Notes on the Unique Extension Theorem. Recall that we were interested in defining a general measure of a size of a set on Ò!ß "Ó.

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1 1. More on measures: Notes on the Unique Extension Theorem Recall that we were interested in defining a general measure of a size of a set on Ò!ß "Ó. Defined this measure T. Defined TÐ ( +ß, ) Ñ œ, +Þ Question: how large a collection of sets Y can we extend this definition to? Decided we cannot consistently extend this definition to all subsets of Ò!ß "ÓÞ Goal: find a general probability measure T on Ò!ß "Ó with TÐ ( +ß, ) Ñ œ, + defined on a 'reasonably large' collection Y of subsets of. Recall we were able to define T on the collection Y! œöall finite unions of disjoint open intervals M œö MlM œintervals (can be open, closed, or half-open)} We claimed we could extend it to a larger collection of sets Y œ 5-field generated by Y!. More generally: We are on a set H, where we have a field of sets Y! on which we have a probability measure T. Recall T is a probability measure on a field of sets if: Y! (i) TÐ9Ñ œ! (ii) TÐHÑ œ "

2 (iii) T is countably additive on À that is, if E is a sequence of disjoint Y!!! sets in Y, and if E Y, then! TÐEÑœTÐEÑÞ We need: Theorem (Unique extension theorem): Any set function T defined on a field Y! of sets and satisfying the properties of a probability measure on Y! extends uniquely to a probability measure on the 5-field Y generated by Y! Þ 2. Outer measures Assume we have a field on a space H. Y! (for example finite unions of open sets) of sets Let T be a measure on Y! (note it does not need to be a probability measure). Let Y be the 5-field generated by Y!. We will show that there is a unique extension of T from Y! to Y. We first define the outer measure as an extension of T. Definition 1: Given a measure T on Y! define its outer measure on all susets of H by T ÐEÑ œ inf " T ÐEÑl E Eà E Y! Ÿ ß T i.e. the smallest sum of measures of a collection of Y! sets containing E. We can also define the inner measure as one minus the largest sum of - measures of a collection of Y! sets contained in E À

3 - T ÐEÑ œ sup " " T ÐEÑl E Eà E Y! Ÿ ß But this is equivalent to the following definition of inner measure: - T ÐEÑ œ " T ÐE ÑÞ Want to extend measure T on Y! to a collection Y of as many sets as possible. How about choosing Y to be the collection of sets E Hwhich have the same inner and outer measure, and then define that to be TÐEÑ? That is, for all sets Esuch that T ÐEÑ œ T ÐEÑ, i.e., - T ÐEÑ T ÐE Ñ œ " (1) define such sets to be in Y and define T ÐEÑ œ T ÐEÑ œ T ÐEÑÞ Is this collection a 5-field? This is what we want, but we want to define a more restrictive condition on E which turns out to be the same. We will define a set to be T -measurable if (extend condition (1)): for every set I H, - T ÐE IÑ T ÐE IÑ œ T ÐIÑÞ (2a) Notice that when IœHthen (2) reduces to (1). Define ` to be the collection of T -measurable sets.

4 Claim: the measure T on ` is essentially the extension of T on Y! that we want. Easy to check: (a) T is monotone, i.e., if E Fthen T ÐEÑŸT ÐFÑ (b) T is sub-additive, i.e., T Œ E Ÿ " T ÐE ÑÞ Thus it is automatically true that so that (2) is equivalent to - T ÐE IÑ T ÐE IÑ T ÐIÑ, - T ÐE IÑ T ÐE IÑ Ÿ T ÐIÑ, (2b). Some properties of the measure T and Lemma 1: ` is a field Proof: Note it suffices to show that (i) ` is closed under complements (easy) and that (ii) ` is closed under intersections. Reason: then if Eß F `, we have ` - E FœÐE F - Ñ, so that E F ` also, and arbitrary finite unions follow. Now to show ` closed under intersections: If Eß F `, then for any set I H, - T ÐIÑ œ T ÐF IÑ T ÐF IÑ

5 œt ÐE F IÑ T - ÐE F IÑ T ÐE F IÑ T ÐE F IÑ T ÐE F IÑ T cðe F IÑ ÐE F IÑ ÐE F IÑd - œt ce F Id T cðe FÑ IÑd which using (2b) implies that E F `, as desired. Lemma 2: If ÖE is a finite or infinite sequence of disjoint sets in `, then if I H, T ÐI E Ñ œ " T ÐE ÑÞ Proof: For the case of two E, we have to replace I by I ÐE" E# Ñ T ci ae E " # - " # " " # " œt eci ÐE E Ñd E f T eci ÐE E Ñd E f For the case of more than two and so on. bd œt ÐI EÑ T " ÐI EÑÞ #, we proceed by induction writing e.g. E ÐE E E ÑœÐE E Ñ E ß " # $ " # $ For a countable sequence of E, we have by monotonicity, for all and then taking the limit in : T ŒI E T ŒI E œ " T ÐI E Ñ so letting Ä we have

6 T ŒI E " T ÐI E Ñ ; () Opposite inequality follows by sub-additivity; thus we have equality in () above, as desired. Corollary: The outer measure T is countably additive on `. Proof: Just let IœH. Lemma : The collection of sets ` is a 5-field. Proof: For E `, we wish to show that F œ E `. It suffices with small changes to assume that the E are disjoint (see text). Letting F œ E, it is clear that F `, since ` is a field. Thus - T ÐIÑ œ T ÐI F Ñ T ÐI F Ñ œ " - T ÐI E Ñ T ÐI F Ñ - "T ÐI E Ñ T ÐI F ÑÞ Letting Ä, we get (using countable additivity, proved earlier, for the equality below) T ÐIÑ - - " T ÐI E Ñ T ÐI F Ñ œ T ÐI FÑ T ÐI F Ñß which with (2b) above proves that F Lemma 4: Y! ` `Þ Proof: We need to show that if E Y and I H, then!

7 Clearly this is true for - T ÐI EÑ T ÐI E Ñ œ T ÐIÑÞ, since this is just finite additivity. I Y! If IÂY, we approximate Iby sets H Y from above.!! Specifically, for %!, let ÖE be a finite collection of sets in Y! such that and I E H, T ÐIÑ Ÿ " TÐE Ñ % Þ " Then clearly G is an approximation of I from above. In this case we have that - T ÐH EÑ T ÐH E Ñ œ T ÐH Ñ. Now take limits as ÄÞ The right side clearly goes to T ÐIÑby the definition of H above, while the left side goes to - T ÐI EÑ T ÐE E Ñ by a simple approximation argument. 4. Completion of the proof of unique extension We now know that is defined on a 5-field `of sets which extends Y and which satisfy equation (2) above. It is easy to show that for E T! Y!, we have T ÐEÑ œ T ÐEÑÞ Final step: Let Y œ 5ÐY Ñ œ 5-field generated by Y. Then we have!! Y! Y `. ` And since T extends T to a countably additive measure on, the restriction of this measure to Y gives us the desired extension. 5. Uniqueness and the 1- - theorem

8 Q: How do we show that the extension of T to T is unique, i.e. that there is no other extension? Definition 2: A collection c of subsets of a space H is a 1-system if it is closed under intersections Definition : A collection of subsets of a space His a --system if (1) H (2) is closed under complements () is closed under countable disjoint unions It is easy to how that if a class V of sets is both a 1-system and a --system then it is a 5-field. Unique extension follows from Dynkin's 1- - theorem: Theorem 1: If c is a 1 -system and is a --system, and if c, then it follows also that 5Ðc Ñ Þ Proof: The proof follows from establishing the following easy statements: 1. Define! to be the 1-system generated by c. Then!, since! is the minimial - system containing V. If we can show that and we will show that 5Ðc Ñ Þ! is also a 1-system, then it is a 5-field and so c 5Ðc Ñ! 2. For a set E, let E œ ÖF À E F! Þ Show that if E c then E is a --system. If E c ß show that Þ! E 4. Show that if F then c.! F 5. Since is a 1 -system, it follows that if F then Þ F!! F

9 'ÞThus if FßG then F G This shows that!!! is a 1-system, which was what was left to be proved. Theorem 2: If c is a 1-system and TßT " # are probability measures on 5c Ð Ñß and if they agree on cß they also agree on 5ÐTÑ. Proof: Let be the class of sets in 5ÐcÑ such that T" ÐEÑ œ T# ÐEÑÞ Then - - if E, then TÐEÑœ" TÐEÑœ" TÐEÑœTÐEÑ " " # #, so that E Þ Thus is closed under complements. Can easily also show that is a --system. Thus c and c is a 1- system, so the 1- - theorem gives 5c Ð Ñ ß as desired.

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