CHAPTER 2. THE LEBESGUE INTEGRAL I 11 2.1.3 Outer measures and construction of measures Recall the construction of Lebesgue measure on the real line. There are various approaches to construct the Lebesgue measure on the real line. For example, consider the collection of finite union of sets of the form ]a; b]; ] − 1; b]; ]a; 1[ R: This collection is an algebra A but not a sigma-algebra (this has been discussed on p.7). A natural measure on A is given by the length of an interval. One attempts to measure the size of any subset S of R covering it by countably many intervals (recall also that R can be covered by countably many intervals) and we define this quantity by 1 1 X [ m∗(S) = inff jAkj : Ak 2 A;S ⊂ Akg (2.7) k=1 k=1 where jAkj denotes the length of the interval Ak. However, m∗(S) is not a measure since it is not additive, which we do not prove here. It is only sigma- subadditive, that is k k [ X m∗ Sn ≤ m∗(Sn) n=1 n=1 for any countable collection (Sn) of subsets of R. Alternatively one could choose coverings by open intervals (see the exercises, see also B.Dacorogna, Analyse avanc´eepour math´ematiciens)Note that the quantity m∗ coincides with the length for any set in A, that is the measure on A (this is surprisingly the more difficult part!). Also note that the sigma-subadditivity implies that m∗(A) = 0 for any countable set since m∗(fpg) = 0 for any p 2 R. In order to get a measure from m∗ one then restricts m∗(S) to a smaller class of subsets which forms a sigma-algebra and contains A where it has the desired property of sigma-additivity. In the following we describe this procedure in an abstract setting. Definition - outer measure. Let Ω be a nonempty set. An outer measure is a set function m∗ : P(Ω) ! R+ [ f1g satisfying the following conditions: m∗(;) = 0 (2.8) A ⊂ B ) m∗(A) ≤ m∗(B) (2.9) 1 1 [ X m∗ An ≤ m∗(An): (2.10) n=1 n=1 For example, m∗ given by (2.7) is an outer measure on P(R). We shall show in the theorem below that in an appropriate abstract setting the set functions as defined by (2.7) are always outer measures. Note, that a covering by a finite number of intervals or more generally cubes or rectangles which gives rise to so called outer Jordan content will not suffice in the definition of an outer measure. For example, the rational numbers in [0; 1] have outer Jordan content 1 but the outer measure m∗ is zero since the outer measure of a single point is equal to zero and the rational numbers are countable. CHAPTER 2. THE LEBESGUE INTEGRAL I 12 Construction of an outer measure from a pre-measure defined on an algebra. Let A ⊂ P(Ω) an algebra. Suppose there is a function m0 : A! R [ f1g with the following properties: 1. m0(;) = 0, S 2. if A = k≥1 Ak 2 A for some countable collection of disjoint Ak 2 A, then X m0(A) = m0(Ak): k≥1 Such a function is called a pre-measure. In many applications we also have the S property that Ω = k≥1 Ak for some countable collection of disjoint Ak 2 A. If m0(Ak) < 1 for all k we call m0 a sigma-finite pre-measure. If m0(Ω) < 1 we call m0 a finite pre-measure. For any S ⊂ Ω define 1 1 X [ m∗(S) = inff m0(Ak): Ak 2 A;S ⊂ Akg: (2.11) k=1 k=1 Theorem. m∗ is an outer measure and m∗(A) = m0(A) for any A 2 A. Proof. Obviously m∗(;) = 0 and if A ⊂ B, then any covering of B is also a covering of A. Hence m∗(A) ≤ m∗(B). In order to prove the sigma-subadditivity let > 0 be arbitrary. We choose for each Ak a covering Ckn 2 A such that 1 1 [ X −k Ak ⊂ Ckn; m0(Ckn) ≤ m∗(Ak) + 2 . n=1 n=1 S1 Now fCkn; k 2 Z+; n 2 Z+g is a countable covering of k=1 Ak by elements of A and therefore it follows from the definition of m∗ that 1 1 1 1 1 [ X X X −k X m∗ Ak ≤ m(Ckn) ≤ m∗(Ak) + 2 ≤ + m∗(Ak): k=1 k=1 n=1 k=1 k=1 To prove that m∗(A) = m0(A) for any A 2 A we note that for any sequence Cn S1 S1 of sets in A such that A ⊂ n=1 Cn we have A = n=1(Cn \ A) and therefore using the fact that m0 is also countably subadditive for general unions in A and monotone on A 1 1 X X m0(A) ≤ m0(Cn \ A) ≤ m0(Cn) n=1 n=1 proving m0(A) ≤ m∗(A) The reversed inequality is obvious since (A; ;; ;;:::) is a countable cover of A and hence m∗(A) ≤ m0(A). A toy example - pre-measure and construction of outer measure. Let Ω = fa; b; cg, A = f;; fag; fb; cg; Ωg, α; β 2 [0; 1] and define a pre-measure m0 (prove this fact!) by m0(;) = 0; m0(fag) = α; m0(fb; cg) = β; m0(Ω) = α + β CHAPTER 2. THE LEBESGUE INTEGRAL I 13 Then m∗(;) = 0; m∗(fag) = α; m∗(fbg) = β; m∗(fcg) = β and m∗(Ω) = α + β; m∗(fa; bg) = α + β; m∗(fa; cg) = α + β; m∗(fb; cg) = β It is easily seen that m∗jA = m0 and that in general m∗ is not additive, that is m∗ is not a measure on P(Ω). Indeed, if β > 0 , then m∗(fb; cg [ fbg) < m∗(fbg) + m∗(fcg). On the other hand, when β = 0 or β = 1 then m∗ defines a measure on P(Ω). An alternative setting for the construction of outer measures. Since in some situations it may be difficult to find an appropriate algebra (or a pre- measure) we present another setting for constructing outer measures. Let Ω be a nonempty set and E ⊂ P(Ω) be a collection of subsets of Ω such that ;; Ω 2 E. Let m0 : E −! [0; 1] a set function such that m0(;) = 0 (so despite its notation m0 is not necessarily a pre-measure or measure defined on E. As before for any S ⊂ Ω define 1 1 X [ m∗(S) = inff m0(Ek): Ek 2 E;S ⊂ Ekg: (2.12) k=1 k=1 Then m∗ is an outer measure and m∗(E) ≤ m0(E) for all E 2 E (the same proof as above). However, in order to prove m∗(E) = m0(E) we need more restrictive hypotheses on E and m0. Example - Outer measure on Rd. Let E the collection of all countable coverings by (closed) cubes Ck with edges parallel to the axes and let jCkj denote the usual d-dimensional size of a cube. The function 1 X m∗(S) = inf jCkj k=1 where the infimum is taken over all sets in is E an outer measure on Rd. This construction differs from the general approach assuming the sigma-additivity on an algebra for countable unions which are in the algebra but corresponds to the alternative setting. We only have finite additivity but the local compactness of d R (or the compactness of closed cubes) guarantees that m∗(A) = jAj for any closed cube A: Indeed, first note that A covers itself hence m∗(A) ≤ jAj. Then consider any countable covering of A. As before we want to prove that 1 X jAj ≤ jCnj: n=1 For an arbitrary > 0 we choose for each n an open cube On which contains Cn and such that jOnj ≤ (1 + )jCnj. Since A is compact we may select a finite subcovering of A. Consequently, N 1 X X jAj ≤ (1 + ) jCnk j ≤ (1 + ) jCnj: k=1 n=1 The monotonicity and the countable subadditivity are shown as above for an abstract pre-measure. CHAPTER 2. THE LEBESGUE INTEGRAL I 14 Definition - m∗-measurable sets. A set A ⊂ Ω is called m∗-measurable if and only if m∗(E) = m∗(E \ A) + m∗(E n A) for every E ⊂ Ω where E n A = E \ Ac. Note that by the monotonicity of the outer measure we always have the inequality with the ≤ sign. Let Σ∗ denote the family of measurable sets. Sets with outer measure equal to zero are always m∗-measurable. When the outer measure is constructed from a pre-measure defined on an algebra A, then all sets A 2 A are measurable. The notion of m∗- measurable sets filters out sets for which the outer measure m∗ is not additive (for disjoint unions) as we see from our previous toy-example. A toy example - m∗-measurable sets. Recall that Ω = fa; b; cg, A = f;; fag; fb; cg; Ωg, α; β 2 [0; 1] and a pre-measure m0 defined by m0(;) = 0; m0(fag) = α; m0(fb; cg) = β; m0(Ω) = α + β We had seen that when β > 0 , m∗(fb; cg [ fbg) < m∗(fbg) + m∗(fcg). Hence fb; cg is not m∗-measurable. In this case a set is m∗-measurable if and only if it is in the algebra A. Carath´eodory's theorem. With the above notations Σ∗is a sigma algebra and the restriction of m∗ to Σ∗ is a measure µ.