MATH 361 Homework 9. Royden Royden Royden

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1 MATH 61 Homework 9 Royde..9 First, we show that for ay subset E of the real umbers, E c + y = E + y) c traslatig the complemet is equivalet to the complemet of the traslated set). Without loss of geerality, assume E ca be writte as a ope iterval e 1, e 2 ), so that E c + y is represeted by the set {x x, e 1 + y) e 2 + y, + )}. This is equal to the set {x x / e 1 + y, e 2 + y)}, which is equivalet to the set E + y) c. Secod, Let B = A y. From Homework 8, we kow that outer measure is ivariat uder traslatios. Usig this alog with the fact that s measurable: m A) = m B) = m B E) + m B E c ) = m B E) + y) + m B E c ) + y) = m A y) E) + y) + m A y) E c ) + y) = m A E + y)) + m A E c + y)) = m A E + y)) + m A E + y) c ) The last lie follows from E c + y = E + y) c. Royde..10 First, sice E 1, E 2 M ad M is a σ-algebra, E 1 E 2, E 1 E 2 M. By the measurability of E 1 ad E 2 : m E 1 ) = m E 1 E 2 ) + m E 1 E2) c m E 2 ) = m E 2 E 1 ) + m E 2 E1) c m E 1 ) + m E 2 ) = 2m E 1 E 2 ) + m E 1 E2) c + m E1 c E 2 ) = m E 1 E 2 ) + [m E 1 E 2 ) + m E 1 E2) c + m E1 c E 2 )] Secod, E 1 E 2, E 1 E c 2, ad E c 1 E 2 are disjoit sets whose uio is equal to E 1 E 2. As above, sice E 1, E 2 M, E c 1, E c 2 M ad hece E c 1 E 2, E 1 E c 2 M. From class, m E ) = m E ) for measurable sets E. Therefore: m E 1 E 2 ) + m E 1 E c 2) + m E c 1 E 2 ) = m E 1 E 2 ) Combiig the two argumets above: Royde..11 Defie E =, + ). m E 1 ) + m E 2 ) = m E 1 E 2 ) + [m E 1 E 2 ) + m E 1 E c 2) + m E c 1 E 2 )] i. Empty itersectio: = m E 1 E 2 ) + m E 1 E 2 ) E = For ay x R, we ca choose a atural umber > x such that x / E. Therefore, there is o x R such that x E for all. This implies that the itersectio stated above is empty. ii. By defiitio, m E ) = +, as each iterval is a ope iterval cotaiig +. 1

2 Royde..12 i. From lecture, we established the followig for a coutable sequece of : m A )) = m A ) For the ifiite case, we use the mootoicity property: A ) A ) m A )) m A m A ) Sice this is true for all N, lettig : A ) m A ) ii. The reverse iequality is true by coutable subadditivity: A ) m A ) From the two iequalities i parts i) ad ii), we ca coclude: A ) = m A ) Royde..1 a. Showig i) ii) vi). )) i) ii): By propositio 5 i Royde, for all sets E, there exists a ope set O such that E O ad m O) m E) + ɛ. Sice s measurable, for such a set O: m O) = m O E) + m O E c ) m O E) + m O E c ) m E) + ɛ m E) + m O \ E) m E) + ɛ m O \ E) ɛ To make the iequality above strict, we ca take ɛ = ɛ 2 for ay give ɛ > 0 ad use the same reasoig above. ii) iv): Sice O is ope, it ca be writte as a coutable disjoit uio of ope itervals. We pick a ope O such that m O\E) < ɛ/2. We cosider two cases: Case 1: O is a ifiite uio of ope itervals: O = m O) 2 I m I )

3 From ii), it is give that m E) <. So: m O) m E) + m O \ E) < Sice the outer measure is fiite, the ifiite sum above must coverge. Therefore, there exists some N such that for all > N, N I < ɛ 2. Defie: U = N Case 2: O is a fiite uio of K itervals. The defie: U = The symmetric measure ca be decomposed ito a uio of disjoit sets: K I I U E = U \ E) E \ U) m U E) m U \ E) + m E \ U) m O \ E) + m O \ U) < m O \ E) + ɛ/2 < ɛ vi) ii) By Propositio 5, there exists some ope set Q such that E \ U) Q ad m Q) m E \ U) + ɛ. Defie O = U Q. The set O covers E, sice U E) U ad E \ U) Q. The: b. Showig i) ii) iv) i) i) ii): Show i part a) above. ii) iv): Defie G as follows: m O \ E) = m U Q) \ E) = m U \ E) Q \ E)) m U \ E) + m Q \ E) m U \ E) + m Q) m U E) + m E\U) + ɛ 2m U E) + ɛ < ɛ G = O such that m O \ E) < 1 for all N. The existece of such O is guarateed by the coditio give i ii). Sice E O for each, E G, ad by the mootoicity property: m G \ E) m O \ E) < 1 Sice the iequality above holds for all N, we must have m G \ E) = 0. iv) i): Sice G is a coutable itersectio of ope sets, it is measurable. All sets with measure zero are measurable, so give that m G \ E) = 0, the set G \ E = G E c is also measurable. Therefore, G E c ) c is measurable, ad G G E c ) c = s also measurable.

4 c. Showig i) iii) v) i) i) iii): Sice s measurable, E c is also measurable. By ii), there exists some O such that m O \ E c ) < ɛ, or equivaletly, m O E) < ɛ. Defie F = O c. The: ɛ > m O E) > m E \ O c ) > m E \ F ) Sice O is ope, F is closed. Sice E c O, we kow O c E, or equivaletly, F E. iii) v): From iii), there exists a closed set F with F E ad m E \ F ) < 1. for all N. Defie the followig: F = F Sice F E for all N, F E. By mootoicity: m E \ F ) m E \ F ) < 1 Sice the iequality holds for all N, we ca coclude m E \ F ) = 0. v) i): Sice F is measurable ad m E \ F ) = 0, E \ F is also measurable. Sice E = F E \ F ), the uio of disjoit, measurable sets, s also measurable. Royde..14 a. Defie E 0 = [0, 1], E 1 = [0, 1 ] [ 2, 1], E = [0, 1 ] [ 1, 1]. The Cator set is equal to the itersectio of E for all N. I particular, E is a descedig sequece of measurable sets, as E +1 E, ad me 1 ) is fiite. By a propositio prove i lecture: m E ) = lim me ) Each E is a uio of disjoit closed itervals I. Sice closed itervals are measurable, ad mi) = li), we kow that me ) = m 2 I ) = 2 mi ) Therefore, it is sufficiet to show that the sum of the itervals which make up E as is equal to zero i order to show that the Cator terary set has measure zero. For ay, E is a uio of 2 closed ivervals each with legth 1. The sum of the legths of each iterval is 2 ) ad therefore me ) = 2 ). For ay ɛ > 0, take to be the first atural umber such that > log 2/ ɛ. This forces me ) < ɛ, which the implies lim me ) = 0. b. F is equal to a coutable uio of closed itervals, ad is therefore a closed set. To show F c is dese, defie F to be the remaiig closed itervals at each stage after the middle iterval of legth α is removed. The F is a coutable uio of disjoit itervals, each with legth strictly less tha 1 2. Therefore, give ay x [0, 1] ad ɛ > 0, choose N > log 1/2 ɛ. This will esure that the iterval x 1 2 ɛ, x ɛ) cotais some poit y that was removed i the th step. Usig the same reasoig as i part a), the measure of F is equal to the sum of the disjoit itervals whose uio is equal to F. At ay stage, there are 2 1 itervals before ay deletios are made. 4

5 Therefore, 2 1 itervals of legth α are removed. Therefore: Questio i Sice A 1 is measurable: mf ) = 1 = 1 α 2 = 1 α ) α 2 1 ) 2 m A 2 ) = m A 2 A 1 ) + m A 2 A c 1) m A 1 ) = m A 1 ) + m A 2 A c 1) m A 2 A c 1) = 0 ii Give that m B) = 0, where B is a subset of R, B must be measurable, sice for ay subset C, m C B) m B), so m C B) = 0. m C B) + m C B c ) = m C B c ) m C) iii From the two parts above, we ca coclude that the set A 2 A c 1 is a measurable set. Hece, we ca write A 2 as the uio of two measurable sets, A 1 ad A 2 A c 1. From the lecture otes, the collectio of measurable sets, M is closed uder takig uios. Therefore, A 2 = A 1 A 2 A c 1) is also measurable. Questio 4 By coutable additivity: ) m B m B ) To show the reverse iequality, defie B = B ad we ote that B [ B [ N By the coutability of : ) m B = m B [ m B [ N m B ) m B ) m B ) 5

Lecture 4: Cauchy sequences, Bolzano-Weierstrass, and the Squeeze theorem

Lecture 4: Cauchy sequences, Bolzano-Weierstrass, and the Squeeze theorem Lecture 4: Cauchy sequeces, Bolzao-Weierstrass, ad the Squeeze theorem The purpose of this lecture is more modest tha the previous oes. It is to state certai coditios uder which we are guarateed that limits