Real Analysis qual study guide - UC Santa Barbara

1 real Analysis qual study guideJames C. TheoryExercise Rand >0show open setsO Rsuch thatm (O) m (A) + .Proof:Let{In}be a countable cover forA, thenA n=1In. Sincem (O) m (A) + . This impliesthatm (O) m (A) wherem (A) = infA In{ n=1l(In)}Ifl(Ik) = for somekthen there is nothing to show, so suppose (an,bn) =Inthenl(In)< , (an+ 2 n ,bn) then we havel(On) =bn an 2 n l(In) l(On) = bn an 2 n = bn an m ( nOn) m (A)So letO= nOn, thenm (O) m (A) O Rstm (O) m (A) + Exercise ,B R,m (A) = 0, thenm (A B) =m (B)Proof:m (A B) m (A) +m (B), andm (B) m (A B), hence we havem (B) m (A B) m (A) +m (B) =m (B) m (A B) =m (B) Exercise Miff >0, O Ropen, such thatE Oandm (O\E)< Proof.

2 ( )O\E=Ec Oimplies thatm (O\E) =m (Ec O), but we havem (O) =m (Ec O) +m (E O)So supposem (E)< m (Ec O) =m (O) m (E O). LetInbe a countable cover forE, soIn= (an,bn). LetOn= (an,bn+ 2 n ) and letO= On. Thenm (O) = l(On) = 2 n +bn an= + bn an,andm (E O) =m (E)sinceE O. So we havem (E O) =m (E) l(In) = bn an m (E O) l(On) l(In)= + bn an bn an= O Ropen, stE Oandm (O\E) ( ) Conversely, suppose >0, O R, such thatE Oandm (O\E)< and thatO M. Thenm (O) =m (Ec O) +m (E O),butm (Ec O) =m (O\E)< This implies thatm (O) =m (E O) + m (O) =m (E) + E M 12 Exercise Miff >0 F Rclosed, such thatF Eandm (E\F)< Proof:( )E\F=Fc Ethis implies thatm (E\F) =m (Fc E), but we havem (F) =m (Fc E) +m (E F)So supposem (E)< m (Fc E) =m (F) m (F E).

3 LetInbe a countable cover forE,whereIn= (an,bn). LetFn= [an,bn 2 n ] and letF= Fn. Then we havem (F) = l(Fn) = bn an 2 n = bn an ,andm (E F) =m (F), sinceF E. Som (E F) =m (F) l(In) = bn an m (E F) l(In) l(Fn) = bn an bn an+ = F RClosed, stF Eandm (E\F) ( ) Conversely, suppose >0, F R, such thatF Eandm (E\F)< and thatF M. Thenm (E) =m (Fc E) +m (E F),butm (Fc E) =m (E\F)< . This implies thatm (E) m (F E) + m (E) m (F) + E M VitaliLetEbe a set of finite outer measure and a collection of intervals that coverEin the senceof Vitali. Then, given >0 there is a finite disjoint collection{IN}of intervals in such that (E\N n=1In)< Exercise there exists a Lebesgue measurable subsetAofRsuch that for every interval(a,b)we have (A (a,b)) = (b a)/2?

4 Proof:First suppose that there is such a mesurable setAsuch that 06= (A (a,b)) = (b a) there exsits an open setOsuch thatA Oand (O\A)< , so let = /2. NowOis open, sothere are disjoint intervals (xk,yk) such thatOis a countable union of these intervals. SoO (a,b) = k=1[(xk,yk) (a,b)] = l(ckl,dkl).Hence (O (a,b)) = ldkl ckl, and we haveA O (a,b) =A (a,b) = l[A (ckl,dkl)]Now = (A (a,b)) =12 l(dkl ckl)but l(dkl ckl) = (O (a,b))= ((O\A) (a,b)) + (A (a,b)) (O\A) +12 l(dkl ckl)< +12 l(dkl ckl)But this implies that /2 = 12 l(dkl ckl) 3So (A) = 0. which implies that (Ac) =.

5 Now if there were to exsits such a setAwe have (Ac) = 0,and sob a= ((a,b)) = (A (a,b)) + (Ac (a,b)) = (Ac (a,b)) =12(b a)So there cannot exist such a set .Exercise thatE [0,1]is measurable and for any(a,b) [0,1]we have (E [a,b]) 12(b a)Show that (E) = :By the previous problem, using the same proof, we know that (Ec) = 0. So the result ,..,Enbe measurable subsets of[0,1]. Suppose almost everyx [0,1]belongsto at leastkof these subsets. Prove that atleast one of theE1,..,Enhas measure of at :Suppose not, then for eachiwe have (Ei)< k/n. Define a functionf(x) as (x) =n i=1 Eiwhere Eidenotes the characteristic function ofEi.

6 Now since all most allx [0,1] are in at leastkof theEiwe havef(x) kalmost everywhere in [0,1]. Nowk= [0,1]k dx [0,1]f(x)dx=n i=1 [0,1] Eidx=n i=1 EiBut this implies thatn i=1 Ei<n i=1kn=kWhich is a contradiction, hence at least oneEihas (Ei) kn .Exercise a measure space(X,A, )and a sequences of measurable setsEn,n N, suchthat n=1 (En)< Show that almost everyx Xis an element of at most finitely manyE :It suffices to show that (x:x Enk) = 0. So consider the followinglimm (x:x m k=1 Enk)If we have shown the above limit is zero, then we re done. To see this look at the following sum, N=1 (x:x N k=1 Enk)< n=1 (En)< and hencelimm (x:x m k=1 Enk)= 0 Therefore almost everyx Xis an element of at most finitely manyE ns.

7 4 Exercise a measure space(X,A, )with (X)< , and a sequencesfn:X Rofmeasurable functions such thatlimn fn(x) =f(x)for allx X. Show that for every >0there existsa setEof measure (E) such thatfnconverges uniformly tofoutside the :This is Ergoroff s theorem. See (Egoroff s)Iffnis a sequence of measurable functions that converge to a on a measurable setEof finite measure, then given >0, there is a subsetAofEwith (A)< such thatfnconverges tofuniformly onE\AProof:Let >0, then for eachn, there exists a setAn Ewith An< 2 n, and there is anNnsuch that for allx / Anandk Anwe have|fk(x) f(x)|<1/n.

8 LetA= An, then byconstructionA Eand A < . Choosen0such that 1/n0< . Now ifx / Aandk Nn0then|fk(x) f(x)|<1/n0< . Thereforefnconverges uniformly onE\ an absolutely continuous monotone function on[0,1]. Prove that ifE [0,1]is a set of Lebesgue measure zero, then the setg(E) ={g(x) :x E} Ris also a set of Lebesguemeasure :LetE [0,1] with zero measure, then for any epsilon >0, there exists an open coverOforE, such that (O\E)< . NowObeing open in [0,1] implies thatO= (an,bn), where (an,bn) aredisjoint. Now by absolutely continuity ofg(x) we have >0 n=1 (In)< n=1|g(In [0,1])|< Nowg(E) |g(In [0,1])|which implies that (g(E))< , so given an there exists a >0 suchthat the above hold, then let =.

9 Since is arbitrary we have (g(E)) = 0 Remark:The above problem ( ) is commonly refered to as Lusin s N Lipschitz continuous in[0,1]. Show that(a) (f(E)) = 0if (E) = 0.(b) IfEis measurable, thenf(E)is also :For part (a) iffis Lipschitz continuous then it is absolutely continuous, and so if (E) = 0,then (f(E)) = 0 (see above proof).For part (b) LetEbe a measurable set and let >0. Now there exists an open setOsuch that (O\E)< , whereOis a disjoint union of intervalsIn= (an,bn). Now sincefis absolutely continiuous,it can be approximated by simple functions, namely In.

10 Choose these functions such that f n=1cn In < Now ( In) =bn an>0, so it is measurable. Let R, then thef(E) is measurable if{x:f(x) }is a measurable set for any R. but we have now{x:f(x) } {x: In+ }We know simple functions are measurable, and our choice of simple functions approximatesf(x),thereforefis measurable .Theorem (Lusin s)Letfbe a measurable real -valued function on an interval [a,b]. Then given >0, there is a continuous function on [a,b] such that {x:f(x)6= (x)}< 5 Proof:Letf(x) be measurable on [a,b] and let >0. For eachn, there is a continuous functionhnon [a,b] such that {x:|hn(x) f(x)| 2 n 2}< 2 n 2 Denote these sets asEn.