Chapter 4 Measurable Functions - LSU Math

1 Chapter 4 Measurable FunctionsIfXis a set andA P(X) is a -field, then (X,A) is called ameasurablespace. If is a countably additive measure defined onAthen (X,A, )is called ameasure space. In this Chapter we will introduce the family ofmeasurable functionsfor which we will seek to define the Lebesgue will prove the very important fact that pointwise limits of measurablefunctions must be Measurable . This is encouraging because pointwise limitsof Riemann integrable Functions need not be Riemann Measurable FunctionsDefinition (X,A, ) be a measure Iff:X Rwe say thatfisA-measurableprovided thatf 1( ,a) ={x X|f(x)< a} Afor alla Iff:X C, the complex numbers, we writef(x) =u(x) +iv(x) forreal-valued functionsu= fandv= f. We sayfisA-measurableprovided that Iff:X S, whereSis atopological space, we say thatfisA- Measurable provided thatf 1(G) Afor everyopensetG Example definition is motivated by Section is necessary to prove that the three parts of the definitionof measura-bility of a function are that iff:X RisA- Measurable , thenf 1(G) Afor every open setG R.

2 Show that the concepts of measurability inDefinition for both real and complex valued Functions are consistentwith the concept of measurability for a topological space valued example, show thatf:X Rsatisfies the definition of measurability ifand only iff 1(G) is Measurable for each open setG reader may note correctly that the concept of measurability has aformal similarity to the definition of continuity. A function between topolog-ical spaces is called continuous provided that the inverse image of each openset is open. Of course for measurability the inverse image ofan open set mustbe Measurable , and measurability is by no means synonymous with open. InRn, every open set is Lebesgue Measurable , but the converse is clearly following exercise gives a very useful equivalent form of measurabilityfor :X Rn, which can be regarded as an example of atopological space.

3 Prove thatfis Measurable if and only iff 1(B) Lforeach Borel setB. (Hint: Show that the familyS={A P(Rn)|f 1(A) L}is a -field.)It will be important to know that many combinations of Measurable func-tions and many Functions of Measurable Functions are again Measurable . Toinvestigate this we need the following (X,A, ) be any measure space arising from the Hopfextension theorem, so that we have a concept of the field of Borel sets gen-erated by a field of elementary sets. Iff:X S, whereSis atopologicalspace, we say thatfis aBaire functionprovided thatf 1(G) is a Borel setfor eachopensetG , every Baire function is Measurable and every continuous function(fromRntoS) is a Baire function. The indicator function of a measurableset that is not a Borel set would be an example of a Measurable function thatis not a Baire each of the functionsf1,f2,..,fnis anA-measurablereal-valued function defined onX.

4 Let :Rn Rbe a Baire function. ThenF= (f1,f2,..,fn)is anA- Measurable function defined We need to show that for each open setG Rwe haveF 1(G) (f1,f2,..,fn) :X Rn. We claim thatfis Measurable . Infact, each open setG Rncan be written as a disjoint union of countablymany open blocks of the formB=nYi=1(ai,bi)which is a Cartesian product ofnopen intervals. Thusf 1(B) =n\i=1f 1i(ai,bi) 1(G) = ( f) 1(G) =f 1 1(G) .Since 1(G) is a Borel set, the theorem is true by the result of remark that in Theorem it would have sufficed to have definedon a setD Rnprovided (f1,..,fn) :X follows from Theorem that such combinations ofmeasurable Functions as the following must be Measurable . c1f1+c2f2wherec1andc2are constants f1 f2, the product of two Measurable Functions f1f2wheref2is nowhere zeroIn particular, f1is also Measurable . Thusfis Measurable if and only if{x|f(x)> } Afor every real-valued Functions , theneach of the following Functions (f1,f2) =f1 f2, wheref1 f2(x) = max(f1(x),f2(x))for eachx (f1,f2) =f1 f2, wheref1 f2(x) = min(f1(x),f2(x))for eachx +=f 0, known as thepositive = ( f) 0, known as thenegative partoff.

5 (The reader shouldnote that the negative part of a real-valued function is positive.)v.|f|=f++f Proof. It suffices to observe that max(x1,x2) and min(x1,x2) are both con-tinuous Functions fromR2toR, making each of these a Baire function. Forthe last part we use the fact that (x1,x2) =x1+x2is continuous and thusa Baire Limits of Measurable FunctionsIn the study of point-wise limits of Measurable Functions and integrable func-tions, we will consider sequences of Functions for whichfn(x) diverges to for some values ofx. Thus it is helpful to extend the concept of real num-bers to the setR =R { }. Measurability for an extended real valuedfunction means that for each Rthe setf 1[ , ] ={x|f(x) } (X,A, )be a measure space and let{fn|n N}beany sequence of Measurable Functions fromXtoR . Then each of the fivefunctions defined as follows (x) = inf{fn(x)|n N}for allx (x) = sup{fn(x)|n N}for allx (x) = lim inffn(x)for allx Xiv.

6 F(x) = lim supfn(x)for allx (x) = limn fn(x)provided the limit exists for allx For the first part, we observe that eacha R,f 1 [ , ] ={x|infnfn(x) }=\n Nf 1n[ , ] isA- Measurable . Sincef (x) infn( fn(x))it follows thatf isA- Measurable as well. Note next that sincein= inf{fk(x)|k n}is an increasing sequence of extended real numbersin, we havelim inffn(x) =f(x) = sup{inf{fk(x)|k n}|n N}so thatfisA- Measurable being the supremum of a sequence of measurablefunctions given as infima. Also,lim supfn(x) = f(x) = inf{sup{fk(x)|k n}|n N}with the result that fisA- Measurable . Finally, we note thatf(x) = limn fn(x)exists if and only if f(x) =f(x). ThusfisA- Measurable provided thepoint-wise limit exists :X Ris a Measurable function for eachn N, where (X,A, ) is a measure space. Prove that the setS={x|limn fn(x) exists}is a Measurable reader should be able to give examples of point-wise convergent se-quences of continuous Functions for which the limit is not continuous, andexamples of point-wise convergent sequences of Riemann integrable functionsfor which the limit is not Riemann integrable.

7 We see alreadythat measura-bility must be a valuable concept for point-wise convergence since point-wiseconvergence does preserve (X,A, ) be a measure space and letNbe the set ofallnull-sets. We say thatxhas some propertyPA almost everywhereif and only if there is a setN Nsuch thatxhas the propertyPfor allx X\N. This is commonly expressed as -almost everywhere( ) orasalmost everywhere( ) provided that there will be no confusion as towhich measure or -algebraAis in a sequence of Measurable Functions on thecom-pletemeasure space(X,A, ). Supposefn(x) f(x) almost everywhereonX. Then the function defined byf(x) = limn fn(x)is defined almosteverywhere and This follows from Theorem It is commonly understoodin thiscontext that although the functionfis defined everywhere by the given limitexcept on a null-setN, we may assignfarbitrary values onNitself.

8 Thenthe completeness of the measure space tells us that the resulting functionfremainsA Measurable regardless of how values are assigned tofwithinthe :X R is a Measurable function that has finitevalues almost everywhere on the measure space (X,A, ) where (X)> that there is a set of positive measure on whichfis the Measurable functionf:Rn Rhas thespecial property that for each fixed vectorc Rn, the translation offgivenbyfc(x) =f(x+c) is equal almost everywhere tof(x) itself. Prove thatf(x) is equal almost everywhere to a constant function. (Hints:Considerboth the sum and the terms ofPn Zl(f 1[n,n+ 1)). Apply Exercise select the special value and conquer!) Simple Functions & Egoroff s TheoremDefinition functionf:X Ris calledA-simple(orsimpleifthere will be no confusion regarding the -fieldAthat is under consideration)if and only iffisA-measurableand{f(x)|x X}is a finite set.]

9 The class of simple Functions is denoted