7. Measure Theory You Have Already Studied Lebesgue Measure in R

Home , Finite measure

7. Measure Theory Example. n Lebesgue measure on Rn, with the σ-algebra of Lebesgue You have already studied Lebesgue measure in R . • Here we consider a more abstract setup. This will al- measurable sets. Counting measure on any set, where every set is low us to develop a unified framework to deal with many • different situations, such as measures supported onsub- measurable. Surface measures. manifolds or fractals, or infinite dimensional spaces, such • as Wiener measure used in the definition of Brownian Hausdorff measures, which are ‘fractional dimen- • n motion. sional’ measures of subsets of R . We will introduce these in due course. Definition 7.1. A measure space is a set X together with a σ-algebra , a collection of subsets of X, and M How to construct measure spaces a measure µ : [0, ], such that M → ∞ Measures can be constructed from ‘exterior measures’. is required to contain the empty set and to •M Let X be any set. An exterior measure µ is a be closed under countable unions and complementation ∗ function from (X) to [0, ] such that (hence also closed under countable intersections). P ∞ (i) µ ( ) = 0; µ is required to be countably additive: if E1,E2,... ∗ ∅ • (ii) If E1 E2, then µ (E1) µ (E2); is a countable family of disjoint measurable sets, then ⊂ ∗ ≤ ∗ (iii) if E1,E2,... is a countable family of sets, then µ E = µ(E ). ∪n n n n µ n En µ (En). ∗ ∪ ≤ ∗ ( ) ∑ n ( ) ∑

There is a general construction, due to Carathéodory, 2. To show that the union of two sets E ,E is 1 2 ∈ M of obtaining a measure from an exterior measure. We in , we use M say that E X is measurable if, for every subset A c ⊂ ⊂ µ (A) = µ (A E1) + µ (A E1) X, we have ∗ ∗ ∩ ∗ ∩ and then write each of these two terms as a sum as c µ (A) = µ (A E) + µ (A E ). follows ∗ ∗ ∩ ∗ ∩ That is, E separates every set efficiently: we have equal- c µ (A E1) = µ (A E1 E2) + µ (A E1 E2) ∗ ∩ ∗ ∩ ∩ ∗ ∩ ∩ ity in the equation above rather than ‘ ’ which holds c c c c ≤ µ (A E1) = µ (A E1 E2) + µ (A E1 E2). automatically. Heuristically, we can think of a non- ∗ ∩ ∗ ∩ ∩ ∗ ∩ ∩ Now using subadditivity of µ , we have measurable set as being one that looks big ‘from the ∗ outside’ but small ‘from the inside’, and the exterior µ (A (E1 E2)) µ ((A E1 E2)) ∗ ∩ ∪ ≤ ∗ ∩ ∩ c c measure as measuring the size of the set as viewed from + µ ((A E1 E2)) + µ ((A E1 E2)) ∗ ∗ c ∩ ∩ ∩ ∩ the outside. But µ (A) µ (A E ) in some sense (Draw a diagram!) Combining the two we get ∗ − ∗ ∩ measures the size of from the inside, so if we get A E c c ∩ µ (A (E1 E2)) + µ (A E1 E2) µ (A) equality, then this indicates that E should be measur- ∗ ∩ ∪ ∗ ∩ ∩ ≤ ∗ able. or equivalently c µ (A) µ (A (E1 E2)) + µ (A (E1 E2) ), Theorem 7.2. Let µ be an exterior measure on X. ∗ ≥ ∗ ∩ ∪ ∗ ∩ ∪ ∗ c c c Then the subset of (X) consisting of measurable since A E1 E2 = A (E1 E2) . Equality in this M P ∩ ∩ ∩ ∪ sets forms a σ-algebra, and µ restricted to is a equation follows from subadditivity, which gives the ‘ ’ ∗ M ≤ measure. direction. 2.’ Additivity for two sets follows directly from the Proof: 1. It is clear that E = Ec . definition of measurability: put A = E E and E = ∈ M ⇒ ∈ M 1 ∪ 2 E1, and obtain The other inequality is automatic, so we have c µ (E1 E2) = µ (E1) + µ (E2). µ (A G) = µ (A) µ (A G ). ∗ ∪ ∗ ∗ ∗ ∩ ∗ − ∗ ∩ Thus G . One then uses induction to obtain finite additivity. ∈ M 3. To show closure under countable unions, let E 3’ Finally we need to show countable additivity. Since i ∈ . Without loss of generality, the are disjoint. Let j∞=1 µ (A Ej) is pinched between two quantities we Ei ∗ ∩ M now know are equal, we also have Gn = E1 En, and let G = En. Then each ∑ ∪ · · · ∪ ∪ ∞ Gn , and we have µ (A Ej) = µ (A G). ∈ M ∗ ∩ ∗ ∩ c j=1 µ (A) = µ (A Gn) + µ (A Gn) ∑ ∗ ∗ ∩ ∗ ∩ c Finally putting A = G we get countable additivity of µ (A Gn) + µ (A G ) ≥ ∗ ∩ ∗ ∩ µ on . □ n ∗ M and since A Gn = j=1A Ej so µ (A Gn) = ∩ ∪ ∩ ∗ ∩ This measure µ has the property of being complete: if µ (A Ej), using finite additivity as proved earlier, ∗ ∩ Z is a measurable set with measure zero, then every sub- we now have ∑ set of is measurable (with measure zero, of course). n Z c Any measure that is not complete can be completed in µ (A Ej) µ (A) µ (A G ). ∗ ∩ ≤ ∗ − ∗ ∩ j=1 a natural way: see exercise 2 in the text. ∑ Hence, taking the limit as n , → ∞ ∞ c µ (A Ej) µ (A) µ (A G ). ∗ ∩ ≤ ∗ − ∗ ∩ j=1 ∑ By countable subadditivity, µ (A G) j∞=1 µ (A ∗ ∩ ≤ ∗ ∩ E ), and so j ∑ µ (A G) µ (A) µ (A Gc) ∗ ∩ ≤ ∗ − ∗ ∩

Measures on metric spaces Proposition 7.4. Let µ be an outer measure on ∗ The theorem above is very nice, but how do you find the metric space X. Suppose that out which sets are measurable? Suppose that X is a (7.1) µ (A B) = µ (A) + µ (B) ∗ ∪ ∗ ∗ metric space. Then we would like at least all open sets whenever dist(A, B) > 0. Then all open sets in X to be measurable. This is guaranteed by the following are measurable, and hence the induced measure µ is condition that relates the metric and measure properties a Borel measure. of X. Before stating it we make the following Such an exterior measure is called a metric exte- • Definition 7.3. The Borel σ-algebra is the inter- rior measure. BX section of all σ-algebras containing all the open sets of Recall that dist(A, B) is defined to be the infimum of • X. (Note that the intersection of any collection of σ- distances d(a, b) where a A and b B. For example ∈ ∈ algebras is itself a σ-algebra.) A measure defined on X the distance in R between (0, 1) and (1, 2) is zero, even B is called a Borel measure on X. though the sets are disjoint.

Said another way, the Borel σ-algebra is the smallest σ-algebra containing all the open sets of X. Proof: Let O be an open set. Define On to be the subset Hence the sum of the tails of these infinite series can of O of points distance > 1/n from the boundary of O. be made arbitrarily small. In particular, for any ϵ > 0

We first claim that µ (O) = limn µ (On). (Note that there exists N such that ∗ ∗ is trivial here). Consider the ‘shells’ µ (Sn) < ϵ. ≥ ∗ 1 1 n>N Sn = On On 1 = x d(x, ∂O) > , ∑ \ − n 1 ≥ n Using subadditivity for { − } O = ON n>N Sn for with . Note that since is open, we ∪ n 2 S1 = O1 O ≥ µ (O) µ (ON ) + ∪µ (Sn) have a formula ∗ ≤ ∗ ∗ n>N∑ O = ON Sn and thus ∪ n>N ∪ µ (ON ) µ (O) ϵ, for each N. ∗ ≥ ∗ − For even n, these sets are a positive distance apart, proving the claim. If on the other hand, µ (O) = , ∗ ∞ by an easy triangle inequality argument. If x S and then we have by subadditivity ∈ n y S we have d(x, ∂O) d(x, y) + d(y, ∂O), and n+2 µ (O) µ (Sn) ∈ 1 ≤ ∗ ≤ ∗ so d(x, y) > . (Similarly for the odd n.) n 1 n(n+1) ∑≥ We now split into two cases, depending on whether and so

µ (O) is finite or infinite. First suppose that µ (O) < µ (Sn) = , ∗ ∗ ∗ ∞ . Then, by condition (??), we have n ∑ ∞ and using the positive distance between the even and µ (S2n) = µ ( nS2n) µ (O) < . ∗ ∗ ∪ ≤ ∗ ∞ odd slices as before, this implies that µ(O ) . n n → ∞ Similarly∑ for the odd shells: We can apply the same argument to µ(A O), for any ∩ set A, obtaining µ (A O) = limn µ (A On). Note µ (S2n+1) µ (O) < . ∗ ∗ ∗ ≤ ∗ ∞ ∩ ∩ n here that even though A O is not open, the identity ∑ ∩

O = O S implies We also give a result about approximation of measur- N ∪ n>N n A O = A O (A S ), able sets by open and closed sets. ∪∩ ∩ N ∪ ∩ n n>N which was all we needed. ∪ Proposition 7.5. Let X be a metric space and µ a Borel measure which is finite on all balls in Now, take any set A. Using condition (??) we see that X of finite radius. Then for each Borel set and for any n E each ϵ > 0 there is a bigger open set O E with µ (A) µ (A On) + µ (A O). ⊃ ∗ ≥ ∗ ∩ ∗ \ µ(O E) < ϵ, and a smaller closed set F E such Taking the limit as n , we obtain \ ⊂ → ∞ that µ(E F ) < ϵ. µ (A) µ (A O) + µ (A O). \ ∗ ≥ ∗ ∩ ∗ \ Let us refer to the two properties as outer regularity The inequality is immediate from subadditivity, so ≤ and inner regularity. we see that O is measurable. □ Proof: We show that closed sets have these properties, and the collection of Borel sets having these properties C is a σ-algebra. The Borel σ-algebra is, by definition, a sub-σ-algebra of , proving our claim. C For a closed set F , inner regularity is trivial. For outer It is easy to check that is closed under complemen- C regularity, we write Fk = F B(x0, k) for each k N, tation, so it remains to check closedness under count- ∩ ∈ where x0 is an arbitrarily chosen point. Each Fk is able unions. Outer regularity is straightforward using k closed and has finite measure, since it is a subset of an ϵ2− argument: Given E = Ei, choose Oi with k ∪ B(x , k). which by hypothesis has finite measure. For µ(O E ) < ϵ2− . Then define O = O , which is 0 i \ i ∪ i each Fk we define a family of open neighbourhoods automatically an open set. Then

l µ(O E) = µ( O E) O = x X d(x, F ) < 2− , i i k,l { ∈ | k } \ ∪ \ µ( Oi Ei) which ‘shrink’ down to Fk. ≤ ∪ \ µ(Oi Ei) Then, since Fk is closed, we have lOk,l = Fk. Ob- ≤ \ ∩ serve that for any l we have the disjoint decomposition < ϵ.∑

O = F O O k,l k ∪ k,i \ k,i+1 i l ∪≥ (where every set here is Borel). Looking at this for l = 0, we see that there is some l(k) so µ( i l(k) Ok,i ≥ \ O ) < ϵ2 k. Then we have k,i+1 − ∪ k µ(Ok,l(k)) < µ(Fk) + ϵ2− . Then define O = O , which is an open set, and ∪k k,N(k) observe

µ(O F ) µ(O F ) µ(O F ) = ϵ. \ ≤ k,N(k)\ ≤ k,N(k)\ k ∑k ∑k

The same argument doesn’t work for inner regular- Premeasures ity, because an infinite union of closed sets need not be We have seen how to construct a measure from an ex- closed. This argument does establish inner regularity terior measure. An exterior measure may in turn be for finite unions, however. constructed from a more basic object called a premea- This observation means that given a countable family sure. This is defined on an algebra, rather than a σ- E in , with E = E , we may assume that it is in- k C ∪ k algebra, of subsets of X, that is, a collection of subsets creasing. Now define E′ = E (B(x , n) B(x , n k,n k ∩ 0 \ 0 − of X containing the empty set and closed under finite and analogously . Since is increasing (in 1)) En′ µ(Ek,n′ ) unions and complementation. Let be an algebra of A k) and kEk,n′ = En′ has finite measure, there exists subsets of X. Then a premeasure µ0 : [0, ] is a ∪ n 1 a k(n) such that µ(E′ ) > µ(E′ ) ϵ2− − . We A → ∞ k(n),n n − function satisfying approximate E′ with a closed set Fn E′ k(n),n k(n),n (i) µ0( ) = 0; n 1 ⊂ to within ϵ2− − . Then the union of the Fn is closed ∅ (ii) If E1,E2,... is a countable collection of disjoint (since at most finitely many intersect in any given ball subsets of , and kEk , then B(x0, n), and F is closed iff F B(x0, n) is closed for A ∪ ∈ A ∩ µ E = µ (E ). all n) and approximates E within ϵ. 0 ∪n n 0 n □ n Given a premeasure( µ ,) we∑ define for any set E X 0 ⊂ ∞ µ (E) = inf µ0(Ej) E jEj, and each Ej . ∗ | ⊂ ∪ ∈ A n=1 { ∑ } (Observe the infimum is over a nonempty set, because X covers any set E.) ∈ A Theorem 7.6. The function µ is an exterior mea- To show that E is measurable, take any set A ∗ ∈ A ⊂ sure, such that X, and any covering Ej of A. Using finite additivity of

(i) µ (E) = µ0(E) for all E ; µ0 on , we have ∗ ⊂ A A (ii) Every A is measurable. c ⊂ A µ0(Ej) = µ0(Ej E) + µ0(Ej E ). Consequently, extends to a measure on the - ∩ ∩ µ0 σ Summing in j, we find that algebra generated by . A µ (E ) = µ (E E) + µ (E Ec) 0 j 0 j ∩ 0 j ∩ This, if you recall, was precisely how Lebesgue mea- j ∑ ∑ ∑ c sure was defined. µ (A E) + µ (A E ). ≥ ∗ ∩ ∗ ∩ Proof: It is straightforward to show that µ is an ex- Since this is true for every cover of A, we find that ∗ c terior measure. To show that µ (E) = µ0(E) for all µ (A) µ (A E) + µ (A E ), ∗ ∗ ≥ ∗ ∩ ∗ ∩ E , we take a cover of E by elements Ej . Then which shows that E is measurable since the inequality ∈ A k 1 ∈ A ≤ define E′ = E (Ek − Ej). Now E = E′ and the is obvious. k ∩ \∪j=1 ∪ k □ sets E′ are disjoint and all in . We then have µ (E) = k A 0 µ (E′ ), and therefore, µ (E) µ (E ), since j 0 j 0 ≤ j 0 j E′ E . Taking the inf of the right hand side shows ∑j ⊂ j ∑ that µ0(E) µ (E). The inequality is easy, by ≤ ∗ ≥ covering E using E . { }

For later use we note the following. Let be the Aσ collection of countable unions of elements of and let A denote the collection of countable intersections of Aσδ elements of . (Of course, if were actually a σ- Aσ A algebra, then we would have = = .) Then A Aσ Aσδ Proposition 7.7. For any set A, and any ϵ > 0, there exists an E such that A E and 1 ∈ Aσ ⊂ 1 µ (E1) < µ (A) + ϵ, and there exists E2 σδ such ∗ ∗ ∈ A that A E2 and µ (A) = µ (E2). ⊂ ∗ ∗ Exercise. What is the relation between the Borel sets of R and the Lebesgue measurable sets of R?