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Measure Theory 1 Measurable Spaces

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Measure Theory 1 Measurable Spaces Measure Theory 1 Measurable Spaces A measurable space is a set S, together with a nonempty collection, , of subsets of S, satisfying the following two conditions: S 1. For any A; B in the collection , the set1 A B is also in . S − S 2. For any A1; A2; , Ai . The elements of ·are· · 2calledS [measurable2 S sets. These two conditions are S summarized by saying that the measurable sets are closed under taking finite differences and countable unions. Think of S as the arena in which all the action (integrals, etc) will take place; and of the measurable sets are those that are \candidates for having a size". In some examples, all the measurable sets will be assigned a \size"; in others, only the smaller measurable sets will be (with the remaining mea- surable sets having, effectively “infinite size"). Several properties of measurable sets are immediate from the definition. 1. The empty set, , is measurable. [Since is nonempty, there exists some measurable set A.;So, A A = is measurable,S by condition 1 above.] − ; 2. For A and B any two measurable sets, A B, A B, and A B are all measurable. [The third is just condition 1\above.[For the second,− apply condition 2 to the sequence A; B; ; ; . For the first, note that ; ; · · · A B = A (A B): Use condition 1 twice.] It follows immediately, by rep\eated application− − of these facts, that the measurable sets are closed under taking any finite numbers of intersections, unions, and differences. 3. For A1; A2; measurable, their intersection, A , is also measurable. · · · \ i [First note that we have the following set-theoretic identity: A1 A2 A3 \ \ \ = A1 (A1 A2) (A1 A3) (A1 A4) . Now, on the right, apply· · · condition− f 1−above[to the−set-differences,[ − and[condition· · ·g 2 to the union.] Thus, measurable sets are closed under taking countable intersections and unions. Here are some examples of measurable spaces. 1. Let S be any set, and let consist only of the empty set . This is a S ; (rather boring) measurable space. 1By A B, we mean A Bc, i.e., the set of all points of A that are not in B. − \ 1 2. Let S be any set, and let consist of all subsets of S. This is a measurable space. S 3. Let S be any set, and let consist of all subsets of S that are countable (or finite). This is a measurableS space. 4. Let S be any set, and fix any nonempty collection of subsets of P S. Let be the collection of subsets of S that result from the following construction.S First set = . Now expand to include all sets that result S P S by taking differences and countable unions of sets in . Next, again expand to include all sets that result by taking differencesSand countable unions ofS sets in (the already expanded) . Continue in this way2, and denote by S S the collection that results. Then (S; ) is a measurable space. Thus, you can generate measurable spaces by startingS with any set S, and any collection of subsets of S (i.e., those that you really want to turn out, in the end, toP be measurable). By expanding that original collection , as described above, you can indeed achieve a measurable space in whichPthe chosen sets are indeed measurable. 5. Let (S; ) be any measurable space, and let K S (not necessar- ily measurable).S Let denote the collection of all subsets⊂ of K that are K -measurable. Then (K; ) is a measurable space. [The two properties forS (K; ) follow immediatelyK from the corresponding properties of (S; ).] K S Thus, each subset of a measurable space gives rise to a new measurable space (called a subspace of the original measurable space). 6. Let (S0; 0) and (S00; 00) be measurable spaces, based on disjoint un- S 0 S00 derlying sets. Set S = S S , and let consist of all sets A S such that A S0 0 and A S00 [ 00. ThenS(S; ) is a measurable⊂space. [The \ 2 S \ 2 S S two properties for (S; ) follow immediately from the corresponding prop- erties of (S0; 0) and (SS00; 00). For instance, the first property follows from: (A B) S0 S= (A S0) S(B S0) and (A B) S00 = (A S00) (B S00).] − \ \ − \ − \ \ − \ 2 Measures Let (S; ) be a measurable space. A measure on (S; ) consists of a S µS nonempty subset, , of , together with a mapping R+ (where R+ M S M ! 2 In more detail: Set 1 = . Let 2 the set that results from applying the above S P S process to 1; then by 3 the set that results from applying the process to 2, etc. Then S S S set = 1 2 . S S [ S [ · · · 2 denotes the set of nonnegative reals), satisfying the following two conditions: 1. For any A and any B A, with B , we have B . 2 M ⊂ 2 S 2 M 2. Let A1; A2; be disjoint, and set A = A1 A2 . Then: This · · · 2 M [ [· · · union A is in if and only if the sum µ(A1) + µ(A2) + converges; and when these holdMthat sum is precisely µ(A). · · · A set A is said to have measure; and µ(A) is called the measure of A. Think2ofMthe collection as consisting of those measurable sets that M actually are assigned a \size" (i.e., of those size-candidates (in ) that were successful); and of µ(A) as that size. Then the first condition abSove says that all sufficiently small measurable sets are indeed assigned size. The second condition says that the only excuse a measurable set A has for not being assigned a size is that \there is already too much measure inside A", i.e., that A effectively has “infinite measure". The last part of condition 2 says that measure is additive under taking unions of disjoint sets (something we would have wanted and expected to be true). Several properties of measures are immediate from the definition. 1. The empty set is in , and µ( ) = 0. [There exists some set A . Set B = and apply;conditionM 1, to conclude; . Now apply condition2 M ; ; 2 M 2 to the sequence ; ; (having union A = ). Since A , we have µ( ) + µ( ) + =;µ;( ·),· ·which implies µ( ) = 0.]; 2 M ; ; · · · ; ; 2. For any A; B , A B, A B, and A B are all in . Furthermore, if A and B are disjoin2 Mt, then\ µ(A [ B) = µ(A−) + µ(B). [TheM first and third follow immediately from condition[1, since A B and A B are both subsets \ − of A. For the second, apply condition 2 to the sequence A B; B; ; ; of disjoint sets, with union A B. Additivity of the measures−also follo; ws; ·from· · [ this, since when A and B are disjoint, A B = A.] 3. For any A; B , with B A,−then µ(B) µ(A). [We have, by the previous item, µ(2A)M= µ(B) + µ⊂(A B).] Thus, \the≤ bigger the set, the − larger its measure". 4. For any A1; A2; , A . [This is immediate from condition · · · 2 M \ i 2 M 1 above, since A and A A1 .] \ i 2 S \ i ⊂ 2 M Thus, the sets that have measure (i.e., those that are in ) are closed under finite differences, intersections and unions; as well as underM countable intersections. What about countable unions? Let A1; A2; be a sequence of sets in , not necessarily disjoint. First note that A =· ·A· can always be M [ i written as a union of a collection of disjoint sets in , namely of A1; A2 M − A1; A3 A2 A1; . If the sum of the measures of the sets in this last list − − · · · 3 converges, then, by condition 2 above, we are guaranteed that A . And if the sum doesn't converge, then we are guaranteed that A is not 2inM . Note, M incidentally, that convergence of this sum is guaranteed by convergence of the sum µ(A1) + µ(A2) + µ(A3) + (but, without disjointness, this last sum may exceed µ(A)). In short, the·sets· · that have measure are not in general closed under countable unions, but failure occurs only because of excessive measure. Here are some examples of measures. 1. Let S be any set, let , the collection of measurable sets, be all subsets of S, let = , and, for AS , let µ(A) = 0. This is a (boring) measure. M S 2 M 2. Let S be any set, all countable (or finite) subsets of S, the collection of all finite subsetsS of S, and, for A , let µ(A) be the nMumber 2 M of elements in the set A. This is is called counting measure on S. Note that the set S itself could be uncountable. 3. Let S be any set and the collection of all subsets of S. Fix a f S nonnegative function S R+ on S. Now let consist of all sets A such that Σ f converges.! Thus, includes allMthe finite subsets of S; 2andS A M possibly some countably infinite subsets (provided there isn't too much f on the subset); and possibly even some uncountable infinite subsets (provided f vanishes a lot on the subset). For A , set µ(A) = ΣAf. This is a measure. For f = 1, it reduces to counting2 measure.M 4. Let (S; ; ; µ) be any measurable space/measure. Fix any K S S M 2 (not necessarily in ). Denote by the collection of all sets in that are subsets of K; and bSy the collectionK of all sets in that areSsubsets of MK M K.
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