IOSR Journal of Mathematics (IOSR-JM) e-ISSN: 2278-5728, p-ISSN: 2319-765X. Volume 12, Issue 1 Ver. II (Jan. - Feb. 2016), PP 34-44 www.iosrjournals.org Extending Baire Measures To Regular Borel Measures Parvinder Singh PG Department of Mathematics, S.G.G.S. Khalsa College, Mahilpur. (Punjab). Abstract: It should be noted that a Borel measure may not be determined by its values on compact 퐺훿 sets. But if some of the conditions are imposed on a Borel measure so that it can be determined by its values on compact 퐺훿 sets. The answer to this question is Regularity. We can discus as in this paper by proving the result that a Borel measure is determined by its values on compact 퐺훿 sets and further that every Baire measure has a unique extension to a regular Borel measure. Definition: Let X be any set, be any 휎-ring on X, 풞 and 풰 be any subclasses of . (1) 휇 be any measure on i.e. (X, , 휇) is a measure space then we say (X, , 휇, 풞, 풰 ) satisfies axiom I. (2) 퐼푓 풞 is closed for finite unions, countable intersections, 휙 ∈ 풞 and 휇 푐 < ∞ ∀ c ∈ 풞, then we say that axiom II is satisfied. i.e. we say (X, , 휇, 풞, 풰 ) satisfies axiom II. (3) 퐼푓 풰 is clsed for countable unions, finite intersections and for every E ∈ there exist U ∈ 풰 s.t. E ⊂ U,then we say that (X, , 휇, 풞, 풰 ) satisfies axiom III. Definition: Suppose that (X, , 휇, 풞, 풰 ) satisfy axiom I, II and III. Let E ∈ (1) If 휇 퐸 = inf{ 휇 푈 / E ⊂ U ∈ 풰}, then E is said to be Outer regular. (2) If 휇 퐸 = sup{ 휇 퐶 / E ⊃ C ∈ 풞 }, then E is said to be Inner regular. (3) The set E is said to be Regular if it is Outer regular as well as Inner regular. (4) The measure 휇 is called Regular if every measurable set E in is Regular. Preposition: Suppose that (X, , 휇, 풞, 풰) satisfies axiom I, II and III. Let E ∈ then (1) E is Outer regular iff for every 휀 > 0 there exist U ∈ 풰 s.t. E ⊂ U and 휇 푈 ≤ 휇 퐸 + 휀. (2) If for every 휀 > 0 there exist C ∈ 풞 s.t. C ⊂ E and 휇 퐸 ≤ 휇 퐶 + 휀, then E is called Inner regular. (3) If E is Inner regular and 휇 퐸 < ∞ then for each 휀 > 0 there exist C ∈ 풞, s.t. C ⊂ E and 휇 퐸 ≤ 휇 퐶 + 휀. Proof: (1) Suppose E is Outer regular and 휀 > 0. Let 휇 퐸 = ∞, By axiom III there exist U ∈ 풰 s.t. E ⊂ U ⇒ 휇 푈 ≥ 휇 퐸 = ∞ ⇒ 휇 푈 = ∞ ⇒ 휇 푈 ≤ 휇 퐸 + 휀. Now suppose that 휇 퐸 < ∞, then we have 휇 퐸 ≤ 휇 퐸 + 휀 and 휇 퐸 = inf{ 휇 푈 / E ⊂ U ∈ 풰} then by definition of infimum there exist U ∈ 풰 s.t. E ⊂ U and 휇 푈 ≤ 휇 퐸 + 휀, shows that the condition is necessary. Conversely: Assume that the condition is satisfied. To show that E is Outer regular, Let n be any natural number, By the condition taking 휀 = 1 we have 푈 ∈ 푛 n 풰 푠. 푡. E ⊂ 푈 and 휇 푈 ≤ 휇 퐸 + 1. n n 푛 n Let Vn = ⋂ 푈i then Vn ∈ 풰, (Vn) is a decreasing sequence and i=1 휇 퐸 + 1 ≥ 휇 푈 ≥ 휇 푉 ∀ n 푛 n n 1 Therefore lim{ 휇 퐸 + } ≥ lim 휇 푈n 푛→∞ 푛 푛→∞ ⇒ 휇 퐸 ≥ inf{ 휇 푈n } ≥ inf { 휇 푉n / E ⊂ V, V ∈ 풰 } ……….(*) On the other hand 휇 퐸 ≤ 휇 푉 for all V s.t. E ⊂ V, V ∈ 풰. ⇒ 휇 퐸 ≤ inf{ 휇 V /E ⊂ V, V ∈ 풰 } …………(**) From (*) and (**) we have 휇 퐸 = inf { 휇 V /E ⊂ V, V ∈ 풰 }, shows that E is Outer regular. (2) And (3) can be proved similarly by using the definition of Supremum. Preposition: (1) If 휇 퐸 = ∞ then E is Outer regular. (2) Every member of 풰 is Outer regular. ∞ (3) If V = ⋂ 푈n, 푈n ∈ 풰, 휇 푈 < ∞ then V is Outer regular. n=1 Proof: (1) Let 휇 퐸 = ∞ and U ∈ 풰 s.t. E ⊂ U, then 휇 푈 = ∞ DOI: 10.9790/5728-12123444 www.iosrjournals.org 34 | Page Extending Baire Measures To Regular Borel Measures ⇒ inf⁡{ 휇 U /E ⊂ U, U ∈ 풰 } = ∞ ⇒ 휇 퐸 = inf 휇 U /E ⊂ U, U ∈ 풰 . Then E is Outer regular. (2) Let W ∈ 풰 then 휇 푊 ≥ inf⁡{휇 U /W ⊂ U, U ∈ 풰} Let U ∈ 풰 s.t. W ⊂ U then 푊 ≤ 휇 U ⇒ 휇 푊 ≤ inf⁡{휇 U /W ⊂ U, U ∈ 풰} ⇒ 휇 푊 = inf⁡{휇 U /W ⊂ U, U ∈ 풰}, shows that W is Outer regular. (3) Let (푈n ) be any sequence of members of 풰 s.t. 휇 푈1 < ∞ ∞ n ∞ ∞ Let V = ⋂ 푈n, Defne 푉n = ⋂ 푈i then 푉n ∈ 풰 and (푉n ) ↓ ⋂ 푉n = ⋂ 푈n=V n=1 i=1 n=1 n=1 ⇒ 휇 V = lim푛→∞ 휇 푉n = inf{ 휇 푉n } ≥ inf{ 휇 U / V ⊂ U ∈ 풰} ⇒ 휇 V ≥ inf{ 휇 U / V ⊂ U ∈ 풰} … … … . (∗) Let U be any member of 풰 s.t. V ⊂ U then 휇 V ≤ 휇 U ⇒ 휇 V ≤ inf{ 휇 U / V ⊂ U ∈ 풰} ……….…..(**) From (*) and (**) we get, 휇 V = inf{ 휇 U / V ⊂ U ∈ 풰} Shows that V is Outer regular. Preposition: (1) If 휇 E = 0 then E is Inner regular. (2) Every member of 풞 is Inner regular. (3) Countable unions of members of 풞 is Inner regular. Proof: (1) Let 휇 E = 0 and C ∈ 풞 and C ⊂ E then 휇 C ≤ 휇 E ⇒ 휇 C = 0 ⇒Sup{ 휇 C / C ⊂ E , C ∈ 풞} = 0 ⇒ 휇 E = Sup{ 휇 C / C ⊂ E , C ∈ 풞}. Hence E is Inner regular. (2) Let D∈ 풞 then D ⊆ D, and D ∈ 풞 Therefore Sup{ 휇 C / C ⊂ D , C ∈ 풞} ≥ 휇 D Also C ⊂ D ⇒ 휇 C ≤ 휇 D ⇒ Sup{ 휇 C / C ⊂ D , C ∈ 풞} ≤ 휇 D Therefore 휇 D = Sup{ 휇 C / C ⊂ D , C ∈ 풞} ⇒ D is Inner regular. ∞ n (3) Let (퐷n) be a sequence of members of 풞 and D = ∪ 퐷n, Define 퐶n= ∪ 퐷j then 퐶n ∈ 풞 for all n and (퐶n) n=1 j=1 ∞ ∞ is monotone increasing sequence with ∪ C= ∪ 퐷n= D n=1 n=1 ⇒(퐶n) ↑ D⇒ 휇 퐶n → 휇 D ⇒ 휇 D = lim푛→∞ 휇 Cn = Sup{ 휇 Cn }≤ Sup{ 휇 C / C ⊂ D , C ∈ 풞} ⇒ 휇 D ≤ Sup{휇 C / C ⊂ D , C ∈ 풞} On the other hand if C ⊂ D , C ∈ 풞 then 휇 C ≤ 휇 D ⇒ Sup{ 휇 C / C ⊂ D , C ∈ 풞} ≤ 휇 D ⇒ 휇 D = Sup{ 휇 C / C ⊂ D , C ∈ 풞} ⇒ D is Inner regular. Hence the proof. Theorem: Countable union of outer regular sets is outer regular. ∞ Proof: Let (En) be any sequence of outer regular sets and E = ∪ 퐸n n=1 If 휇 E = ∞, then E is outer regular as proved earlier. Now suppose that 휇 E < ∞, Let 휀 > 0, since En are outer regular we can find a set U ∈ 풰 s.t. E ⊂ U , and 휇 U ≤ 휇 E + 휀 n n n n n 2푛 ∞ Let U = ∪ 푈n, then U ∈ 풰 and E ⊂ U then n=1 ∞ ∞ 휇 U − E = 휇 ( ∪ 푈n ) − ( ∪ 퐸n) n=1 n=1 ∞ ∞ ∞ ≤ 휇 ∪ (푈n − 퐸n) ≤ 푛=1 휇 푈n − 퐸n = [휇 푈n) − 휇(퐸n ] [Because 휇 En < ∞] n=1 푛=1 ∞ ≤ [ 휀 ] = 휀 푛=1 2푛 ⇒ 휇 U − E ≤ 휀 ⇒ 휇 U) ≤ 휇 E + 휀 ⇒ E is outer regular. Theorem: Finite union of outer regular sets is outer regular. k Proof: Let 퐸1,퐸2, … … … . , 퐸k be k outer regular sets and let E= ∪ 퐸i, i=1 Define 퐸n = 퐸k for n > k. Then (퐸n) is a sequence of outer regular sets ∞ k and ∪ 퐸n = ∪ 퐸n = E. n=1 n=1 Hence E is outer regular set by the proceeding theorem. DOI: 10.9790/5728-12123444 www.iosrjournals.org 35 | Page Extending Baire Measures To Regular Borel Measures Theorem: Finite intersection of outer regular sets of finite measure is outer regular. Proof: Suppose E and F are outer regular sets and 휇 E < ∞푎푛푑 휇 F < ∞. Let 휀 > 0 푏푒 given, by the outer regularity of E, we can find a set U ∈ 풰 s.t. 퐸 ⊂ U and 휇 U ≤ 휇 E + 휀 2 휀 Similarly we can find a set V ∈ 풰 s.t. 퐹 ⊂ V and 휇 V ≤ 휇 F + 2 Then U∩V ∈ 풰, E ∩ F ⊂ U∩V and 휇[ U ∩ V − E ∩ F ] ≤ 휇[(푈 − 퐸) ∪ (V − F)] ≤ 휇 푈 − 퐸 + 휇 V − F = [ 휇 푈) − 휇(퐸 ] + [휇(V) − 휇(F)]= 휀 + 휀 = 휀 2 2 ⇒ 휇[ U ∩ V − E ∩ F ] ≤ 휀 ⇒ E ∩ F is outer regular. Theorem: The countable intersection of outer regular sets of finite measure is outer regular. ∞ Proof: Let (En) be any sequence of outer regular sets of finite measure and E = ∩ 퐸n . n=1 To show that E is outer regular. n ∞ Let 휀 > 0, Define 퐹n = ∩ 퐸j , then (퐹n) is a decreasing sequence of outer regular sets and lim푛→∞ Fn = ∩ 퐹n= j=1 n=1 ∞ ∩ 퐸n = E. Thus (퐹n) ↓ E and 휇(퐹i) < ∞ for all i. n=1 By continuity of measure for decreasing sequences we obtain that 휇 E = lim푛→∞ 휇 Fn i.e. 휇 Fn → 휇 E ⇒ There exist k s.t. 휇 F ≤ 휇 E + 휀 , Since F is outer regular and 휇(퐹 ) < ∞, we can find k 2 k k 휀 U ∈ 풰 s.t. F ⊂ U and 휇 U ≤ 휇 F + , k k 2 Thus E ⊂ U and 휇 U − E = 휇[ U − Fk ∪(Fk − E)] ≤ 휇 U − F + 휇(F − E) ≤ 휇 U) − 휇(F + 휇(F − 휇(E) = 휀 + 휀 = 휀 k k k k) 2 2 ⇒ 휇 U − E < 휀, Shows that E is outer regular.