[PDF] Lecture 6 The Dynkin ? ? Theorem - LSU

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Lecture 6. The Dynkinπ-λTheorem.

It is often the case that two measures which agree on a certain class of sets actually agree on all sets in the relevantσ-algebra. There are a couple of standard tools to prove that the measures are the same: the Monotone Class lemma and the Dynkinπ-λtheorem. They are essentially equivalent devices and it is largely a matter of taste which one to take as standard equipment. We shall do theπ-λtheorem and use it in the case of Lebesgue measure. Suppose thatμandμ?are translation invariant measures on the Borelσ-algebra of R dboth assigning the same (finite) measure to the unit box [0,1]d. We will show that thenμ=μ?. LetLdenote the set of all Borel setsA?Rnfor whichμ(A) =μ?(A). By hypothesis, [0,1]d?L. It seems reasonable to conclude from this that the setPof all boxes [a1,b1]×···×[ad,bd], with rationalaiandbi, would belong toL. Let us accept this for now; i.e. supposeP?L. Now if we can show from this thatLcontains theσ-algebra generated byPthen we would be done, because theσ-algebra generated byπis the Borel σ-algebra. (This follows from two observations : (i) each box inPis the intersection of open sets : [a1,b1]× ··· ×[ad,bd] =? k≥1? a 1-1 k,b1+1k? a d-1k,bd+1k? and (ii) every openUsubset ofRdis the union of small boxes [a1,b1]× ··· ×[ad,bd] with rational endpoints and centered at the rational points inU.) ThusLwould in fact be the whole Borelσ-algebra. That is,μ(A) =μ?(A) for every Borel setA. Thus the key tool would be the result thatLcontains theσ-algebra generated byP. This will, essentially, be proved by theπ-λtheorem. There are some technical problems involved which will be settled later.

6.0. Definition

. LetPandLbe collections of subsets of a setX. The collectionP is called aπ-system if it is closed under finite intersections; i.e. ifA,B?Pthen

A∩B?P:

Pis aπ-system ifA∩B?Pfor allA,B?P

The collectionLis called aλ-system if the following hold : (L1)∅ ?L; (L2) ifA?LthenAc?L; (L3)Lis closed under countabledisjointunions; i.e. ifA1,A2,...?Land ifAi∩Aj=∅ for everyi?=j, then?∞ j=1Aj?L.

6.1.Dynkin"sπ-λTheorem

.LetPbe aπ-system of subsets ofX,andLa λ-system of subsets ofX.Suppose also thatP?L.Then:

σ(P)?L,

i.e.Lcontains theσ-algebraσ(P)generated byP. We will do the proof later but let us apply it to prove the uniqueness of Lebesgue measure. 49

6.2. Proposition.Every translation-invariant Borel measure onRwhich assigns finite

measure to the unit interval is a constant multiple of Lebesgue measure. Proof . Letμbe a translation-invariant Borel measure onRwhich assigns finite measure to the unit interval. Letmbe Lebesgue measure onR. Our first objective will be to check that

μ?[a,b)?=km?[a,b)?

for every rationala,b, wherekis the finite constant k=μ?[0,1)? Then we shall show by aπ-λargument thatμ(A) =km(A) holds for all setsAin the sigma-algebra generated by the intervals [a,b), i.e. it holds for all Borel setsA. For any positive integerp, the interval [0,p) is the union ofpdisjoint translates of [0,1), and so by translation-invariance ofμ, we have ?[0,p)?=pμ?[0,1)?=pk By the same argument, for any positive integerp, we also have ?[0,p)?=qμ?[0,p/q)?

Combining these two relations we have

?[0,p/q)?=kp q=km?[0,p/q)?

Then by translation-invariance it follows that

?[a,b)?=km?[a,b)? for all intervals [a,b) for whichb-ais rational. Thusμis a constant multiple of Lebesgue measuremon intervals [a,b) with rational endpoints. Now we use the π-λtheorem to jazz this up to all Borel sets. The first idea would be takePto be the collection of all intervals [a,b) with rational endpoints, andLto be the class of all Borel setsAfor whichμ(A) =km(A) holds. But there is a problem with this: the collectionLsatisfies all properties of being aλsystem except that we cannot establish closure under complements, as the argument

μ(Ac) =μ(R)-μ(A) =∞ -km(A) =km(Ac)

is non-sense. The way to get around this problem with infinite measure is to focus down to a finite interval [-N,N) and then letN↑ ∞at the end.

So, fix any positive integerN, and let

L N={all Borel setsAfotr whichμ?A∩[-N,N)?=km?A∩[-N,N)?} 50
and

P={all intervals [a,b) witha,brational}

What we have proven before shows that

P?LN It is clear thatPis aπ-system. It is also clear thatLcontains the empty set and is closed under countable disjoint unions. To check closure under complements, consider any

A?LN. Then

?Ac∩[-N,N)?=μ?[-N,N)?-μ(A) =km?[-N,N)?-km(A) =km?Ac∩[-N,N)? (The subtraction works becausem([-N,N)) = 2Nis finite.) This shows thatAc?LN.

ThusLNis aλ-system.

By Dynkin"s theorem we conclude then thatLN?σ(P). Butσ(P) is the entire Borel sigma-algebra. So, in fact,LNis the entire Borel sigma-algebra, and this means that for any Borel setAwe have the relation ?Ac∩[-N,N)?=km?Ac∩[-N,N)?

Now letN↑ ∞. Since

N≥1?Ac∩[-N,N)?=A

we conclude that

μ(A) =km(A)

for every Borel setA. We can now move this result up to higher dimensions.

6.3. Proposition

.Supposeμis a translation-invariant measure on the Borel subsets ofRd,for whichkdef=μ([0,1]d)<∞.Then

μ=km

Proof . We will use essentially the same argument as in the one-dimensional case. Let p

1,...,pdbe positive integers. Note that [0,pj) is the union ofpjtranslates [n,n+1), with

n? {0,1,...,pj-1}, of [0,1). Taking products of these intervals we see that [0,p1)× ··· ×[0,pd) is the union of of the boxes [n1,n1+ 1)× ··· ×[nd,nd+ 1) wheren1,...,nd? {0,1,...,p-1}. Any pair of distinct boxes in this collection have at least one 'side" disjoint, and so these boxes are disjoint. There are p

1···pd

51
of these boxes. So, by translation-invariance of the measureμ, we have ?[0,p1)× ··· ×[0,pd)?=p1···pdμ?[0,1)d?=kp1···pd Again, by the same reasoning, for any positive integersq1,...,qd, we have ?[0,p1)× ··· ×[0,pd)?=q1···qdμ?[0,p1/q1)× ···[0,pd/qd)?

Combining the preceding relations we have

μ(B) =km(B)

for the boxB= [0,p1/q1)× ··· ×[0,pd/qd). Translation invariance then shows that the above equality holds for every boxBwith rational sides. LetPbe the collection of all boxes with rational corners. This is aπ-system which generates the Borel sigma-algebra ofRd. Fix any positive integerNand letLNbe the collection of all Borel setsA?Rdfor which

μ(A∩BN) =km(A∩BN)

where B

N= [-N,N)d

ThenLNis aλ-system and, by what we have proven above,LN?P. Therefore, by the π-λtheorem,LN?σ(P). Sinceσ(P) is the Borel sigma-algebra, it follows that

μ(A∩BN) =km(A∩BN)

for every Borel setA. Now letN↑ ∞to conclude that

μ(A) =km(A)

for every Borel setA?Rd. Finally, we turn to the proof of theπ-λtheorem. There are a couple of simple observations we will need :

6.4. Lemma

.LetXbe a set, and considerλandπsystems of subsets ofX. (i)Aλ-system which also aπ-system(i.e. is closed under finite intersections)is a

σ-algebra.

(ii)Aλsystem is closed under proper differences: ifLis aλ-system andA,B?L withB?A,thenA-B?L. Proof . (i) Suppose that theλ-systemLis also aπsystem; i.e. ifA,B?Lthen A∩B?L. The only point we have to check is thatLis closed under countable unions. So letE1,E2,...?L, and setE=?jEj. Since we know thatLis closed under countable disjoint unions we need to writeEas a disjoint union of sets inL. This is achieved as follows. LetHjbe the set of all points inEjwhich do not already belong to any of the 'previous" setsE1,...,Ej-1. Then clearly the setsHjare disjoint andE=?jHj. To see 52
thatHjis inLwe note thatHj=Ej∩Ecj-1∩...∩Ec1(this makes sense forj >1; for j= 1,H1=E1is already inL.) SinceLis closed under complements (being aλsystem) and since we have also assumed thatLis closed under finite intersections, we haveHj?L.

HenceE?L, as required.

(ii) IfA,B?L, andA?B, thenA-B=A∩Bccan be expressed in the following way :

A∩Bc= (Ac?B)c.

Notice that the setsAcandBare disjoint becauseA?B. So, sinceLis closed under complements and finite disjoint unions, we see thatA-B?L. For the next step towards theπ-λtheorem, observe that the intersection of any family ofλsystems is again aλsystem. IfXis a set andP? P)(X), definel(P) to be the intersection of allλsystems which containPas subset (for example, the power setP(X) is aλsystem?P). Thus: l(P)def=∩{Λ : Λ is aλsystem and Λ?P} Thusl(P)is the smallestλsystem containingPas a subset.

6.4. Lemma

.LetPbe aπ-system of subsets of a setX. Thenl(P)is aσ-algebra.

Proof.

We have seen that a collection of subsets ofXwhich is both aπsystem and a λsystem is actually aσ-algebra. Thus, it will suffice to prove thatl(P) is aπ-system. LetA,B?l(P). Our goal is to prove thatA∩Bis also inl(P). We will do this by establishing the following: (1) for any fixedA?l(P), the collection of allB?Xfor whichA∩B?l(P) is aλ system; (2)A∩B?l(P) wheneverA?PandB?l(P); (3) ifA?l(P) andB?l(P) thenA∩B?l(P). This would show thatl(P) is aπsystem.

For (1), fix anyA?l(P), and let

l

A={B?X:A∩B?l(P)}

It is clear thatlAcontains∅and is closed under countable disjoint unions. Next ifB?lA then

A∩Bc=A-(A∩B)

is a proper difference of sets in theλ-systeml(P), and so is inl(P). Thus,lAis aλ-system.

This establishes (1).

Now supposeA?P. ThenA∩B?P?l(P) for everyBin theπsystemP. So B?lA. Thus,lA?P. Then, by definition oflPas the smallestλsystem containingP, we have l

A?l(P)

Looking back at the definition oflA, we see that this means simply thatA∩B?l(P) for everyB?l(P). This establishes (2). 53
Now we"ll bootstrap ourselves up one level: using what has been established in (2) and applying again essentially the same argument we will reach our goal (3). To this end, now fixB?l(P), and think of l

B={A?X:A∩B?l(P)}

This is aλ-system, as we have already seen. In (2) we proved essentially that l B?P Then, by definition ofl(P) as the smallestλ-system containingP, we have l

B?l(P)

Glancing at the definition oflBwe see that this means

A∩B?l(P) for everyB?l(P)

SinceAis any element ofl(P) we have thus established our goal (3), i.e.l(P) is aπsystem.

Proof of theπ-λtheorem

LetXbe a set, and

P?L? P(X)

withPaπsystem andLaλsystem. Then, by definition ofl(P) as the smallestλsystem containingP, we have l(P)?L

Butl(P) is aσ-algebra, andl(P)?P. Therefore,

σ(P)?l(P)

This completes the argument, sincel(P)?L. Note that since every sigma-algebra is also aλsystem it follows thatl(P)?σ(P). Thus, in fact, l(P) =σ(P)

Problem Set

1. Let (X,F,μ) be a measure space withμ(X) = 1, and letA? F. Show that the set

of allB? Fwhich satisfy

μ(A∩B) =μ(A)μ(B)

is aλ-system.

2. Find an example of aλsystem which is not aσ-algebra.

3. Letk?RandEa Borel subset ofRd. Prove that

m(kE) =|k|dm(E), 54
quotesdbs_dbs19.pdfusesText_25

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