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Measures and Measure Spaces Chapter 2 Measures and Measure Spaces In summarizing the flaws of the Riemann integral we can focus on two main points: 1) Many nice functions are not Riemann integrable. 2) The Riemann integral does not behave well with respect to limits of sequence of functions. That is, we cannot interchange the limit and the integral. In the core part of this course we will see how it is possible to develop a theory of integration that goes a long way to repair these flaws. To do so we will start by looking at how we go about defining an abstract notion of length for sets. 2.1 Algebras and σ-algebras Definition 2.1.1. Let X be a non-empty set. We denote by P(X) = fA j A ⊆ Xg: P(X) is called the power set of X An algebra of subsets of X is a collection A ⊆ P(X) such that 1. ; 2 A 2. E1;E2 2 A implies E1 [ E2 2 A 3. E 2 A implies Ec = X n E 2 A A is said to be a σ-algebra if 1) ; 2 A 1 S1 2) fEngn=1 ⊆ A implies n=1 En 2 A 3) E 2 A implies Ec = X n E 2 A Remark 2.1.2. 1) Every σ-algebra is an algebra. c c c 2) E1 \ E2 = (E1 [ E2) , so algebras are closed under intersection as well. A similar use of DeMorgan's Laws shows that a σ-algebra is closed under countable intersections. 3) Let A be an algebra. Let fFng ⊆ A. Let E1 = F1, E2 = F2 n F1, E3 = F3 n (F1 [ F2) and then proceed recursively to define En+1 = Fn+1 n (F1 [ F2 [···[ Fn). Then fEng ⊆ A, fEng is pairwise disjoint and 1 1 [ [ En = Fn: n=1 n=1 16 4) Let A ⊂ P(X) be an algebra (σ-algebra) and let E 2 A. Let AjE = fA \ EjA 2 Ag: Then AjE is an algebra (σ-algebra) in P(E). Example 2.1.3. P(X) is a σ-algebra. T Proposition 2.1.4. If fAαgα2I is a collection of algebras (resp. σ-algebras) then α2I Aα is an algebra (resp. σ-algebra). Proof. This proof is left as an exercise. The previous proposition leads us immediately to the following important observation. Observation. Given any set S ⊆ P(X) then there exists a smallest algebra (resp. σ-algebra) that contains S, namely \ fA j S ⊆ Ag: Notation 2.1.6. Given S ⊆ P, let A(S), the algebra generated by S, be the smallest algebra containing S, and σ(S), the σ-algebra generated by S, be the smallest σ-algebra containing S. Example 2.1.7. The collection of all finite unions of sets of the form fR; (−∞; b]; (a; b]; (a; 1)g is an algebra, but it is not a σ-algebra. Definition 2.1.8. Let S = fU ⊆ R j U is openg. The σ-algebra generated by S is called the Borel σ-algebra of R and is denoted B(R). More generally, we may take the Borel σ-algebra of any topological space (X; τ) which we denote by B(X). Of course open and closed sets are certainly the most important elements elements of B(X). However, there are other classes of sets in B(X) which also play an important role. In particular, a set A ⊆ X is called T1 S1 Gδ if A = n=1 Un where each Un is open. A set A ⊆ X is called Fσ if A = n=1 Fn where each Fn is S1 T1 closed. A set is A ⊆ X called Gδσ if A = n=1 An where each An is Gδ and it is Fσδ if A = n=1 An where each An is Fσ. Additional subscripts may be appended in the most obvious way. Remark 2.1.9. The Gδ sets are exactly the complements of the Fσ sets, and vice versa. Notice that in a metric space (X; d) closed sets are always Gδ. To see why suppose that F is closed. For each n 2 N let [ 1 U = B(x; ): n n x2F where B(x; r) = fy 2 X j d(x; y) < rg. Then Un is open and 1 \ F = Un: n=1 Problem 2.1.10. Are there sets in R which are neither Fσ or Gδ? The next fact, which is proved using transfinite induction, shows that the answer to the previous problem is clearly yes. 17 Fact 2.1.11. jB(R)j = c, the cardinality of R. Problem 2.1.12. Since singletons are closed in R it is clear that every countable set is an Fσ set. But is Q a Gδ set in R ? To answer this problem we first recall the notion of category for a topological space. Definition 2.1.13. A set A in a topological space (X; τ) is said to be nowhere dense if A has empty interior. It is of first category in (X; τ) if 1 [ A = An n=1 where An is nowhere dense for each n 2 N. A is of second category in (X; τ) if it is not of first category. A is residual if Ac is of first category. 1 Theorem 2.1.14. [Baire's Category Theorem]. Let X be a complete metric space. If fUngn=1 is a T1 collection of dense open sets then n=1 Un is dense. Proof. Let W be open and non-empty. Then there exists an x1 2 X and 0 < r1 ≤ 1 such that B(x1; r1) ⊆ B[x1; r1] ⊆ W \ U1: 1 Next we can find x2 2 X and 0 < r2 < 2 such that B(x2; r2) ⊆ B[x2; r2] ⊆ B(x1; r1) \ U2: 1 We can then proceed recursively to find sequences fxng ⊆ X and frng ⊂ R with 0 < rn ≤ n , and B(xn+1; rn+1) ⊆ B[xn+1; rn+1] ⊆ B(xn; rn) \ Un+1: Since rn ! 0 and B[xn+1; rn+1] ⊆ B[xn; rn], Cantor's Intersection Theorem implies that there exists an 1 \ x0 2 B[xn; rn:] n=1 But then x0 2 B[x1; r1] ⊆ W and x0 2 B[xn; rn] ⊆ Un for each n 2 N. This shows that 1 \ x0 2 W \ ( Un): n=1 Remark 2.1.15. The Baire Category Theorem shows that if fUng is a sequence of open dense sets, then 1 1 T T Un is dense in X. We also know that Un is a Gδ. These dense Gδ subsets of a complete metric space n=1 n=1 are always residual, as we will see below, and as such are topologically fat. Our next corollary shows the connection between the Baire Category Theorem and our notion of category. Corollary 2.1.16. [Baire Category Theorem II] Every complete metric space (X; d) is of second category in itself. 18 Proof. Assume that X is of first category. Then there exists a sequence An of nowhere dense sets so that 1 1 [ [ X = An = An: n=1 n=1 c Now let Un = (An) . Then Un is open and dense. But 1 \ Un = ; n=1 which is impossible. Corollary 2.1.17. Q is not Gδ in R. T1 S1 c Proof. Assume that Q = n=1 Un, where Un is open for each n. Then R=Q = n=1 Fn, where Fn = Un 0 is closed and nowhere dense. Let Q = fr1; r2;:::g. Then Fn = Fn [ frng is closed and nowhere dense and S1 0 R = n=1 Fn. This is a contradiction since R is not of first category. The next example is a set in R which is not Borel. Observe that the construction of this set requires the Axiom of Choice. While this is not necessary to construct non-Borel sets it turns out that this set is even more pathological than simply being non-Borel. We will see later that it is an example of a set that is also not Lebesgue measurable. Example 2.1.18. Consider [0; 1]. Define an equivalence relation on [0; 1] by x ∼ y if x − y 2 Q. Use the Axiom of Choice to choose one element from each equivalence class and denote this set by S. We will show later that S is not Borel. Remark 2.1.19. There is nothing special about using the open sets of R to define B(R). B(R) = σf(a; b) j a; b 2 Rg = σf(a; b] j a; b 2 Rg = σf[a; b) j a; b 2 Rg = σf[a; b] j a; b 2 Rg 2.2 Measures and Measure Spaces In trying to find a notion of length or measure for an arbitray subset A ⊆ R we are looking for a function m : P(R) ! R∗ = R [ {±∞} with the following properties: 1. m(I) = the usual length of I for all intervals I 2. If fAng is a sequence of pairwise disjoint subsets of R, then 1 1 [ X m( An) = m(An): n=1 n=1 3. m(x + A) = m(A) for all A 2 P(R) and x 2 R. The problem is that no such measure exists. If we weaken our conditions and only look for a measure on some sufficiently large sub-σ-algebra A of P(R) then we may succeed. More specifically, we would like A to contain B(R). The measure function that we get is called the Lebesgue measure on R.
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