The Hahn-Banach and Radon-Nikodym Theorems
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1 The Hahn-Banach and Radon-Nikodym Theorems 1. Signed measures Definition 1. Given a set Hand a 5-field of sets Y, we define a set function. À Y Ä to be a signed measure if it has all the properties of a measure but can be negative; in particular we require À (a). ÐEÑœ!. ÐEÑif E are disjoint (b) There are never two sets Eß F with. ÐEÑ œ _ and. ÐFÑ œ _ n particular, a set can exist with measure _ or one with measure _, but not both types of sets (since then we would sometimes have to compute, which is impossible). A signed measure is called an additive set funtion if it is required to have finite measure on all sets. Note that: Lemma 1: f Å or Æ then. Ð Ñ Ä. ÐÑ Ä_ Proof: This follows from the fact that if _ œ! Ð5 Ñ 5, so œ! Å then _. ÐÑ œ. Ð! Ñ. Ð5 5 Ñ œ lim. Ð! Ñ. Ð5 5Ñ Ä_ 5œ! 5œ! œ lim. Ð Ñ, Ä_ with a similar proof for the second statement. n fact it can be shown that such a signed measure is always a difference of two standard measures: 12
2 2. Hahn Decomposition: preliminaries Below we assume we are in a general measure space ÐHY ß ß. Ñ. Recall that the symmetric difference E? FœaEFbaFEb. Note also that. ÐE? FÑ œ ( lme MFl Recall Fatou's lemma for non-negative 0 : ( lim inf 0 Ÿ lim inf ( 0.. Lemma 2 (alternative Fatou Lemma): Proof: ( lim sup 0.. lim sup ( 0.. ( lim sup 0 œ ( lim inf Ð0Ñ lim inf ( Ð0Ñ œ lim sup ( 0 Lemma 3: The symmetric difference measure metric, i.e., Ð+Ñ.ÐEß EÑ œ! Ð,Ñ.ÐEß FÑ œ.ðfß EÑ Ð-Ñ.ÐEß GÑ Ÿ.ÐEß FÑ.ÐFß GÑ Proof: To prove (c), note that.ðeß FÑ œ. ÐE? FÑ.ÐEß GÑ œ ( lme MGl.. Ÿ ( lme MFl.. ( lmf MGl.. is a 13
3 œ.ðeßfñ.ðfßgñþ Lemma 4: The collection of measurable sets is complete with respect to the distance measure d, in the sense that if ÖE form a Cauchy sequence, i.e.,.ðeß E7Ñ ß 7ÒÄ _!, then there is a set E such that.ðeß EÑ ÒÄ _!. The set E is unique up to sets of measure!, and up to sets of measure!, we have E œ lim sup E œ lim inf E œ lim E Þ Proof: t suffices to show that a subsequence ÖE 5 of the sequence 5 converges to some set E, since then we will have (letting and 5 be any elements of the sequence and subsequence).ðe ß EÑ Ÿ.ÐE ß E Ñ.ÐE ß EÑ!, Ò 5 5, 5 Ä_ showing then that the full sequence also converges to E. Let 5 be a 5 subsequence of the natural numbers such that.ðe ß EÑ Ÿ # for any 5 (possible since the E form a Cauchy sequence). Then 5.ÐE ß E Ñ Ÿ # (1) since 5 5Þ For simplicity of notation assume the subsequence E Ä E 5 5 is the the full sequence, with (1) now becoming 5.ÐE5ß E5 Ñ Ÿ # Þ We will show that up to sets of measure!, lim inf E5 œ lim sup E5 œ lim E5ß (3) with the last limit being defined as the common value of the lim sup and lim inf sets (again, now up to sets of measure 0). To show (2) above, let F œ E 5 5, and Fœ F œlim F (since the F are increasing). Let G œ E, and GœG œ 5 Ä_ 5 lim GÞ Then we claim that Ä_.ÐFß GÑ Ÿ.ÐFß FÑ.ÐFß EÑ.ÐEß GÑ.ÐGß GÑ Ä_ Ò!. (4) This follows since F? F Æ 9 (the empty set) and similarly G? G Æ 9. n addition, 14
4 .ÐEß FÑ œ. E? Œ E5 œ. Œ E5 E _ 5 5 _ œ. ÐE E Ñ Ÿ. ÐE E Ñ Ÿ # œ # Ò! ß Ä_ where the third inequality follows because we are removing the set E from the union of E (on the left), so that we may equally remove it from each E individually (on the right). Similarly,.ÐE ß G Ñ!Þ This proves (4). 5 5 Ä_ Ò But since Fœlim supe and Gœlim sup E, (4) shows that FœGexcept on a set of measure 0, proving (3). But now it follows as in (4) that.ðeß FÑ Ÿ.ÐEß FÑ.ÐFß FÑ Ä_ Ò!, showing that the limit of the sets E5 exists and equals F œ lim sup E5 (or equivalently Gœlim inf EÑ, up to sets of measure!. 3. Hahn decomposition theorem 5 5 Now assume we have a signed measure. on the measurable space ÐHßYÑ. For simplicity we assume that. is finite everywhere (i.e., it's an additive set function), though the results are easily extended to the general case of a signed measure. Hahn Decomposition Theorem: There exist disjoint sets E and E such that. ÐÑ! for E, and. ÐÑ Ÿ! for E A. Proof: f we can find a set E such that. ÐE Ñ. ÐÑ for all Y, then letting E œ H E we will be done. This is because for any set E, we must have.ðñ!, since otherwise we would have. ÐE Ñ œ. ÐE Ñ. ÐÑ. ÐE Ñ, which is impossible, and a similar argument shows that if E.ÐÑ Ÿ!. then Thus let! œ sup. ÐEÑ. We wish to find a set E such that. ÐE Ñ œ! Þ E Y 15
5 To find such an E, select E Y such that!. ÐEÑ Ÿ Î#. Then we claim that Eœlim sup E œ E has measure!. To this end, note that.ðe E Ñ Ÿ # 5 5, since otherwise we would have. ÐE E Ñ. ÐE Ñ #! #!. Thus the sets E form a Cauchy sequence in the sense above, so that E Ä_ Ò E, i.e.,. ÐE? EÑ Ä_ Ò!. Thus since. ÐEÑ Ä_ Ò!, it follows easily that. ÐEÑ œ! ÑÞ Thus set E œ E. An easy consequence of the above theorem is that where. 3 are positive (ordinary) measures, and. ÐÑ œ. Ð E Ñ. œ.. #, (1). ÐÑ œ. Ð E ÑÞ # The decomposition (1) of. is called the Jordan decomposition. 4. Absolute continuity and singularity Definition 2. Given two measures. and / on a measureable space ÐHßYÑ, we say that. and / are mutually singular or. ¼ /, if they have disjoint - supports, i.e., there exists a set Esuch that. ÐE Ñ œ / ÐEÑ œ!. 16
6 Definition 3. Given two measures. and / we say that / is absolutely continuous (ac) with respect to.(or / <<. Ñif / ÐEÑ œ! whenever. ÐEÑ œ! for E YÞ Note that it is known that if 0! is a measureable function, we can define a new measure / from. through Clearly / << / ÐEÑ œ ( 0ÐBÑ.. ÐBÑ. (1) E We can ask the reverse question: if / <<., is it true there is a function 0 such that (1) holds? The answer is yes: 5. Radon-Nikodym theorem Given two 5 -finite measures / and. on Y in H, let Z be the class of nonnegative functions 0 such that ' 0 Ÿ / ÐÑ. Assume that. and / are not mutually singular (i.e. there is some with positive. and / measures) Lemma 5: There is at least one 0 Z such that ' 0..!. 17
7 Proof: Find a set with postive. and / measure, by taking subsets we may assume is finite in.-measure. Choose 0 so that! ' 0 / Ð Ñ (we can replace 0 by % 0 if this is false). Now just on the set the signed measure 7ÐÑ œ / ÐÑ ' 0 must be purely positive when restricted to some #, by Hahn decomposition. Thus for any set Y ( 0 ÐBÑM ÐBÑ œ ( 0 ÐBÑ Ÿ / Ð # Ñ Ÿ / ÐÑ. # # Choosing 0ÐBÑ œ 0 ÐBÑ M # ÐBÑ shows there is an 0 Z with ' 0!. Lemma 6: f 0 Z for œ ß #ß á ß then sup 0 Z. Proof: For any set, ( sup 0 œ ( lim sup 0 œ lim ( sup 0 Ä_ 5 5 5Ÿ Ä_ 5Ÿ œ lim ( 05 Ÿ lim / Ð5Ñ œ lim / ÐÑ œ / ÐÑÞ Ä_ Ä_ Ä_ 5œ 5œ 5 where 5 œ œb À 05ÐBÑ sup 03ÐBÑ (if there is a tie between two 3Ÿ different 0 just choose one of them so the are disjoint). 5 5 Thus sup 0 Z and so Z is closed under suprema of sequences of functions. Theorem 7. f / and. are 5-finite measures on Y in H ß+. / <<., then there is a unique function 0 such that for all E Y, / ÐEÑ œ ( 0ÐBÑ.. ÐBÑÞ Ð#Ñ E 1
8 Proof: First assume that Hhas finite measure. Consider the class Z of all positive functions 0 such that 0.. Ÿ / ÐÑfor all H. Let! œ sup ' ' H 0Þ Choosing 0 Z so that ( H then clearly letting 0! œ sup 0 We now claim that for any, For if not, the measure 0! Îß Z, we have ( 0! œ ( sup0 sup( 0 œ! Þ H H H ( 0! œ / ÐÑÞ 0 Z 7ÐÑ œ / ÐÑ ( 0! (2a) would be positive somewhere, so that by the Lemma above there would be an 0! with ' H 0! and for all, Thus we would have ( 0 Ÿ 7ÐÑ.! 7ÐÑ ( 0 œ / ÐÑ ( 0! ( 0 while œ / ÐÑ ( Ð0! 0 Ñ which gives a contradiction. ( H Ð0 0 Ñ! ß! 19
9 Uniqueness of 0 (up to sets of measure!) follows from Ð#Ñabove, since we know a function 0 is uniquely determined from knowing its integral over every set E. For example, if there were another function 0 which made (2) true, and 0 0 on a set of positive measure, then (2) would be different for 0 and for 0 integrated over ). Finally, if His 5-finite, write it as a countable union of finite pieces H 3, and use the above proof on each H3 to find a function 0 on Hwhich on each piece is constructed as above. This completes the proof. H 3 Definition 4: The function 0 above is the Radon-Nikodym derivative of / with respect to.ß Thus we can write 0œ./ Þ./ / ÐEÑ œ ( ÐBÑ ÐBÑÞ We also have (easily shown by starting with simple functions 1): 6. Lebesgue decomposition E ( 1ÐBÑ./ ÐBÑ œ ( 1ÐBÑ./ ÐBÑ ÐBÑ E Notice in the proof of the Radon-Nikodym theorem, the function 0 ÐBÑ which maximizes ' H 0ÐBÑ can be defined even if / <<. fails to hold. n that case, we will write E! Z / ÐÑ œ ( 0 ÐBÑ.. / = ÐÑ Ð Ñ! 4 / +- ÐÑ / = ÐÑ where / +- ÐÑ is defined as the integral above, and where / = ÐÑ is defined as the difference of the other two terms in (4). 190
10 But notice the remaining measure / = must satisfy / =ÐÑ œ! if. ÐÑ!, since otherwise by the same argument as in the proof of the theorem, there would be a function 0! (not all 0) such that for all ( 0./ Ÿ. ÐÑÞ = But then the function 0! defined above can be increased to 0! 0 and still be in Z, which contradicts that is maximal in Z. 0! However, the conclusion that / = ÐÑ œ! if. ÐÑ! means that / = and. are mutually singluar. Thus (4) says that given a fixed measure., any other measure / can be decomposed into a measure / +- (which is absolutely continuous with respect to.) and a measure / = which is singluar with respect to t is easy to show that this decomposition is unique. Definition 5: The decomposition (4) above is called the Lebesgue decomposition of the measure /. 191
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