MAA6616 COURSE NOTES FALL 2012

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1 MAA6616 COURSE NOTES FALL σ-algebras Let X be a set, and let 2 X denote the set of all subsets of X. We write E c for the complement of E in X, and for E, F X, write E \ F = E F c. Let E F denote the symmetric difference of E and F : E F := (E \ F ) F \ E) = (E F ) \ (E F ). (1.1) Definition 1.1. Let X be a set. A Boolean algebra is a nonempty collection A 2 X which is closed under finite unions and complements. A σ-algebra is a Boolean algebra which is also closed under countable unions. A measurable space is a pair (X, M ) where M 2 X is a σ-algebra. A function f : X Y from one measurable spaces (X, M ) to another (Y, N ) is called measurable if f 1 (E) M whenever E N. If E α is any collection of sets in X, then since ( ) c = E α, (1.2) α E c α a Boolean algebra (resp. σ-algebra) is automatically closed under finite (resp. countable) intersections. It follows that a Boolean algebra (and a σ-algebra) on X always contains and X. (Proof: X = E E c and = E E c.) Remark 1.2. There is a superficial resemblance between measurable spaces and topological spaces; and between measurable functions and continuous functions. In particular, a topology on X is a collection of subsets of X closed under arbitrary unions and finite intersections, whereas for a σ-algebra we insist only on countable unions, but require complements also. For functions, recall that a function between topological spaces is continuous if and only if pre-images of open sets are open; the definition of measurable function is plainly similar. The two categories are related by the Borel algebra construction, which we will examine later. The disjointification trick in the next Proposition is often useful. Proposition 1.3 (Disjointification). A nonempty collection M 2 X is a σ-algebra if and only if M is closed under complementation and countable disjoint unions. Proof. The proof amounts to the observation that if {G n } is a sequence of subsets of X, then the sets ( n 1 ) F n = G n \ G k (1.3) are disjoint, and N F n = N G n for all N (and thus F n = G n). Date: January 6, α k=1

2 Examples 1.4. a) Of course, for any set X its power set 2 X is a σ-algebra. At the other extreme, for any set X the collection {, X} is a σ-algebra. In general, if M N 2 X are σ-algebras, we say M is coarser than N ; likewise N is finer than M. So {, X}, 2 X are respectively the coarsest and finest σ-algebras on X. b) If X is any set, the collection M = {E X : E is countable or X \ E is countable } (1.4) is a σ-algebra (the proof is left as an exercise). c) If X is a set, M 2 X a σ-algebra, and E is any nonempty subset of X, then M E := {A E : A M } 2 E (1.5) is a σ-algebra on E (exercise). d) If {M α : α A} is a collection of σ-algebras on X, then their intersection α A M α is also a σ-algebra (checking this is a simple exercise). This means that given any set E 2 X, we can define the σ-algebra M (E ) = M (1.6) where the intersection is taken over all σ-algebras M containing E (this collection is nonempty, since it contains 2 X ). M (E ) is called the σ-algebra generated by E. e) An important instance of the construction in (d) is when X is a topological space and E is the collection of open sets of X. In this case the σ-algebra generated by E is called the Borel σ-algebra and is denoted B X. We will study the Borel σ-algebra over R more closely in Subsection 1.1. f) If X is a set, (Y, N ) a measurable space, and f : X Y a function, then the collection f 1 (N ) = {f 1 (E) : E N } 2 X (1.7) is a σ-algebra on X (check this); called the pull-back σ-algebra. The pull-back σ- algebra is the smallest σ-algebra on X for which the function f : X Y is measurable. g) More generally, if we have a set X, a family of measurable spaces (Y α, N α ) where α ranges over some index set A, and functions f α : X Y α, then we can construct a σ-algebra M on X by taking E = {f 1 α (E α ) : α A, E α N α } (1.8) and letting M = M (E ). Unlike the case of a single f, E by itself will not be a σ-algebra in general. M is characterized by the property that it is the smallest σ- algebra on X such that each of the functions f α is measurable. An important special case of this construction is the product σ-algebra (see Subsection 1.2). The following proposition is trivial but useful. Proposition 1.5. If M 2 X is a σ-algebra and E M, then M (E ) M. The proposition is used in the following way: suppose we want to prove that a particular statement is true for every set in some σ-algebra M (say, the Borel σ-algebra B X ), which we know is generated by a collection of sets E (say, the open sets of X). Then it suffices to prove that 1) the statement is true for every set in E, and 2) the collection of sets for which 2

3 the statement is true forms a σ-algebra. (The monotone class theorem (see Theorem?? below) is typically used in a similar way.) A function f : X Y between topological spaces is said to be Borel measurable if it is measurable when X and Y are equipped with their respective Borel σ-algebras. Proposition 1.6. If X and Y are topological spaces, then every continuous function f : X Y is Borel measurable. Proof. Problem 7.5. (Hint: follow the strategy described after Proposition 1.5.) We will defer further discussion of measurable functions to the next chapter The Borel σ-algebra over R. Before going further, we take a closer look at the Borel σ-algebra over R. We begin with a useful lemma on the structure of open subsets of R: Lemma 1.7. Every nonempty open subset U R is an (at most countable) disjoint union of open intervals. Here we allow the degenerate intervals (, a), (a, + ), (, + ). Proof. Let {I n : n N} denote the set of all intervals with rational endpoints such that I n U. This collection is evidently countable, and I n = U. Write I n = (a n, b n ). For each n define α n = inf{x R : x a n and (x, b n ) U} (1.9) β n = sup{x R : x b n and (a n, x) U}. (1.10) After removing possible repetitions, the collection of open intervals (α n, β n ) is disjoint, at most countable, and n (α n, β n ) = U. Proposition 1.8 (Generators of B R ). Each of the following collections of sets E 2 R generates the Borel σ-algebra B R : the open intervals E 1 = {(a, b) : a, b R} the closed intervals E 2 = {[a, b] : a, b R} the (left or right) half-open intervals E 3 = {[a, b) : a, b R} or E 4 = {[a, b) : a, b R} the (left or right) open rays E 5 = {(, a) : a R} or E 6 = {(a, + ) : a R} the (left or right) closed rays E 7 = {(, a] : a R} or E 8 = {[a, + ) : a R} Proof. We prove only the open and closed interval cases, the rest are similar and left as exercises. The proof makes repeated use of Proposition 1.5. To prove M (E 1 ) = B R, first note that since each interval (a, b) is open, M (E 1 ) B R by Proposition 1.5. Conversely, each open set U R is a countable union of open intervals, so M (E 1 ) contains all the open sets of R, and since the open sets generate B R by definition, we have B R M (E 1 ) by Proposition 1.5 again. Thus M (E 1 ) = B R. For the closed intervals E 2, first note that each closed set is a Borel set, since it is the complement of an open set; thus E 2 B R so M (E 2 ) B R by Proposition 1.5. Conversely, each open interval (a, b) is a countable union of closed intervals [a+ 1, b 1 ]. More precisely, n n since a < b we can choose an integer N such that a + 1 < b 1 ; then N N (a, b) = [a + 1 n, b 1 n ] (1.11) n=n 3

4 It follows that E 1 M (E 2 ), so by Proposition 1.5 and the first part of the proof, B R = M (E 1 ) M (E 2 ). (1.12) Sometimes it is convenient to use a more refined version of the above Proposition, where we consider only dyadic intervals. Definition 1.9. A dyadic interval is an interval of the form ( I = m + k 2, m + k + 1 ] n 2 n where m, n, k are integers, n 0, and 0 k 2 n 1. (1.13) In other words, the dyadic intervals are the intervals of length 2 n (n = 0, 1, 2...) whose endpoints are dyadic rationals (rationals of the form j/2 n ). (Draw a picture of a few of these to get the idea). A key property of dyadic intervals is the nesting property: if I, J are dyadic intervals, then either they are disjoint, or one is contained in the other. Dyadic intervals are often used to discretize analysis problems. The other key property is Proposition Every open subset of R is a countable disjoint union of dyadic intervals. Proof. Problem 7.3. It follows (using the same idea as in the proof of Proposition 1.8) that the dyadic intervals generate B R. The use of half-open intervals here is only a technical convenience, to allow us to say disjoint in the above proposition instead of almost disjoint. Naturally, one can define dyadic cubes in R n and prove similar results; we defer these to the discussion of Lebesgue integration on R n in the next chapter Product σ-algebras. Given a collection of measurable spaces (X α, M α ), one would like to construct a σ-algebra on the Cartesian product X = α X α that is compatible with the σ-algebras M α. In particular, from the definition of the product of sets there are coordinate functions f α : X X α, and we would like each of these functions to be measurable. For simplicity, at present we will only consider the case of finitely many factors. The product X = n j=1 X j is identified with the set of tuples {(x 1,... x n ) : x j X j }. For x = (x 1, x 2,... x n ) n j=1 X j, the coordinate projections π j : X X j are defined by π j (x) = x j. If each X j is equipped with a σ-algebra N j, then we can form a σ-algebra on X as in Example 1.4(g): Definition If (X j, N j ), j = 1,... n are measurable spaces, the product σ-algebra n j=1n j is the σ algebra on X = n j=1 X j generated by {π 1 (E j ) : E j N j, j = 1,... n}. (1.14) The definition guarantees that each coordinate projection π j is a ( k N k, N j ) measurable function. (We will return to this point in the discussion of product measure in the next chapter.) Of course, one can view R n as the n-fold Cartesian product R n = R R, and the product topology coincides with the usual metric topology on R n. Thus, we now have two 4

5 ways of constructing σ-algebras on R n : the Borel σ-algebra B R n, or the product σ-algebra obtained by giving each copy of R the Borel σ-algebra B R and forming the product σ-algebra n 1B R. It is reasonable to suspect that these two σ-algebras are the same, and indeed this is the case. Proposition B R n = n j=1b R. Proof. We use Proposition 1.5 to prove inclusions in both directions. By definition, the product σ-algebra n j=1b R is generated by the sets π 1 j (E j ), j = 1,... n where E j R is open and π j (x 1,... x n ) = x j is the projection map. But π 1 j (E j ) = R E j R, where E j is the j th factor. This is an open subset of R n, hence Borel, so by Proposition 1.5 we have n j=1b R B R n. For the reverse inclusion, first note that every open subset U R n is equal to a countable union of open boxes B = (a 1, b 1 ) (a n, b n ) (just take all the open boxes contained in U having rational vertices). Arguing as in the proof of Proposition 1.8, the open boxes generate the Borel σ-algebra B R n. If B = n j=1 (a j, b j ) is an open box and we put E j = (a j, b j ), then from the description of π 1 j (E j ) in the first part of the proof we obtain B = n j=1 π 1 j (E j ). (1.15) So, each open box belongs to the product σ-algebra n j=1b R, and hence B R n n j=1b R by Proposition Measures Definition 2.1. Let X be a set and M a σ-algebra on X. A measure on M is a function µ : M [0, + ] such that µ( ) = 0, If {E j } j=1 is a sequence of disjoints sets in M, then ( ) µ E j = j=1 µ(e j ) j=1 If µ(x) <, then µ is called finite; if X = j=1x j with µ(x j ) < for each j then µ is called σ-finite. Almost all of the measures of importance in analysis (and certainly all of the measure we will work with) are σ-finite. A triple (X, M, µ) where M is a σ-algebra and µ a measure on M is called a measure space. At this point, we cannot give any really non-trivial examples (such as Lebesgue measure) of measures; this will have to wait until we have proved the Caratheodory and Hahn- Kolmogorov theorems in the following sections. For now we describe some simple measures and some procedures for producing new measures from old. 5

6 Examples 2.2. a) Let X be any set, and for E X let E denote the cardinality of E. The function µ : 2 X [0, + ] defined by { E if E is finite µ(e) = + if E is infinite (2.1) is a measure on (X, 2 X ), called counting measure. It is finite if and only if X is finite, and σ-finite if and only if X is countable. b) Let X be a set and M the σ-algebra of countable and co-countable sets (Example 1.4(b)). For E M define µ(e) = 0 if E is countable and µ(e) = + is E is co-countable. Then µ is a measure. c) Let (X, M, µ) be a measure space and E M. Recall M E from Example 1.4(c). The function µ E (A) := µ(a E) is a measure on (E, M E ). (Why did we assume E M?) d) (Linear combinations) If µ is a measure on M and c > 0, then (cµ)(e) : c mu(e) is a measure, and if µ 1,... µ n are measures on the same M, then (µ 1 + µ n )(E) := µ 1 (E) + µ n (E) (2.2) is a measure. Likewise an infinite sum of measures µ n is a measure. (The proof of this last fact requires a small amount of care; see Problem 7.7.) One can also define products and pull-backs of measures, compatible with the constructions of product and pull-back σ-algebras; these examples will be postponed until the next chapter, after we have built up some more machinery of measurable functions. Theorem 2.3 (Basic properties of measures). Let (X, M, µ) be a measure space. a) (Monotonicity) If E, F M and E F, then µ(e) µ(f ). b) (Subadditivity) If {E j } j=1 M, then µ( j=1 E j) j=1 µ(e j) c) (Monotone convergence for sets) If {E j } j=1 M and E j E j+1 j, then µ( E j ) = lim µ(e j ). d) (Dominated convergence for sets) If {E j } j=1 M and E j E j+1 j, and µ(e 1 ) <, then µ( E j ) = lim µ(e j ). Proof. a) By additivity, µ(f ) = µ(f \ E) + µ(e) µ(e). b) We use the disjointification trick (see Proposition 1.3): for each j 1 let F j = E j \ ( j 1 k=1 E k ). (2.3) Then the F j are disjoint, and F j E j for all j, so by countable additivity and (a) ( ) ( ) µ E j = µ F j = j=1 j=1 6 µ(f j ) j=1 µ(e j ). (2.4) j=1

7 c) The sets F j = E j \E j 1 are disjoint sets whose union is j=1 E j, and for each j, j k=1 F k = E j. So by countable additivity, ( ) ( ) µ E j = µ F j j=1 = j=1 µ(f k ) k=1 = lim j j µ(f k ) k=1 = lim j µ ( j ) F k k=1 = lim j µ(e j ). d) The sequence µ(e j ) is decreasing (by (a)) and bounded below, so lim µ(e j ) exists. Let F j = E 1 \ E j. Then F j F j+1 for all j, and j=1 F j = E 1 \ j=1 E j. So by (c) applied to the F j, and since µ(e 1 ) <, µ(e 1 ) µ( E j ) = µ(e 1 \ E j ) j=1 j=1 = lim µ(f j ) = lim(µ(e 1 ) µ(e j )) = µ(e 1 ) lim µ(e j ). Again since µ(e 1 ) <, it can be subtracted from both sides. Remark 2.4. Note that in item (d) of Theorem 2.3, the hypothesis µ(e 1 ) < can be replaced by µ(e j ) < for some j. However the finiteness hypothesis cannot be removed entirely. To see this, consider (N, 2 N ) equipped with counting measure, and let E j = {k : k j}. Then µ(e j ) = for all j but µ( j=1 E j) = µ( ) = 0. For any set X and subset E X, there is a function 1 E : X {0, 1} defined by { 1 if x E 1 E (x) = 0 if x E, (2.5) called the characteristic or indicator function of E. For a sequence of subsets (E n ) of X, say E n E pointwise if 1 En 1 E pointwise 1. This notion allows the formulation of a more refined version of the dominated convergence theorem for sets, which foreshadows (and is a special case of) the dominated convergence theorem for the Lebesgue integral. See Problems 7.10 and What would happen if we asked for uniform convergence? 7

8 Definition 2.5. Let (X, M, µ) be a measure space. µ(e) = 0. A null set is a set E M with It follows immediately from countable subadditivity that a countable union of null sets is null. The contrapositive of this statement is a measure-theoretic version of the pigeonhole principle: Proposition 2.6 (Pigeonhole principle for measures). If (E n ) is a sequence of sets in M and µ( E n ) > 0, then µ(e n ) > 0 for some n. It will often be tempting to assert that if µ(e) = 0 and F E, then µ(f ) = 0, but one must be careful: F need not be a measurable set. This is not a big deal in practice, however, because we can always enlarge the σ-algebra on which a measure is defined so as to contain all subsets of null sets, and it will usually be convenient to do so. Definition 2.7. If (X, M, µ) has the property that F M whenever E M, µ(e) = 0, and F E, then µ is called complete. Theorem 2.8. Let (X, M, µ) be a measure space. Let N := {N M µ(n) = 0}. Define Then M is a σ-algebra, and M := {E F E M, F N for some N N } µ(e F ) := µ(e) defines a complete measure on M such that µ M = µ. Proof. First note that M and N are both closed under countable unions, so M is as well. To see that M is closed under complements, consider E F M with E M, F N N. We may assume that E and N are disjoint (replace F, N by F \ E, N \ E if necessary). Then we have (E F ) c = (E N) c N \ F, and by inspection E c \ N M and N \ F N N. Thus M is a σ-algebra. The proof that µ is a measure on M, and extends µ, is left as an exercise (Problem 7.12). 3. Outer measures and the Caratheodory Extension Theorem 3.1. Outline. The point of the construction of Lebesgue measure on the real line is to extend the naive notion of length for intervals to a suitably large family of subsets of R. We will accomplish this via the Caratheodory Extension Theorem (Theorem 3.3) Definition 3.1. Let X be a nonempty set. A function µ : 2 X [0, + ] is called an outer measure if: µ ( ) = 0, (Monotonicity) µ (A) µ (B) whenever A B, (Subadditivity) if {A j } j=1 2 X, then ( ) µ A j j=1 8 µ(a j ) j=1

9 Definition 3.2. If µ is an outer measure on X, then a set E X is called outer measurable (or µ -measurable) if µ (A) = µ (A E) + µ (A E c ) (3.1) for every A X. The significance of outer measures and outer measurable sets stems from the following theorem: Theorem 3.3 (Caratheodory Extension Theorem). If µ is an outer measure on X, then the collection M of outer measurable sets is a σ-algebra, and the restriction of µ to M is a complete measure. The proof will be given later in this section; we first explain how to construct outer measures. In fact, all of the outer measures we consider will be constructed using the following proposition: Proposition 3.4. Let E 2 X and µ 0 : E [0, + ] be such that, X E and µ 0 ( ) = 0. For every A X, define { } µ (A) = inf µ 0 (E n ) : E n E and A E n (3.2) Then µ is an outer measure. Note that we have assumed X E, so there is at least one covering of A by sets in E (take E 1 = X and all other E j empty), so the definition (3.2) makes sense. Proof of Proposition 3.4. It is immediate from the definition that µ ( ) = 0 (cover the empty set by empty sets) and that µ (A) µ (B) whenever A B (any covering of B is also a covering of A). To prove countable subadditivity, we make our first use of the ɛ/2 n trick. Let (A n ) be a sequence in 2 X and let ɛ > 0. Then for each n 1 there exists a countable collection of sets (En) k k=1 in E such that A n k=1 Ek n and µ 0 (En) k < µ (A n ) + ɛ2 n. (3.3) k=1 But now the countable collection (En) k n,k=1 covers A = A n, and µ (A) µ 0 (En) k < (µ (A n ) + ɛ2 n ) = ɛ + µ (A n ). (3.4) k, Since ɛ > 0 was arbitrary, we conclude µ (A) µ (A n ). Example 3.5 (Lebesgue outer measure). Let E 2 R be the collection of all open intervals (a, b) R, with < a < b < +, together with and R. Define µ 0 ((a, b)) = b a, the length of the interval; and µ 0 ( ) = 0, µ 0 (R) = +. For any subset A R, its Lebesgue outer measure m (A) is defined as in (3.2): { } m (A) = inf (b n a n ) : A (a n, b n ) (3.5) 9

10 where we allow the degenerate interval R = (, + ). In the next section we will construct Lebesgue measure from m via the Caratheodory Extension Theorem; the main issues will be to show that the outer measure of an interval is equal to its length, and that every Borel subset of R is outer measurable. The other desirable properties of Lebesgue measure (such as translation invariance) will follow from this construction. Before proving Theorem 3.3 we make one observation, which will be used repeatedly: to prove that a set E is outer measurable for a given outer measure µ, it suffices to prove that µ (A) µ (A E) + µ (A \ E) (3.6) for all A X, since the opposite inequality for all A is immediate from the subadditivity of µ. We also need one lemma, which will be used to obtain completeness. A set E X is called µ -null if µ (E) = 0. Lemma 3.6. Every µ -null set is µ -measurable. Proof. Let E be µ -null and A X. By monotonicity, A E is also µ -null, so by monotonicity again, µ (A) µ (A \ E) = µ (A E) + µ (A \ E) (3.7) so the lemma follows from the observation immediately above. Proof of Theorem 3.3. We first show that M is a σ-algebra. It is immediate from Definition 3.2 that M contains and X, and since (3.1) is symmetric with respect to E and E c, M is also closed under complementation. Next we check that M is closed under finite unions (which will prove that M is a Boolean algebra). So, let E, F M and fix an arbitrary A X. To unclutter the notation, partition A into four disjoint sets A 00 = A \ (E sup F ), A 01 = (A \ E) F, A 10 = (A \ F ) E, A 11 = A E F. (3.8) (Draw a Venn diagram to see what is going on.) In this notation, to prove that E F is outer measurable we must prove that µ (A 00 A 01 A 10 A 11 ) = µ (A 01 A 10 A 11 ) + µ (A 00 ) (3.9) Since E is outer measurable, we have µ (A 00 A 01 A 10 A 11 ) = µ (A 00 A 01 ) + µ (A 10 A 11 ) (3.10) and µ (A 01 A 10 A 11 ) = µ (A 01 ) + µ (A 10 A 11 ). (3.11) Similarly, by the outer measurability of F we have µ (A 00 A 01 ) = µ (A 00 ) + µ (A 01 ). (3.12) Combining the last three equations gives (3.9). By induction, M is closed under finite unions. Now we show that M is closed under countable disjoint unions. (It then follows from Proposition 1.3 that M is a σ-algebra.) Let (E n ) be a sequence of disjoint outer measurable sets, and let A X. We must show that µ (A) = µ (A E n ) + µ (A \ E n ). (3.13) 10

11 As noted above, we only need to prove that µ (A) µ (A E n ) + µ (A \ E n ). (3.14) since the opposite inequality follows from the subadditivity of µ. For each N 1, we have already proved that N E n is outer measurable, and therefore µ (A) µ (A N E n ) + µ (A \ N E n ). (3.15) By monotonicity, we also know that µ (A \ N E n) µ (A \ E n), so it suffices to prove that µ (A By the outer measurability of N E n µ (A N+1 E n ) lim N µ (A E n ) = µ (A N E n ) (3.16) N E n ) + µ (A E N+1 ) (3.17) for all N 1 (the disjointness of the E n was used here). Iterating this identity, we get and taking limits, µ (A N+1 lim N µ (A E n ) = N µ (A E k+1 ) (3.18) k=0 N E n ) = µ (A E N+1 ) (3.19) But also, by countable subadditivity, we have µ (A E n ) µ (A E N+1 ) (3.20) which proves (3.16), and thus M is a σ-algebra. To prove that µ is a measure on M, note that it is immediate that µ ( ) = 0. To prove countable additivity, suppose that (E n ) is a disjoint sequence of outer measurable sets. It follows from outer measurability that for every N 1, N+1 µ ( E n ) = µ ( N=0 N=0 N E n ) + µ (E N+1 ). (3.21) By iterating this we see that µ is finitely additive, that is N µ ( E n ) = 11 N µ (E n ). (3.22)

12 By monotonicity of µ, we get by taking limits µ ( E n ) µ (E n ). (3.23) But the reverse inequality is immediate by the countable subadditivity of µ, so we are done. Finally, that µ is a complete measure on M is an immediate consequence of Lemma Construction of Lebesgue measure In this section, by an interval we mean any set I R of the from (a, b), [a, b], (a, b], [a, b), including, open and closed half-lines and R itself. We write I = b a for the length of I, interpreted as + in the line and half-line cases and 0 for. Recall the definition of Lebesgue outer measure of a set A R from Example 3.5: { } m (A) = inf I n : A I n (4.1) where the I n are open intervals, or empty. Theorem 4.1. If I R is an interval, then m (I) = I. Proof. We first consider the case where I is a finite, closed interval [a, b]. For any ɛ > 0, the single open interval (a ɛ, b + ɛ) covers I, so m (I) (b a) + 2ɛ = I + 2ɛ, and thus m (I) I. For the reverse inequality, again choose ɛ > 0, and let (I n ) be a cover of I by open intervals such that I n < m (I) + ɛ. Since I is compact, there is a finite subcollection (I nk ) N k=1 of the I n which covers I. Then N I nk > b a = I. (4.2) k=1 To see this, observe that by passing to a further subcollection, we can assume that none of the intervals I nk is contained in another one. Then re-index I 1,... I N so that the left endpoints a 1,... a N are listed in increasing order. Since these intervals cover I, and there are no containments, it follows that a 2 < b 1, a 3 < b 2,...a N < b N 1. (Draw a picture.) Therefore N N N 1 I k = (b k a k ) = b N a 1 + (b k a k+1 ) b N a 1 > b a = I. (4.3) k=1 k=1 k=1 From the inequality (4.2) we conclude that I < m (I) + ɛ, and since ɛ was arbitrary we have proved m (I) = I. Now we consider the cases of bounded, but not closed, intervals (a, b), (a, b], [a, b). If I is such an interval and I = [a, b] its closure, then since m is an outer measure we have by monotonicity m (I) m (I) = I. On the other hand, for all ɛ > 0 sufficiently small we have I ɛ := [a + ɛ, b ɛ] I, so by monotonicity again m (I) m (I ɛ ) = I 2ɛ and letting ɛ 0 we get m (I) I. Finally, the result is immediate in the case of unbounded intervals, since any unbounded interval contains arbitrarily large bounded intervals. 12

13 Theorem 4.2. Every Borel set E B R is m -measurable. Proof. By the Caratheodory extension theorem, the collection of m -measurable sets is a σ-algebra, so by Propositions 1.8 and 1.5, it suffices to show that the open rays (a, + ) are m -measurable. Fix a R and an arbitrary set A R. We must prove m (A) m (A (a, + )) + m (A (, a]). (4.4) To simplify the notation put A 1 = A (a, + ), A 2 = A (, a]. Let (I n ) be a cover of A by open intervals. For each n let I n = I n (a, + ) and I n = I n (, a]. The families (I n), (I n) are intervals (not necessarily open) that cover A 1, A 2 respectively. Now I n = I n + I n (4.5) = m (I n) + m ( I n) + m ( m (I n) (by Theorem 4.1) (4.6) I n) (by subadditivity) (4.7) m (A 1 ) + m (A 2 ) (by monotonicity). (4.8) Since this holds for all coverings of A by open intervals, taking the infimum we conclude m (A) m (A 1 ) + m (A 2 ). Definition 4.3. A set E R is called Lebesgue measurable if m (A) = m (A E) + m (A E c ) (4.9) for all A R. The restriction of m to the Lebesgue measurable sets is called Lebesgue measure, denoted m. By Theorem 3.3, m is a measure; by Theorem 4.2, every Borel set is Lebesgue measurable, and by Theorem 4.1 the Lebesgue measure of an interval is its length. It should also be evident by now that m is σ-finite. So, we have arrived at the promised extension of the length function on intervals to a measure. (We have not yet proved anything about the uniqueness of m; this will have to wait until we have proved the Hahn Uniqueness Theorem. See Corollary 5.5.) Next we prove that m has the desired invariance properties. Given E R, x R, and t > 0, let E + x = {y R : y x E}, E = {y R : y E}, and te = {y R : y/t E}. (4.10) Theorem 4.4. If E R is Lebesgue measurable, x R, and t > 0, then the sets E + x, E, and te are Lebesgue measurable. Moreover m(e + x) = m(e), m( E) = m(e), and m(te) = tm(e). Proof. We give the proof for E + x; the others are similar and left as exercises. First, it is immediate that for any interval I, I + x = I. Since the map (I n ) (I n + x) is a bijection between coverings of E and coverings of E+x, it follows that m (A) = m (A+x) for every set 13

14 S R. Finally, observe that A (E+x) = ((A x) E)+x and A (E+x) c = ((A x) E c )+x, so if E is Lebesgue measurable, then for any A R m (A) = m (A x) (4.11) = m ((A x) E) + m ((A x) E c ) (4.12) = m (A (E + x)) + m (A (E + x) c ) (4.13) Thus E + x is Lebesgue measurable and m(e + x) = m(e). The condition (4.9) does not make clear which subsets of R are Lebesgue measurable. We next prove two fundamental approximation theorems, which say that 1) if we are willing to ignore sets of measure zero, then every Lebesgue measurable set is a G δ or F σ, and 2) if we are willing to ignore sets of measure ɛ, then every set of finite Lebesgue measure is a union of intervals. (Recall that a set in a topological space is called a G δ -set if it is a countable intersection of open sets, and an F σ -set if it is a countable union of closed sets.) Theorem 4.5. Let E R. The following are equivalent: a) E is Lebesgue measurable. b) For every ɛ > 0, there is an open set U E such that m (U \ E) < ɛ. c) For every ɛ > 0, there is a closed set F E such that m (E \ F ) < ɛ. d) There is a G δ set G such that E G and m (G \ E) = 0. e) There is an F σ set F such that E F and m (E \ F ) = 0. Proof. The equivalence of (b) and (c), and of (d) and (e), is an easy consequence of the fact that the complement of an open or G δ set is a closed or F σ set respectively. The details are left as an exercise. (a) = (b): Let E be measurable, and suppose for the moment that m(e) <. Let ɛ > 0. Since E is measurable there is a covering of E by open intervals I n such that I n < m(e) + ɛ. Put U = I n. By subadditivity of m, m(u) m(i n ) = I n < m(e) + ɛ. (4.14) Since U E and m(e) <, we conclude (from the additivity of m) that m (U \ E) = m(u \ E) = m(u) m(e) < ɛ. To remove the finiteness assumption on E, we apply the ɛ/2 n trick: for each n Z let E n = E (n, n + 1]. The E n are disjoint measurable sets whose union is E, and m(e n ) < for all n. For each n, by the first part of the proof we can pick an open set U n so that m(u n \ E n ) < ɛ/2 n. Let U be the union of the U n ; then U is open and U \ E (U n \ E n ). From the subadditivity of m, we get m (U \ E) = m(u \ E) < n Z ɛ2 n = 3ɛ. (b) = (d): Let E R be given and for each n 1 choose (using (b)) an open set U n E such that m (U n \ E) < 1. Put G = n U n; then G is a G δ containing E, and G \ E U n \ E for every n. By monotonicity of m we see that m (G \ E) < 1 for every n n and thus m (G \ E) = 0. (Note that in this portion of the proof we cannot (and do not!) assume E is measurable.) (d) = (a): Since G is a G δ, it is a Borel set and hence Lebesgue measurable by Theorem 4.2. By Lemma 3.6, every m -null set is Lebesgue measurable, so G \ E, and therefore also E = G \ (G \ E), is Lebesgue measurable. 14

15 Remark 4.6. Some books take the condition in Theorem 4.5(b) as the definition of Lebesgue measurability. It is possible to prove from this (without using Caratheodory extension) that the collection of sets with this property is a σ-algebra, and the restriction of m to this σ-algebra is a measure. See Problem Theorem 4.7. If E is Lebesgue measurable and m(e) <, then for each ɛ > 0 there exists a set A which is a finite union of intervals such that m(e A) < ɛ. Proof. Let (I n ) be a covering of E by open intervals such that I n < m(e) + ɛ/2. (4.15) Since the sum is finite there exists an integer N so that I n < ɛ/2. (4.16) n=n+1 Let U = I n and A = N I n. Then A \ E U \ E, so m(a \ E) m(u) m(e) < ɛ/2 by (4.15). Similarly E \ A U \ A n=n+1 I n, so m(e \ A) < ɛ/2 by (4.16). Therefore m(e A) < ɛ. Thus, while the typical measurable set can be quite complicated in the set-theoretic sense (i.e. in terms of the Borel hierarchy), for most questions in analysis this complexity is irrelevant. In fact, Theorem 4.7 is the precise expression of a useful heuristic: Littlewood s First Principle of Analysis: Every measurable set E R with m(e) < is almost a finite union of intervals. Theorem 4.8. If E R is Lebesgue measurable, then That is, m is a regular Borel measure. m(e) = inf{m(u) : U E and U is open} = sup{m(k) : K E and K is compact} Proof. The first equality follows from monotonicity in the case m(e) = +, and from Theorem 4.5(b) (together with the additivity of m) in the case m(e) <. For the second equality, let ν(e) be the value of the supremum on the right-hand side. By monotonicity m(e) ν(e). For the reverse inequality, first assume m(e) < and let ɛ > 0. By Theorem 4.5(c), there is a closed subset F E with m(e \ F ) < ɛ/2. Since m(e) <, we have by additivity m(e) < m(f ) + ɛ/2, so m(f ) > m(e) ɛ/2. However this F need not be compact. To fix this, for each n 1 let K n = F [ n, n]. Then the K n are an increasing sequence of compact sets whose union is F. By monotone convergence for sets (Theorem 2.3(c)), there is an n so that m(k n ) > m(f ) ɛ/2. It follows that m(k n ) > m(e) ɛ, and thus ν(e) m(e). The case m(e) = + is left as an exercise. 15

16 Remark 4.9. In the preceding theorem, for any set E R (not necessarily measurable) the infimum of m(u) = m (U) over open sets U E defines the Lebesgue outer measure of E. One can also define the Lebesgue inner measure of a set E R as m (E) = sup{m (K) : K E and K is compact}. (4.17) By monotonicity we have m (E) m (E) for all E R. One can then prove that if E R and m (E) <, then E is Lebesgue measurable if and only if m (E) = m (E), in which case this quantity is equal to m(e). You are asked to prove this in Problem (Some care must be taken in trying to use this definition if m (E) = +. Why?) Example 4.10 (The Cantor set). Recall the usual construction of the middle thirds Cantor set. Let E 0 denote the unit interval [0, 1]. Obtain E 1 from E 0 by deleting the middle third (open) subinterval of E 0, so E 1 = [0, 1] [ 2, 1]. Continue inductively; at the 3 3 nth step delete the middle thirds of all the intervals present at that step. So, E n is a union of 2 n closed intervals of length 3 n. The Cantor set is defined as the intersection C = n=0 E n. It is well-known (though not obvious) that C is uncountable; we do not prove this here. It is clear that C is a closed set (hence Borel) but contains no intervals, since if J is an interval of length l and n is chosen so that 3 n < l, then J E n and thus J C. The Lebesgue measure of E n is (2/3) n, which goes to 0 as n, and thus by monotonicity m(c) = 0. So, C is an example of an uncountable, closed set of measure 0. Another way to see this is that at the n th stage (n 1) we have deleted a collection of 2 n 1 disjoint open intervals, each of length 3 n. Thus the Lebesgue measure of [0, 1] \ C is 2 n 1 3 n = = 1. (4.18) 3 Thus m(c) = 0. Example 4.11 (Fat Cantor sets). The standard construction of the Cantor set can be modified in the following way: fix a number 0 < c < 1. Imitate the construction of the Cantor set, except at the n th stage delete, from each interval I present at that stage, an open interval centered at the midpoint of I of length 3 n c. (In the previous construction we had c = 1.) Again at each stage we have a set E n which is a union of 2 n closed intervals; let F = n=0 E n. One can prove (in much the same way as for C) that 1) F is an uncountable, closed set; 2) F contains no intervals; and 3) m(f ) = 1 c > 0. Thus, F is a closed set of positive measure, but which contains no intervals. Example 4.12 (A Lebesgue non-measurable set). Define an equivalence relation on R by declaring x y if and only if x y Q. This relation partitions R into disjoint equivalence classes whose union is R. In particular, for each x R its equivalence class is the set {x + q : q Q}. Since Q is dense in R, each equivalence class C contains an element of the closed interval [0, 1]. By the axiom of choice, there is a set E [0, 1] that contains exactly one member x C from each class. We claim the set E is not Lebesgue measurable. To see this, let y [0, 1]. Then y belongs to some equivalence class C, and hence y differs from x C by some rational number in the interval [ 1, 1]. This means that [0, 1] (E + q). (4.19) q [ 1,1] Q 16

17 On the other hand, since E [0, 1] and q 1 we see that (E + q) [ 1, 2]. (4.20) q [ 1,1] Q Finally, by the construction of E the sets E + p and E + q are disjoint if p, q are distinct rationals. So if E were measurable, the the sets E + q would be also, and we would have by the countable additivity and monotonicity of m 1 m(e + q) 3 (4.21) q [ 1,1] Q But by translation invariance, all of the m(e + q) must be equal, which is a contradiction. Remark: The above construction can be modified used to show that if F is any Lebesgue set with m(f ) > 0, then F contains a nonmeasurable subset. See Problem Premeasures and the Hahn-Kolmogorov Theorem Definition 5.1. Let A 2 X be a Boolean algebra. A premeasure on A is a function µ 0 : A [0, + ] satisfying µ 0 ( ) = 0 If {A j } j=1 is a sequence of disjoint sets in A and 1 A j A, then ( ) µ 0 A j = j=1 µ 0 (A j ) Finiteness and σ-finiteness are defined for premeasures in the same way as for measures. Example 5.2. By an h-interval we mean a (finite or infinite) interval of the form (a, b]. (By convention (a, + ) is an h-interval but (, b) is not.) The collection A 2 R of finite unions of h-intervals is a Boolean algebra, and if I A is a disjoint union I = n j=1 (a j, b j ] then the function n µ 0 (I) = b j a j (5.1) j=1 is a premeasure on A. (Warning: this is immediate if we have already constructed Lebesgue measure, but it is not as obvious as it seems to prove from scratch there are many different ways to decompose a given I as a countable disjoint union of h-intervals, so verifying the countable additivity condition is somewhat delicate. See Section 6.) The only case of the proposition we will really care about is when E is an algebra and µ 0 is a premeasure. Theorem 5.3 (Hahn-Kolmogorov Theorem). If µ 0 is a premeasure on a Boolean algebra A 2 X and µ is the outer measure on X defined by (3.2) then every set A A is outer measurable and µ A = µ 0. In particular every premeasure µ 0 : A [0, + ] on a Boolean algebra A can be extended to a measure µ : B [0, + ] on a σ-algebra B A. 17 j=1

18 Proof. If A 2 X is a Boolean algebra and µ 0 : A [0, + ] is a premeasure, then µ 0 determines an outer measure µ by Proposition 3.4. We will prove that 1) µ A = µ 0, and 2) every set in A is outer measurable. The theorem then follows from Theorem 3.3. To prove 1), let E A. It is immediate that µ (E) µ 0 (E), since for a covering of E we can take A 1 = E and A j = for all other j. For the reverse inequality, let (A j ) be any covering of E by sets A j A, and define sets B n A by B n = E (A n \ n 1 j=1 A j ). (5.2) The B n are disjoint sets in A whose union is E, and B n A n for all n. Thus by the countable additivity and monotonicity of µ 0, µ 0 (E) = µ 0 (B n ) µ 0 (A n ) (5.3) Since the covering was arbitrary, we get µ 0 (E) µ (E). For 2), let A A, E X, and ɛ > 0. Then there exists a sequence of sets (B j ) A such that E j=1 B j and j=1 µ 0(B j ) µ (E) + ɛ. By additivity and monotonicity of µ 0 again, we have µ (E) + ɛ µ 0 (B j A) + µ 0 (B j A c ) (5.4) Since ɛ was arbitrary, we are done. j=1 j=1 µ (E A) + µ (E A c ) (5.5) We will refer to the measure µ constructed in Theorem 5.3 as the Hahn-Kolmogorov extension of the premeasure µ 0. The relationship between premeasures, outer measures, and measures in this construction is summarized in the following table: domain additivity condition bear premeasure Boolean algebra A countably additive, when possible mama outer measure all of 2 X monotone, countably papa subadditive measure σ-algebra containing A countably additive baby The premeasure µ 0 has the right additivity properties, but is defined on too few subsets of X to be useful. The corresponding outer measure µ constructed in Proposition 3.4 is defined on all of 2 X, but we cannot guarantee countable additivity. By Theorem 5.3, restricting µ to the σ-algebra of outer measurable sets is just right. We have established that every premeasure µ 0 on an algebra A can be extended to a measure on the σ-algebra generated by A. The next theorem addresses the uniqueness of this extension: 18

19 Theorem 5.4 (Hahn uniqueness theorem). Given any σ-finite premeasure µ 0 on a Boolean algebra A, there is a unique extension of µ 0 to a measure µ on the σ-algebra generated by A. Proof. Let M be the σ-algebra generated by A, let µ denote the Hahn-Kolmogorov extension of µ 0, but restricted to M, and let µ be any other extension of µ 0 to M. To prove that µ = µ, we first show that µ (E) µ(e) for all E M. Let E M and let (A n ) be a sequence in A such that E A n. Then µ (E) µ (A n ) = µ 0 (A n ) (5.6) Taking the infimum over all such coverings of E, it follows that µ (E) µ(e). (Recall the definition of µ.) Also notice that by monotone convergence for sets, if A = A n, then N µ (A) = lim N µ ( A n ) = lim µ( N A n ) = µ(a). (5.7) N Next we show that equality holds in the case µ(e) <. In this case, let ɛ > 0 and choose a covering (A n ) of E by sets in A such that for A = A n, we have µ(a) < µ(e) + ɛ, and consequently µ(a \ E) < ɛ. Thus µ(e) µ(a) = µ (A) = µ (E) + µ (A \ E) (5.8) µ (E) + µ(a \ E) (5.9) < µ (E) + ɛ. (5.10) Since ɛ was arbitrary, we conclude µ(e) = µ (E). (Observe that up to this point we have not used the σ-finiteness assumption; we use it only now to move to the µ(e) = + case.) Now write X = A n with the A n mutually disjoint and µ 0 (A n ) < for all n. Then for all E M µ(e) = µ(e A n ) = µ (E A n ) = µ (E). (5.11) Corollary 5.5 (Uniqueness of Lebesgue measure). If µ is a Borel measure on R such that µ(i) = I for every interval I, then µ(e) = m(e) for every Borel set E R. Uniqueness can fail in the non-σ-finite case. An example is outlined in Problem Lebesgue-Stieltjes measures on R Let µ be a Borel measure on R. (This means that the domain of µ contains all Borel sets, though we allow that the domain of µ may be larger.) Say µ is locally finite if µ(i) < for every compact interval I. Given a locally finite Borel measure, define a function F : R R by 0 if x = 0, F (x) = µ((0, x]) if x > 0, (6.1) µ((x, 0]) if x < 0 19

20 It is not hard to show that F is nondecreasing and continuous from the right (that is, F (a) = lim x a + F (x) for all a R). In this section we prove the converse: given any increasing, right-continuous function F : R R, there is a unique locally finite Borel measure µ such that (6.1) holds. This will be accomplished using the Hahn-Kolmogorov extension theorem. Let A 2 R denote the Boolean algebra generated by the half-open intervals (a, b]. (We insist that the interval be open on the left and closed on the right; this is only a convention chosen to be compatible with our definition of F.) More precisely, A consists of all finite unions of intervals of the form (a, b] (with (, b] and (a, + ) allowed). Fix a nondecreasing, right-continuous function F : R R. Since F is monotone, the limits F (+ ) := lim x + F (x) and F ( ) := lim x F (x) exist (possibly + or respectively). For each interval I = (a, b] in A, we define its F -length by I F := F (b) F (a). (6.2) Given a set A A, we can write it as a disjoint union of intervals A = N I n with I n = (a n, b n ]. Define µ 0 (A) = N I n F = N F (b n ) F (a n ). (6.3) Proposition 6.1. The expression (6.3) is a well-defined premeasure on A. Proof. That µ 0 is well-defined is left as an exercise; note that once this is established it is immediate from the definition that µ 0 is finitely additive. To prove that µ 0 is a premeasure, let (I n ) be a disjoint sequence of intervals in A and suppose I = A. By finite additivity we may assume I is an interval. Then ( N ) µ 0 (I) = µ 0 I n + µ 0 (I \ ) ( N N ) I n µ 0 I n = N µ 0 (I n ). (6.4) Taking limits, we conclude µ( I n) µ 0(I n ). For the reverse inequality, we employ a compactness argument similar to the one used in the proof of Theorem??. However, the situation is more complicated since we are dealing with half-open intervals. The strategy will be to shrink I to a slightly smaller compact interval, and enlarge the I n to open intervals, using the right-continuity of F and the ɛ/2 n trick to control their F -lengths. So, first suppose that I = (a, b] is a finite interval and fix ɛ > 0. By right continuity of F, there is a δ > 0 such that F (a + δ) F (a) < ɛ. Moreover if I n = (a n, b n ] then there exist δ n > 0 such that F (b n + δ n ) F (b n ) < ɛ2 n. Shrink I to the compact interval J = [a + δ, b] and enlarge the I n to open intervals J n = (a n, b n + δ n ). Then finitely many of the J n cover J, and these may be chosen so that none is contained in another, and they may be relabeled 20

21 J 1,... J N so their left endpoints are listed in increasing order. It follows that µ 0 (I) F (b) F (a + δ) + ɛ (6.5) F (b N + δ N ) F (a 1 ) + ɛ (6.6) N 1 = F (b N + δ N ) F (a N ) + (F (a j+1 F (a j )) + ɛ (6.7) j=1 N 1 F (b N + δ N ) F (a N ) + (F (b j + δ j ) F (a j )) + ɛ (6.8) j=1 µ 0 (I n ) + 2ɛ. (6.9) Since ɛ was arbitrary, we are done in the case I = (a, b] is finite. The infinite cases are left as an exercise. By the Hahn-Kolmogorov Theorem, µ 0 extends to a Borel measure µ F, called the Lebesgue- Stieltjes measure associated to F. It is immediate from the definition that µ 0 is σ-finite (each interval (n, n + 1] has finite F -length), so the restriction of µ F to the Borel σ-algebra is uniquely determined by F. In particular we conclude that the case F (x) = x recovers Lebesgue measure. Examples 6.2. a) (Dirac measure) Define the Heaviside function { 1 if x 0 H(x) = 0 if x < 0 (6.10) Then for any interval I = (a, b], µ H (I) = 1 if 0 I and 0 otherwise. Since Dirac measure δ 0 is a Borel measure and also has this property, and the intervals (a, b] generate the Borel σ-algebra, it follows from the Hahn Uniqueness Theorem (Theorem 5.4) that µ H (E) = δ 0 (E) for all Borel sets E R. More generally, for a finite set x 1,... x n in R and positive numbers c 1,... c n, let F (x) = n j=1 c jh(x x j ). Then µ F = n j=1 c jδ xj. b) (Infinite sums of point masses) Even more generally, if (x n ) is an infinite sequence in R and (c n ) is a sequence of positive numbers with c n <, define F (x) = n:x c n<x nh(x x n ). Then µ F (E) = n:x c n E n. A particularly interesting case is when the x n enumerate the rationals; the resulting function F is continuous precisely on the irrationals. We will return to this example after we have proved the Radon- Nikodym theorem. c) (Cantor measure) Recall the construction of the Cantor set C from Example Each number x [0, 1] has a base 3 expansion, of the form x = a n3 n, where a n {0, 1, 2} for all n. The expansion is unique if we insist that every terminating expansion (a n = 0 for all n sufficiently large) is replaced with an expansion ending with an infinite string of 2 s (that is, a n = 2 for all n sufficiently large). With these conventions, it is well-known that C consists of all points x [0, 1] such that the base 3 expansion of x contains only 0 s and 2 s. (Referring again to the construction of C, x belongs to E 1 if and only if a 1 is 0 or 2, x belongs to E 2 if and only if both 21

22 a 1, a 2 belong to {0, 2}, etc.) Using this fact, we can define a function F : C [0, 1] by taking the base 3 expansion x = a n3 n, setting b n = a n /2, and putting F (x) = b n2 n. (The ternary string of 0 s and 2 s is sent to the binary string of 0 s and 1 s.) If x, y C and x < y, then F (x) < F (y) unless x, y are the endpoints of a deleted interval, in which case F (x) = p2 k for some integers p and k, and F (x) and F (y) are the two base 2 expansions of this number. We can then extend F to have this constant value on the deleted interval (x, y). The resulting F is monotone and maps [0, 1] onto [0, 1]. Since F is onto and monotone, it has no jump discontinuities, and again by monotonicity, F is continuous. This function is called the Cantor-Lebesgue function, or in some books the Devil s Staircase. Finally, if we extend F to be 0 for x < 0 and 1 for x > 1, we can form a Lebesgue-Stieltjes measure µ F supported on C (that is, µ F (E) = 0 if E C = ). This measure is called the Cantor measure. It will be an important example of what is called a singular continuous measure on R. One can prove that the Lebesgue-Stieltjes measures µ F have similar regularity properties as Lebesgue measure; since the proofs involve no new ideas they are left as exercises. Lemma 6.3. Let µ F be a Lebesgue-Stieltjes measure. For any Borel set E R, µ F (E) = inf{ µ F (a n, b n ) : E (a n, b n )} (6.11) Proof. Problem Theorem 6.4. Let µ F be a Lebesgue-Stieltjes measure. Then for every Borel set E R, Proof. Problem µ F (E) = inf{µ F (U) : E U, U open} (6.12) = sup{µ F (K) : K E, K compact} (6.13) 7. Problems Problem 7.1. Prove the exercise claims in Example 1.4. Problem 7.2. a) Let X be a set and let A = (A n ) be a sequence of disjoint, nonempty subsets whose union is X. Prove that the set of all finite or countable unions of members of A (together with ) is a σ-algebra. (A σ-algebra of this type is called atomic.) b) Prove that the Borel σ-algebra B R is not atomic. (Hint: there exists an uncountable family of mutually disjoint Borel subsets of R.) Problem 7.3. a) Prove Proposition (First prove that every open set U is a union of dyadic intervals. To get disjointness, show that for each point x U there is a unique largest dyadic interval I such that x I U.) b) Prove that the dyadic intervals generate the Borel σ-algebra B R. Problem 7.4. Fix an integer n 1. Prove that the set of finite unions of dyadic subintervals of (0, 1] of length at most 2 n (together with ) is a Boolean algebra. Problem 7.5. Prove that if X, Y are topological spaces and f : X Y is continuous, then f is Borel measurable. 22

23 Problem 7.6. Let (X, M ) be a measurable space and suppose µ : M [0, + ] is a finitely additive measure which satisfies the condition in part (c) of Theorem 2.3. Prove that µ is a measure. Problem 7.7. Prove that an infinite sum of measures is a measure (Example 2.2(d)). (You will need the following fact from elementary analysis: if (a mn ) m, is a doubly indexed sequence of nonnegative reals, then m=1 a mn = m=1 a mn. Prove this by proving that both sums are equal to sup F (m,n) F where the supremum is taken over all finite subsets of N N.) a mn (7.1) Problem 7.8. Let A be an atomic σ-algebra generated by a partition (A n ) of a set X (see Problem 7.2). a) Fix n 1. Prove that the function δ n : A [0, 1] defined by { 1 if A n A δ n (A) = 0 if A n A (7.2) is a measure on A. b) Prove that if µ is any measure on (X, A ), then there exists a unique sequence (c n ) with each c n [0, + ] such that µ(a) = c n δ n (A) (7.3) for all A A. Problem 7.9. Let (X.M, µ) be a measure space. Prove the following: a) If E, F M and µ(e F ) = 0 then µ(e) = µ(f ). b) Define E F if and only if µ(e F ) = 0; then is an equivalence relation on M. c) For E, F M define d(e, F ) = µ(e F ). Then d defines a metric on the set of equivalence classes M /. Problem Let X be a set. For a sequence of subsets (E n ) of X, define lim sup E n = E n, lim inf E n = E n. (7.4) N=1 n=n N=1 n=n a) Prove that lim sup 1 En = 1 lim sup En and lim inf 1 En = 1 lim inf En (thus justifying the names). Conclude that E n E pointwise if and only if lim sup E n = lim inf E n = E. (Hint: for the first part, observe that x lim sup E n if and only if x lies in infinitely many of the E n, and x lim inf E n if and only if x lies in all but finitely many E n.) b) Prove that if the E n are measurable, then so are lim sup E n and lim inf E n. Deduce that if E n E pointwise and all the E n are measurable, then E is measurable. Problem 7.11 (Fatou theorem for sets). Let (X, M, µ) be a measure space, and let (E n ) be a sequence of measurable sets. 23