MA40042 Measure Theory and Integration 4.2 Set-up Let X be a nonempty set. We start with Lecture 4 • A collection R of subsets of X. We shall assume throughout that ? 2 R. Constructing measures I: • A function ρ: R! [0; 1]. We shall assume throughout that ρ(?) = 0. The idea is that R is a collection of `simple' sets, and we want the measure of Outer measure R 2 R to be ρ(R). We then need to `grow' ρ into a measure, by defining a measure on some σ-algebra on X that has been built up from R and ρ. An important case is that of the Lebesgue measure. On X = R, we take R to • How to construct measures be the collection of intervals I, and define the length ρ by • The Lebesgue measure ρ(?) = 0 ρ [a; b) = b − a ρ [−∞; b) = ρ [a; 1) = ρ(R) = 1: on intervals and boxes d On X = R , we take R to be the collection of interval boxes Id and ρd to be the • Constructing an outer measure (hyper)volume ρd(I1 × I2 × · · · × Id) = ρ(I1)ρ(I2) ··· ρ(Id): • Properties of outer measures (Remember our convention that 0 × 1 = 0.) In particular, a Cartesian product of semi-open intervals has volume d ! d Y Y 4.1 How to construct measures: a roadmap ρd [ai; bi) = (bi − ai): So far we have only seen simple examples of measures, such as the counting measure i=1 i=1 or some discrete measures. To create more powerful measures, such as the Lebesgue (As ever, we suppress the dimension d when it's clear by context.) measure of length/area/volume, we have to build them up from basic pieces. Suppose we are working on a set X. 4.3 Constructing µ∗ 0. We begin with a collection R of subsets of X, and a function ρ on R. The Given X, R, and ρ, we will construct a function µ∗ : P(X) ! [0; 1], which gives idea is that R is the collection of sets we `know', and, for R 2 R, we `want' a concept of size to all subsets of X. We shall see later that µ∗ has many of the the measure of R to be ρ(R). properties of a measure. 1. The first step is to construct an outer measure µ∗ from R and ρ. This gives The idea is that, for any set A ⊂ X, we approximate A from the outside by a a `size' to all subsets of X, and has many of the properties of a measure. countable union of sets from R, and we know what we want their sizes to be. 2. The outer measure µ∗ is (in general) not a measure. But if we restrict ourselves Definition 4.1. Let X be a nonempty set, R a collection of subsets containing to the σ-algebra of measurable sets satisfying a certain `splitting condition', the empty set, and ρ a function from R to [0; 1] satisfying ρ(?) = 0. this does give a measure µ on those measurable sets. Given a countable subcollection C ⊂ R, we write X 3. The measure µ does not necessarily `extend' (that is, agree with) our original ρ(C) = ρ(R): set-up: we can't guarantee that every R 2 R is a measurable set, and even if R R2C is measurable, there's no guarantee that µ(R) = ρ(R). However, Carath´eo- We call C a covering of a set A ⊂ X if A ⊂ S R. dory's extension theorem tells us that if R and ρ have certain properties, R2C We define a function µ∗ : P(X) ! [0; 1] by then µ is a proper extension of ρ. In certain other conditions, the µ is the unique extension of ρ. µ∗(A) := inf fρ(C): C is a covering of Ag : (By convention, inf ? = 1, meaning we take µ∗(A) = 1 if there is no covering of Proof of Theorem 4.2. For the first point, note that one covering of ? is C = f?g, A.) and ρ(C) = 0. No covering has a negative size, so µ∗(?) = 0. Monotonicity is immediate, since any covering of B is also a covering of A. The infimum in the definition means we want to take the tightest outer approx- ∗ Now for countable subadditivity. First note that if for some An we have µ (An) = imation { or at least the limit of increasingly tight approximations { of the set ∗ 1, then there's nothing to prove. So assume µ (An) is finite for all n. A. Since the definition of µ∗ includes an infimum, we will want to show that, for arbitrary > 0, 1 ! ∗ [ X ∗ µ An ≤ µ (An) + , n=1 n which would prove the theorem. Let's pause for a moment, and look at a failed attempt to prove the statement. By definition, for each An there exists a covering Cn of An with ∗ ρ(Cn) ≤ µ (An) + . S1 S1 Now the union C = n=1 Cn is countable and is a covering of n=1 An. Hence 1 ! 1 1 1 ∗ [ X X ∗ X ∗ µ An ≤ ρ(C) ≤ ρ(Cn) ≤ µ (An) + = µ (An) + 1 × . When R = I and ρ is the length/volume as above, we write λ∗ for the function n=1 n=1 n=1 n=1 obtained from this covering construction, and call it the Lebesgue outer measure. Had we been left with `5' at the end, we go back to the proof and replace all our `'s with `/5's. But we can't go back and divide through by 1. But, we can 4.4 Properties of outer measures use `the /2n trick'. The function µ∗ has many of the properties of a measure. Proof continued. By definition, for each An there exists a covering Cn of An with ∗ ∗ Theorem 4.2. Let X, R, ρ, and µ be as above. Then ρ(Cn) ≤ µ (An) + : 2n 1. µ∗( ) = 0 S1 S1 ? Now the union C = n=1 Cn is countable and is a covering of n=1 An. Hence 2. µ∗ is monotone, in that for A ⊂ B ⊂ X we have µ∗(A) ≤ µ∗(B). 1 ! 1 1 1 ∗ [ X X ∗ X ∗ µ An ≤ ρ(C) ≤ ρ(Cn) ≤ µ (An) + n = µ (An) + , ∗ 2 3. µ is countably subadditive, in that for a countably infinite sequence A1, n=1 n=1 n=1 n=1 A ;::: of subsets of X, we have 2 since 1 X 1 1 1 ∗ [ X ∗ = + + + ··· = × 1 = . µ An ≤ µ (An): 2n 2 4 8 n=1 n n The statement is proved. Definition 4.3. Any function that satisfies points 1, 2, and 3 from the previous theorem is called an outer measure on X. Matthew Aldridge Note that `an outer measure' is any function satisfying these properties, and the [email protected] construction of µ∗ above is one way of forming an outer measure..