclusion of Proposition 1.12 can fail when the sum of the measures diverges.

    Hint

    (a) The measures are summable, so Proposition 1.12 applies directly. (b) List the natural numbers so that each one recurs infinitely often, for example \(1;\ 1,2;\ 1,2,3;\ 1,2,3,4;\dots\) , and let \(A_n\) be the singleton of the \(n\) -th entry of the list.