We know the length of an interval and want the length of an arbitrary Borel set. A Borel set can be extremely complicated, and there is no formula that produces its length from the lengths of the intervals used to build it, because there is no usable description of how it was built. Lebesgue’s idea was to stop building from inside and to approximate from outside instead. Cover the set by countably many intervals, add up their lengths, and take the smallest total that any cover achieves. This assigns a number to every subset of the line.

The price is that the number so obtained, the outer measure, is not additive on arbitrary sets. It cannot be: Chapter 1 opened with the fact that no translation-invariant, countably additive length exists on all subsets. Carathéodory found the way out. He gave a criterion, stated purely in terms of the outer measure, that singles out a family of sets on which the outer measure is countably additive, and he proved that this family is always a \(\sigma\) -algebra. For length, the family contains every Borel set.

The chapter has two layers. The concrete layer constructs Lebesgue measure on \(\mathbb R\) . The abstract layer shows that the same construction turns any reasonable assignment of mass on an algebra into a measure on the generated \(\sigma\) -algebra, and that the result is unique under a finiteness condition. It closes with the set that cannot be measured.

2.1 Covering a set by intervals

Because the empty interval is allowed, finite covers are included. The infimum is over a nonempty collection, since the intervals \((-k,k)\) cover everything, and it may be \(\infty\) . Three properties of \(\lambda^*\) are all that the general theory uses, and they are worth isolating.

The trick of spending \(\varepsilon2^{-n}\) on the \(n\) -th set, so that countably many small errors total \(\varepsilon\) , appears in almost every argument of this chapter. Taking \(\mathcal E\) to be the open intervals and \(\rho=\ell\) shows that \(\lambda^*\) is an outer measure. What is not yet clear is that it gives the right answer on the one kind of set whose length we already know. A cover might, conceivably, be cleverer than the interval it covers.

The only non-formal ingredient is compactness, and it enters at exactly one point: to replace a countable cover by a finite one, for which lengths can be compared by elementary means. Every construction of Lebesgue measure contains this step in some guise.

Example 2.2: The Cantor set is uncountable and has outer measure zero

Let \(C_0=[0,1]\) . Remove the open middle third to get \(C_1=[0,\tfrac13]\cup[\tfrac23,1]\) . Remove the open middle third of each remaining interval to get \(C_2\) , a union of \(4\) intervals of length \(\tfrac19\) , and continue. At stage \(n\) the set \(C_n\) is a union of \(2^n\) closed intervals, each of length \(3^{-n}\) . The Cantor set is \(C=\bigcap_nC_n\) .

By monotonicity, subadditivity, and Theorem 2.4,

\[ \lambda^*(C)\le\lambda^*(C_n)\le 2^n\cdot3^{-n}=\bigl(\tfrac23\bigr)^n\qquad\text{for every } n . \]

At \(n=10\) the bound is \((2/3)^{10}=1024/59049\approx0.0173\) , and it tends to \(0\) . Hence \(\lambda^*(C)=0\) .

The set \(C\) is not small in the sense of cardinality. A point of \([0,1]\) lies in \(C\) exactly when it has a base-\(3\) expansion using only the digits \(0\) and \(2\) : removing the middle third at stage \(n\) removes the numbers whose \(n\) -th ternary digit is forced to be \(1\) . Replacing each digit \(2\) by \(1\) and reading the result in base \(2\) maps \(C\) onto \([0,1]\) . So \(C\) has as many points as the whole interval, and still it can be covered by intervals of arbitrarily small total length. The left panel of Figure 2.1 shows the first stages.

2.2 The sets that split every set additively

An outer measure is subadditive. On which sets is it additive? Carathéodory’s answer is a definition that looks, at first sight, as if it had been reverse-engineered from the proof it makes possible. It was.

The inequality \(\mu^*(A)\le\mu^*(A\cap E)+\mu^*(A\setminus E)\) holds for all sets by subadditivity, so only the reverse inequality needs checking, and only for test sets \(A\) with \(\mu^*(A)<\infty\) . The definition says that \(E\) is a clean knife: it cuts every set, however wild, into two pieces whose outer measures add up. Demanding this of all test sets, including non-measurable ones, is what gives the criterion its strength. A weaker demand, such as additivity only when \(A\) is the whole space, is not enough: the first outer measure of the next example is one for which every set passes that weaker test, and yet \(\mu^*\) is not additive on the sets so admitted.

2.3 Lebesgue measure

Theorem 2.6 applied to \(\lambda^*\) gives a complete measure on some \(\sigma\) -algebra. The theorem alone does not say the \(\sigma\) -algebra is large; Example 2.3 shows it could be trivial. For \(\lambda^*\) it is not.

So \(\mathcal B(\mathbb R)\subseteq\mathcal L\) , the measure \(\lambda\) is complete, and \(\lambda(I)\) is the length of \(I\) for every interval. It is \(\sigma\) -finite because \(\mathbb R=\bigcup_n[-n,n]\) . On the Borel sets it is the only measure that gives intervals their lengths: the half-open intervals \((a,b]\) together with \(\emptyset\) form a \(\pi\) -system that generates \(\mathcal B(\mathbb R)\) , the sets \((-n,n]\) belong to it, have finite measure, and increase to \(\mathbb R\) , so Theorem 1.17(b) applies. The question that closed Chapter 1 is answered: length exists on the Borel sets, and it is unique.

How far is a Lebesgue measurable set from an open set, a closed set, or a finite union of intervals? Not far, and the precise statement is one of the most used tools in the subject.

The proof of the second implication shows more than was asked: every Lebesgue measurable set is a Borel set minus a set of measure zero. So \(\mathcal L\) is exactly the completion of \(\mathcal B(\mathbb R)\) with respect to \(\lambda\) , in the sense of Theorem 1.14.

2.4 Premeasures and the extension theorem

Nothing in the construction of \(\lambda\) was special to length except Theorem 2.4. The general statement starts from mass assigned consistently on an algebra.

A premeasure is finitely additive, hence monotone on \(\mathcal A_0\) . The italicised clause is the whole difficulty: an algebra need not contain countable unions, but when it does, the premeasure must already be countably additive there. The set function of Example 1.4 is finitely additive on an algebra and is not a premeasure.

Lebesgue measure fits this pattern, with \(\mathcal A_0\) the algebra of finite disjoint unions of half-open intervals \((a,b]\) and unbounded intervals \((-\infty,b]\) , \((a,\infty)\) , and \(\mu_0\) the total length. So does a much larger family of measures on the line.

Taking \(F(x)=x\) gives Lebesgue measure on the Borel sets. A function \(F\) with a jump produces a measure with an atom.

2.5 A set that cannot be measured

The proof used three properties of \(\lambda\) and nothing else: countable additivity, invariance under translation, and \(\lambda([0,1))=1\) . It therefore proves the claim made at the start of Chapter 1: no function defined on all subsets of \(\mathbb R\) has those three properties. The set \(V\) was not constructed but chosen, and the axiom of choice is known to be unavoidable here; one cannot exhibit a non-measurable set by a formula. In practice this means that every set and every function that arises from countable operations on explicit objects is measurable, and that measurability hypotheses, though they must be stated, are rarely the obstacle.

There are thus three families on the line, strictly nested: \(\mathcal B(\mathbb R)\subsetneq\mathcal L\subsetneq\mathcal P(\mathbb R)\) . The second inclusion is strict by Vitali’s theorem. For the first, every subset of the Cantor set is Lebesgue measurable, being a subset of a null set in a complete measure space, and the Cantor set has as many subsets as there are sets of real numbers; but it can be shown that there are only as many Borel sets as real numbers. So most subsets of the Cantor set are Lebesgue measurable and not Borel.

Exercises

1. Show directly from Definition 2.1 that a one-point set has Lebesgue outer measure zero, and that \(\lambda^*([0,1]\setminus\mathbb Q)=1\) , using only subadditivity and Theorem 2.4.

   Hint

   Cover \(\{x\}\) by the single interval \((x-\varepsilon,x+\varepsilon)\) . For the second claim, \([0,1]\) is the union of \([0,1]\cap\mathbb Q\) , which has outer measure zero by Example 2.1, and \([0,1]\setminus\mathbb Q\) . Subadditivity gives \(1\le0+\lambda^*([0,1]\setminus\mathbb Q)\) , and monotonicity gives the reverse inequality.

2. Let \(A,B\subseteq\mathbb R\) be arbitrary sets with \(\inf\{\lvert a-b\rvert:a\in A,\ b\in B\}>0\) . Prove that \(\lambda^*(A\cup B)=\lambda^*(A)+\lambda^*(B)\) , without assuming that either set is measurable.

   Solution

   Let \(\delta>0\) be the infimum of the distances. Subadditivity gives \(\lambda^*(A\cup B)\le\lambda^*(A)+\lambda^*(B)\) , so we prove the reverse inequality, assuming \(\lambda^*(A\cup B)<\infty\) .

   A refinement step. An open interval \(I\) of length \(\ell\) can be covered by finitely many open intervals, each of length less than \(\delta\) , with total length at most \(\ell+\eta\) , for any \(\eta>0\) : choose \(m\) so large that \((\ell+\eta)/m<\delta\) , cut the closure of \(I\) into \(m\) equal closed pieces, and enlarge each piece to an open interval longer by \(\eta/m\) .

   The argument. Let \(\varepsilon>0\) and choose open intervals \(I_k\) covering \(A\cup B\) with \(\sum_k\ell(I_k)\le\lambda^*(A\cup B)+\varepsilon\) . Refine each \(I_k\) as above with \(\eta=\varepsilon2^{-k}\) . The result is a countable cover of \(A\cup B\) by open intervals of length less than \(\delta\) and total length at most \(\lambda^*(A\cup B)+2\varepsilon\) . An interval shorter than \(\delta\) cannot contain a point of \(A\) and a point of \(B\) . So the cover splits into the intervals that meet \(A\) , which cover \(A\) , and the remaining ones, which cover \(B\) . Therefore

   \[ \lambda^*(A)+\lambda^*(B)\le\lambda^*(A\cup B)+2\varepsilon , \]

   and \(\varepsilon\) was arbitrary. Notice that no measurability was used: positive distance is a geometric substitute for the Carathéodory criterion.

3. Let \(E\subseteq\mathbb R\) satisfy \(\lambda^*(E)=0\) . Prove that the set of squares \(\{x^2:x\in E\}\) also has outer measure zero, by treating \(E\cap[-n,n]\) first.

   Solution

   Fix \(n\) and put \(E_n=E\cap[-n,n]\) , which has outer measure zero by monotonicity. Given \(\varepsilon>0\) choose open intervals \(I_k\) covering \(E_n\) with \(\sum_k\ell(I_k)<\varepsilon\) ; intersecting each with \((-n-1,n+1)\) we may assume \(I_k\subseteq(-n-1,n+1)\) . For \(x,y\) in that range,

   \[ \lvert x^2-y^2\rvert=\lvert x+y\rvert\,\lvert x-y\rvert\le2(n+1)\lvert x-y\rvert . \]

   The image of the interval \(I_k\) under the continuous map \(x\mapsto x^2\) is an interval \(J_k\) , and by the estimate its length is at most \(2(n+1)\ell(I_k)\) . By Theorem 2.4, \(\lambda^*(J_k)\) is that length. The sets \(J_k\) cover \(\{x^2:x\in E_n\}\) , so subadditivity gives

   \[ \lambda^*(\{x^2:x\in E_n\})\le\sum_k\lambda^*(J_k)\le2(n+1)\varepsilon . \]

   Since \(\varepsilon\) is arbitrary, each of these sets has outer measure zero, and \(\{x^2:x\in E\}\) is their countable union. The squaring map is not Lipschitz on the whole line, which is why the set had to be cut into bounded pieces first.

4. On \(X=\{1,2,3,4\}\) let \(\mu^*\) be the outer measure of Proposition 2.3 for the covering sets \(\{1,2\}\) , \(\{2,3\}\) , \(\{3,4\}\) , \(\{4,1\}\) , each at cost \(1\) . Compute \(\mu^*\) on every subset and determine the \(\mu^*\) -measurable sets.

   Solution

   A single covering set costs \(1\) and two covering sets cost \(2\) ; two of them, for instance \(\{1,2\}\) and \(\{3,4\}\) , cover \(X\) . So \(\mu^*(E)\in\{0,1,2\}\) , and \(\mu^*(E)=1\) exactly when \(E\) is a nonempty subset of one covering set. The covering sets are the four “adjacent” pairs, so:

   \[ \mu^*(E)=\begin{cases}0,&E=\emptyset,\\ 1,&E\ \text{a singleton or one of}\ \{1,2\},\{2,3\},\{3,4\},\{4,1\},\\ 2,&E=\{1,3\},\ E=\{2,4\},\ \text{or}\ E\ \text{has at least three points}.\end{cases} \]

   Now test measurability. A singleton \(\{i\}\) fails with the test set \(A=\{i,j\}\) for an adjacent \(j\) : the pieces have outer measures \(1+1=2\) , but \(\mu^*(A)=1\) . An adjacent pair such as \(E=\{1,2\}\) fails with \(A=\{2,3\}\) : the pieces \(\{2\}\) and \(\{3\}\) give \(1+1=2\ne1\) . The set \(E=\{1,3\}\) fails with \(A=\{1,2\}\) , whose pieces are \(\{1\}\) and \(\{2\}\) . A set is measurable exactly when its complement is, which disposes of the three-point sets and of \(\{2,4\}\) . Hence the only \(\mu^*\) -measurable sets are \(\emptyset\) and \(X\) : of the \(16\) subsets, \(2\) are measurable. The overlapping covering sets leave no nontrivial set that splits every test set additively.

5. Prove that for every \(E\subseteq\mathbb R\) there is a countable intersection of open sets \(G\supseteq E\) with \(\lambda(G)=\lambda^*(E)\) , and that when \(\lambda^*(E)<\infty\) the set \(E\) is measurable if and only if \(\lambda^*(G\setminus E)=0\) .

   Hint

   If \(\lambda^*(E)=\infty\) take \(G=\mathbb R\) . Otherwise choose open \(U_n\supseteq E\) with \(\lambda(U_n)<\lambda^*(E)+1/n\) , as in the first step of the proof of Theorem 2.10, and let \(G=\bigcap_nU_n\) . If \(E\) is measurable, subtract. If \(\lambda^*(G\setminus E)=0\) , then \(G\setminus E\) is measurable by completeness and \(E=G\setminus(G\setminus E)\) .

6. Show that \(\tfrac14\) belongs to the Cantor set although it is not an endpoint of any interval removed or kept during the construction, by finding its ternary expansion.

   Answer

   \(\tfrac14=\sum_{k\ge1}2\cdot3^{-2k}\) , since the series sums to \(2\cdot\tfrac19/(1-\tfrac19)=\tfrac14\) . So \(\tfrac14=0.020202\ldots\) in base \(3\) , an expansion using only the digits \(0\) and \(2\) , and \(\tfrac14\in C\) . Endpoints of the construction intervals have expansions that are eventually constant (\(0\) or \(2\) repeated), and this one is not.

7. Modify the Cantor construction so that at stage \(n\) a middle open interval is removed from each remaining interval in proportion \(\theta_n\in(0,1)\) of its length. (a) Show the limiting set has measure \(\prod_{n}(1-\theta_n)\) . (b) Evaluate this for \(\theta_n=(n+1)^{-2}\) and for constant \(\theta_n=\theta\) .

   Solution

   (a) Let \(K_n\) be the set remaining after stage \(n\) ; it is a finite union of closed intervals. Stage \(n\) removes the proportion \(\theta_n\) of each remaining interval, hence of the total remaining length, so \(\lambda(K_n)=(1-\theta_n)\lambda(K_{n-1})\) and by induction \(\lambda(K_n)=\prod_{j=1}^n(1-\theta_j)\) . The sets \(K_n\) decrease to \(K=\bigcap_nK_n\) and \(\lambda(K_0)=1<\infty\) , so continuity from above gives \(\lambda(K)=\lim_n\prod_{j\le n}(1-\theta_j)\) .

   (b) For \(\theta_n=(n+1)^{-2}\) the partial products telescope:

   \[ \prod_{k=2}^{N}\Bigl(1-\frac1{k^2}\Bigr)=\prod_{k=2}^N\frac{(k-1)(k+1)}{k\cdot k}=\frac{1}{N}\cdot\frac{N+1}{2}=\frac{N+1}{2N}\longrightarrow\frac12 . \]

   So this set has measure \(\tfrac12\) . For constant \(\theta\) the product is \((1-\theta)^n\to0\) , however small \(\theta\) is: removing a fixed proportion at every stage always leaves a null set, and positive measure requires the proportions to shrink fast enough that their sum converges.

8. A student claims: “The rationals in \([0,1]\) are dense, so every open set containing them contains \([0,1]\) , and therefore \(\lambda^*(\mathbb Q\cap[0,1])=1\) .” Find the false step, and explain what the argument does prove if only finite covers by intervals are allowed.

   Solution

   The false step is “every open set containing the rationals of \([0,1]\) contains \([0,1]\) ”. Density of a set \(Q\) means that every nonempty open set meets \(Q\) ; it does not mean that an open set containing \(Q\) must contain every point near \(Q\) . Example 2.1 exhibits an open set \(U\) containing all rationals with \(\lambda^*(U)\le\varepsilon\) , obtained by giving the \(k\) -th rational an interval of length \(\varepsilon2^{-k}\) . The correct value is \(\lambda^*(\mathbb Q\cap[0,1])=0\) .

   With finite covers the argument is nearly right. If finitely many open intervals cover \(\mathbb Q\cap[0,1]\) , then the complement of their union in \([0,1]\) is a closed set containing no rational, hence containing no interval, and the complement in \([0,1]\) of a finite union of intervals is a finite union of intervals and points; so it is a finite set. The intervals therefore cover \([0,1]\) up to finitely many points, and their total length is at least \(1\) . The “outer content” defined by finite covers assigns \(\mathbb Q\cap[0,1]\) the value \(1\) , the same as its closure. Countable covers are what allow each rational its own, ever smaller, interval.

9. Let \(F(x)=x+\lfloor x\rfloor\) , where \(\lfloor x\rfloor\) is the greatest integer not exceeding \(x\) . Verify that \(F\) is nondecreasing and right-continuous, and compute \(\mu_F([0,2])\) , \(\mu_F((0,2))\) , \(\mu_F(\{1\})\) , and \(\mu_F(\mathbb Q\cap[0,2])\) .

   Solution

   Both \(x\) and \(\lfloor x\rfloor\) are nondecreasing, so \(F\) is. The function \(\lfloor x\rfloor\) is constant on each \([n,n+1)\) , hence right-continuous, and \(x\) is continuous; so \(F\) is right-continuous, with a jump of size \(1\) at each integer: \(F(n^-)=2n-1\) and \(F(n)=2n\) .

   Using the formulas of Example 2.4:

   \[ \mu_F([0,2])=F(2)-F(0^-)=4-(-1)=5,\qquad \mu_F((0,2))=F(2^-)-F(0)=3-0=3,\qquad \mu_F(\{1\})=F(1)-F(1^-)=2-1=1 . \]

   These agree with the description \(\mu_F=\lambda+\sum_{n\in\mathbb Z}\delta_n\) , which holds because both sides give \((a,b]\) the mass \(b-a\) plus the number of integers in \((a,b]\) : the closed interval has length \(2\) and contains the three integers \(0,1,2\) , and the open interval has length \(2\) and contains only the integer \(1\) . For the last set, \(\lambda(\mathbb Q\cap[0,2])=0\) and the point masses contribute one for each integer in the set, so \(\mu_F(\mathbb Q\cap[0,2])=3\) .

10. Prove that \(E\subseteq\mathbb R\) is Lebesgue measurable if and only if \(E=H\cup N\) , where \(H\) is a countable union of closed sets and \(\lambda^*(N)=0\) .

    Hint

    Use part (c) of Theorem 2.10 with \(\varepsilon=1/n\) to get closed sets \(F_n\subseteq E\) with \(\lambda^*(E\setminus F_n)<1/n\) ; let \(H=\bigcup_nF_n\) and \(N=E\setminus H\) . The converse is immediate, since closed sets are Borel and sets of outer measure zero are measurable.

11. Let \(D\) be the set of numbers in \([0,1]\) that have a decimal expansion in which the digit \(7\) never appears. Prove that \(D\) is a Borel set with \(\lambda(D)=0\) , and find the smallest \(n\) for which the stage-\(n\) cover has total length below \(\tfrac1{100}\) .

    Solution

    For a string \(d_1\dots d_n\) of digits none of which is \(7\) , let \(J(d_1\dots d_n)\) be the closed interval of numbers having a decimal expansion that begins \(0.d_1\dots d_n\) ; it has length \(10^{-n}\) . Let \(D_n\) be the union of these \(9^n\) intervals. Then \(D_n\) is closed, \(D_{n+1}\subseteq D_n\) , and \(D=\bigcap_nD_n\) : a point of \(D\) lies in every \(D_n\) by definition. Conversely let \(x\) lie in every \(D_n\) , and for each \(n\) choose an admissible string \(s_n\) of length \(n\) with \(x\in J(s_n)\) . These strings need not extend one another, since a point can lie in two adjacent intervals of the same level, so select by pigeonhole: some digit \(d_1\) is the first digit of \(s_n\) for infinitely many \(n\) ; among those \(n\) , some digit \(d_2\) is the second digit for infinitely many; and so on. For each \(k\) the string \(d_1\dots d_k\) begins some \(s_n\) with \(n\ge k\) , so \(x\in J(s_n)\subseteq J(d_1\dots d_k)\) , an interval of length \(10^{-k}\) whose left endpoint is the number with finite expansion \(0.d_1\dots d_k\) . Hence \(x\) is the sum of the series with digits \(d_1,d_2,\dots\) , an expansion with no digit \(7\) . So \(D\) is a countable intersection of closed sets, hence Borel.

    By monotonicity and subadditivity

    \[ \lambda(D)\le\lambda(D_n)\le9^n\cdot10^{-n}=\bigl(\tfrac9{10}\bigr)^n\longrightarrow0, \]

    so \(\lambda(D)=0\) . For the last question, \((9/10)^{43}\approx0.01078\) and \((9/10)^{44}\approx0.00970\) , so the smallest \(n\) with total length below \(\tfrac1{100}\) is \(n=44\) . The set \(D\) is uncountable, like the Cantor set, of which it is a base-\(10\) cousin.

12. Let \(V\) be the set in the proof of Theorem 2.15. Prove that every Lebesgue measurable subset of \(V\) has measure zero, and deduce that \(\lambda^*(V)>0\) .

    Hint

    Let \(E\subseteq V\) be measurable. Its translates modulo one by the rationals of \([0,1)\) are measurable, pairwise disjoint subsets of \([0,1)\) , each of measure \(\lambda(E)\) , exactly as in the proof of Theorem 2.15. Countably many copies of \(\lambda(E)\) sum to at most \(1\) , which forces \(\lambda(E)=0\) . If \(\lambda^*(V)\) were \(0\) , then \(V\) would be measurable by completeness.

13. Prove that \(\lambda^*\) is continuous from below on arbitrary sets: if \(E_1\subseteq E_2\subseteq\cdots\) are subsets of \(\mathbb R\) , measurable or not, then \(\lambda^*\bigl(\bigcup_nE_n\bigr)=\lim_n\lambda^*(E_n)\) .

    Hint

    Monotonicity gives one inequality. For the other, assume the limit is finite and use Exercise 2.5 to choose measurable \(G_n\supseteq E_n\) with \(\lambda(G_n)=\lambda^*(E_n)\) . The sets \(H_n=\bigcap_{k\ge n}G_k\) are measurable, increase, and satisfy \(E_n\subseteq H_n\subseteq G_n\) . Apply continuity from below to \(H_n\) .

14. Let \(\mu^*\) be an outer measure on \(X\) and let \(E\) be \(\mu^*\) -measurable. Prove that \(\mu^*(A\cup E)+\mu^*(A\cap E)=\mu^*(A)+\mu^*(E)\) for every set \(A\subseteq X\) , measurable or not.

    Solution

    Use the Carathéodory criterion for \(E\) twice, with two different test sets. With the test set \(A\cup E\) , whose part inside \(E\) is \(E\) and whose part outside \(E\) is \(A\setminus E\) ,

    \[ \mu^*(A\cup E)=\mu^*(E)+\mu^*(A\setminus E). \]

    With the test set \(A\) ,

    \[ \mu^*(A)=\mu^*(A\cap E)+\mu^*(A\setminus E). \]

    Add \(\mu^*(A\cap E)\) to both sides of the first equation and substitute the second:

    \[ \mu^*(A\cup E)+\mu^*(A\cap E)=\mu^*(E)+\mu^*(A\setminus E)+\mu^*(A\cap E)=\mu^*(E)+\mu^*(A). \]

    No subtraction was performed, so the identity is valid in \([0,\infty]\) even when some of the terms are infinite. This is Proposition 1.9(b) with only one of the two sets required to be measurable.

15. Apply the construction of Theorem 2.13 to the finitely additive set function \(\mu_0\) of Example 1.4 on the algebra of finite or cofinite subsets of \(\mathbb N\) . Compute the resulting outer measure \(\mu^*\) , and identify the step in the proof of Theorem 2.13(a) that fails.

    Solution

    Every singleton \(\{n\}\) belongs to the algebra and has \(\mu_0(\{n\})=0\) . Any \(E\subseteq\mathbb N\) is covered by the countably many singletons of its points, at total cost \(0\) . Hence \(\mu^*(E)=0\) for every \(E\subseteq\mathbb N\) . In particular \(\mu^*(\mathbb N)=0\) while \(\mu_0(\mathbb N)=1\) : the outer measure does not extend \(\mu_0\) , and conclusion (a) of Theorem 2.13 fails.

    The proof of (a) breaks at the line \(\mu_0(A)=\sum_k\mu_0(B_k)\) . There \(A\in\mathcal A_0\) is written as a countable disjoint union of members \(B_k\) of the algebra, and the equality is exactly the premeasure property, countable additivity for unions that lie in the algebra. For \(A=\mathbb N\) and \(B_k=\{k\}\) the equality would read \(1=0\) . Theorem 2.6 still holds, trivially: all sets are measurable and the measure is identically zero. A finitely additive set function that is not a premeasure has no countably additive extension at all, because any extension would itself be countably additive on the algebra.

16. Let \(\mu\) be a measure on \(\mathcal B(\mathbb R)\) that is invariant under translation and satisfies \(\mu((0,1])=1\) . Prove that \(\mu\) is Lebesgue measure on the Borel sets, using Theorem 1.17.

    Hint

    The \(n\) translates \((\tfrac{j-1}n,\tfrac jn]\) , \(j=1,\dots,n\) , are disjoint with union \((0,1]\) , so each has \(\mu\) -measure \(\tfrac1n\) ; hence \(\mu((0,\tfrac kn])=\tfrac kn\) , and by translation \(\mu((a,b])=b-a\) whenever \(b-a\) is rational. Half-open intervals with rational endpoints form a \(\pi\) -system generating the Borel sets, and \((-n,n]\) increases to \(\mathbb R\) with finite measure.

17. Construct an open dense subset \(U\) of \(\mathbb R\) with \(\lambda(U)<\varepsilon\) , and state what can be said about the interior, the closure, and the measure of its complement.

    Answer

    Take \(U=\bigcup_kI_k\) with \(I_k\) the open interval of length \(\varepsilon2^{-k-1}\) centred at the \(k\) -th rational; then \(\lambda(U)\le\varepsilon/2\) . The complement is closed, has empty interior because \(U\) is dense, and has infinite measure, since \(\lambda([-n,n]\setminus U)\ge2n-\varepsilon\) .