Measures Example 1.19 Let ( , E) be a measurabel space. If D is a paving on define X X ′ ′ ′ c Definition: A measure on ( , E) is a set-function µ : E [0, ] satisfying D = D P( ) D = D for D D . X → ∞ { ∈ X | ∈ } µ( ) = 0. ′ ∅ Then σ(D)= σ(D ). µ is σ-additive, that is, for any sequence A1, A2,... E of pairwise ∈ ′ k ′ disjoint sets (Ai Aj = for i = j) we have O is the paving of closed sets in R , then Bk = σ(O ), that is, the ∩ ∅ 6 k k Borel-algebra is also generated by the closed sets. ∞ ∞ µ An = µ(An). n=1 ! n=1 Actually, the Borel-algebra is generated by virtually any sensible paving. [ X Theorem 1.24 (and 1.22 for k = 1): The Borel-algebra, Bk, is generated by the open boxes, Ik, (the intervals for k = 1) The high-point is the existence of the Lebesgue measure, m, on the measurable space (R, B) – where B is the Borel-algebra. The Lebesgue (a ,b ) ... (ak,bk). 1 1 × × measure is uniquely determined by the specification that k In fact, Bk, is generated by the open boxes with rational corners, I . m((a, b)) = b a a<b. 0 − . – p.1/25 . – p.3/25 k Generated sigma-algebras Open sets are covered by I0 Lemma: With (Ei)i∈I a family of σ-algebras on then X x G = Ei G i∈I B(x, r) \ is a σ-algebra – the “minimum” of the σ-algebras Ei, i I. ∈ For any open set G Rk and any point x G, there is an ordinary ball of ⊂ ∈ radius r> 0 and center x such that B(x, r) G. Observation: The sigma-algebra generated by D is denoted σ(D) and is ⊂ defined as the smallest σ-algebra containing D – the “minimum” of all There exists an y Qk and an open box σ-algebras containing the paving D. ∈ x (y q, y + q) ... (y q, y + q) B(x, r) k 1 1 k k Let Ok denote the paving of open sets in R . ∈ − × × − ⊂ k Definition: The Borel-algebra, Bk, on R is the smallest σ-algebra for some q <r/(2√k). Hence containing the open sets, that is, Bk = σ(Ok). G = ∈Ik ⊂ J. ∪J 0 ,J G . – p.2/25 . – p.4/25 Constructable sets The Borel-algebra on Rk Definition: The enlargement of the paving D on is X ⋄ c D = D P( ) D = A or D = nAn for A, A1, A2,... D . { ∈ X | ∪ ∈ } P(Rk) ⋄ D⋄n = D⋄(n−1) , but D⋄n does not need to be a σ-algebra. Definition: The constructable sets are ∞ D⋄n. n=1 ∪ Ok Constructable Bad news: Obviously ∞ I⋄n B but in fact ∪n=1 0 ⊂ ∞ I⋄n = B. ∪n=1 0 6 Sets outside of the constructable sets are far outside of what we can possibly imagine! . – p.5/25 . – p.7/25 The Borel-algebra on Rk The Borel-algebra on Rk P(Rk) P(Rk) Bk O Ok Constructable . – p.6/25 . – p.8/25 Measures Measures on finite sets Let ( , E) be a measurabel space. X Let be a finite set with the σ-algebra P( ). X X Definition: A measure on ( , E) is a set-function µ : E [0, ] satisfying X → ∞ µ( ) = 0. Let p : [0, ] be a function and define ∅ X → ∞ µ is σ-additive, that is, for any sequence A1, A2,... E of pairwise µ(A)= p(x) for A ∈ ⊂X disjoint sets (Ai Aj = for i = j) we have x∈A ∩ ∅ 6 X ∞ ∞ Then µ is finitely additive, µ An = µ(An). n=1 ! n=1 [ X µ(A B)= p(x)= p(x)+ p(x)= µ(A)+ µ(B) ∪ x∈A∪B x∈A x∈B X X X k The Lebesgue measure, mk, on the measurable space (R , Bk) is if A B = . ∩ ∅ uniquely specified by requiring that for for ai <bi, i = 1,...,k, Since is finite, finite additivity and σ-additivity is the same – any infinite k X sequence of disjoint sets are from some point necessarily empty. mk((a1,b1) ... (ak,bk)) = (bi ai). × × − . – p.9/25 . – p.11/25 i=1 Y Sigma-additivity means: Some simple measures ① If µ(An)= for at least one n, then Definition: For an arbitray set define on P( ) the one-point measure, ǫx, ∞ X X in x by ∞ ∈X µ An = . 1 hvis x A ∞ ∈ n=1 ! ǫx(A)= [ 0 hvis x / A ∈ ∞ Then ǫ is a measure. ② If µ(An) < for all n while µ ( An)= , then x ∞ n=1 ∞ N S Definition: The empirical measure, ǫx1,...,xn , in x1,...,xn is defined as µ(An) for N . ∈X → ∞ → ∞ n=1 n X 1 ǫ (A)= ǫ (A) x1,...,xn n xi i=1 ③ ∞ X If µ(An) < for all n while µ ( n An) < , then ∞ =1 ∞ for A P( ). An empirical measure is a Probability measure since ∈ X N S∞ n µ(An) µ An for N . 1 → → ∞ ǫx1,...,xn ( )= 1 = 1. n=1 n=1 ! X n X [ . – p.10/25 i=1 . – p.12/25 X More examples Nullsets Definition: For an arbitrary set we define the counting measure X (tællemålet) τ on P( ) by X Consider a measure µ on ( , E). X τ(A)= number of elements in A Definition: A set A is a µ-nullset if there is a B E such that ⊂X ∈ k The Lebesgue measure, mk, on the measurable space (R , Bk) is A B, µ(B) = 0 ⊂ uniquely specified by requiring that k c mk((a ,b ) ... (ak,bk)) = (bi ai) Definition: A set A is µ-almost everywhere if A is a µ-nullset. 1 1 × × − ⊂X i=1 Y for ai <bi, i = 1,...,k, any finite real numbers. Existence of mk: A little hard and lengthy – not this course. Uniqueness of mk: Proved next week. – p.13/25 . – p.15/25 Lebesgue measure Union of nullsets Lemma: For all x Rk ∈ mk ( x ) = 0 { } Lemma: Let µ be a measure on ( , E). If A1, A2,... is a sequence of ∞ X µ-nullsets the n=1 An is a µ-nullset. Proof for k = 1: Observe that S ∞ 1 1 x = x , x + . Proof: For any A we have B E such that A B and µ(B ) = 0. { } − n n n n n n n n=1 ∈ ⊂ \ Hence ∞ ∞ Then downward continuity implies that An Bn. ⊂ n=1 n=1 1 1 2 [ [ m ( x )= lim m x , x + = lim = 0 According to Boole’s inequality { } n→∞ − n n n→∞ n ∞ ∞ µ Bn µ(Bn) = 0. Consequence: All intervals ≤ n=1 ! n=1 [ X (a, b) [a, b) (a, b] [a, b] have the same Lebesgue measure. – p.14/25 . – p.16/25 Lebesgue measure Cantors set k Q is a mk-nullset. Proof: Let Qk = x , x ,..., . Using Boole’s inequality { 1 2 } ∞ ∞ k mk(Q )= mk xn mk( xn ) = 0. { } ≤ { } n=1 ! n=1 [ X All lower-dimensional sets in Rk have Lebesgue measure 0. Closed and open boxes have the same mk-measure; k mk([a ,b ] ... [ak,bk]) = (bi ai) 1 1 × × − i=1 Y and so do all semi-closed boxes. – p.17/25 . – p.19/25 Cantors set Cantors set Divide a closed interval – the unit interval [0, 1], say – into three equal parts. Remove the central open part. Repeat the procedure on the remaining two intervals. Continue forever. The Cantor set is what is left when we are done. C . – p.18/25 . – p.20/25 Cantors set Cantors set . – p.21/25 . – p.23/25 Cantors set Cantors set . – p.22/25 . – p.24/25 Cantors set In the n’th step we have a collection of 2n−1 disjoint, closed intervals each of length 3−(n−1). ∞ Let Cn denote the union of these intervals, then = Cn. C n=1 T Remark: Since Cn is closed, is closed. Especially, B. C C ∈ Lemma: m( ) = 0. C Proof: n−1 n−1 n−1 1 2 m( ) m(Cn) = 2 = 0 for n C ≤ 3 3 → → ∞ Remark: There exists a bijective map from Cantors set to the set of 0-1 sequences – hence the Cantor set is uncountable. – p.25/25.