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# ANALYSIS II MATH 105

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This 2 page Class Notes was uploaded by Kavon Feest on Thursday October 22, 2015. The Class Notes belongs to MATH 105 at University of California - Berkeley taught by Staff in Fall. Since its upload, it has received 15 views. For similar materials see /class/226615/math-105-university-of-california-berkeley in Mathematics (M) at University of California - Berkeley.

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Date Created: 10/22/15

Mathematics 105 Spring 2004 7 M Christ Midterm Exam 2 Comments Distribution of scores There were 50 points possible The highest score was 50 and the third highest was 39 The median was 25 the 75th percentile was 35 and the 25th percentile score was 21 Final course grades will be based1 15 on MTl 20 on this exam 20 on problem sets and 45 on the nal exami 1a In de ning a Lebesgue measurable set a few students wrote GA lt 5 instead of GAe lt e The distinction is that G A is not de ned unless one already knows that G A is measurable In testing measurability one uses exterior measure which is de ned for all sets 1b If f is measurable then E f dc is de ned if and only if the two numbers E fJr dc and E f dc are not both equal to 00 A few students also mentioned the case where both integrals equal 700 But this can never arise since f f are by de nition nonnegative functions The question did not call for a de nition of E fdc so there was no need for you to reproduce the chain of de nitions 1C A few people forgot to put absolute value signs around det Clearly these are needed the Lebesgue measure of TA canlt be negative 2a The relevant example is the fat or generalized Cantor set which we discussed in a homework problemi Given a 6 01 this set is constructed by deleting from 01 a subinterval of length 13 leaving two subintervals of equal lengthsi From the middle of each of those we delete a subinterval of length 132 leaving four subintervals of equal lengthsi From the middle of each of those we delete a subinterval of length 133 and so forth the fat Cantor set is the set of all points which are never deleted if this process is continued through in nitely many steps 2b Q N 01 is an example as was shown in a homework problemi It is an essential part of the de nition of outer exterior measure and hence of Lebesgue measure that countably in nite covers are allowed using nite covers by rectangles to de ne outer measure would yield a different notion under which for instance Q N 01 would have outer measure equal to 1 rather than equal to 0 3 Please see Stroock or lecture notes for solution The most common problem here and it turned out to be quite a bit more common than I had expected was a misunderstanding of the relationship between S z z gt a and the sets 5 z z gt a given that A for all z E E A typical error was to assert that S UnSn or nSni To refute the rst assertion consider the example 100 and 71 for all n 2 2 with a 0 To refute the second set a 0 and ln for all n Then I 6 Sn for all n but 0 so z 5 A few people complicated these assertions by introducing sets like Un UqEQ z fn gt a 4 without affecting the essential difficulty described above A good method of attack was to note that limnn00 fn lim supngDo for all z there7s no need to bring the lim inf into the discussion in this problem since existence of the limit for all z is given Setting gnz supkgtn it is true that z gnz gt a Uk2nz gt a so the set on the left is a countable intersection of measurable sets and hence is measurable Now a subtler mistake was sometimes made namely the assertion that z z gt a nz gnz gt a Here 9 decreases to limsup for all I by its de nition However it could happen that 9 gt a for all n but the limit equals a A true statement is that z z 2 a nz gnz 2 a and this leads to a solution This problem illustrates nicely how limsup and lim inf are often useful in discussing genuine limits they7re not merely substitutes to be used when the limit fails to exist 11f you ever plan to be teaching and grading in your future career you might nd it instructive to re ect on what on earth this sentence could actually mean particularly in light of concepts like standard deviation 4 This problem caused quite a bit more dif culty than lld anticipated 1 expected problem 5 to be the tricky one but on the average people had more trouble with number Let7s fix rmly in our minds the fact that this result is false if it is only given that 0 S A XAI for all 1 Indeed consider fj XMHH and A 0 with EnLa ORE Then 0 S A XAI for all 1 yet dA 1 while lAl 0 Another example is ilj for all z E R Now all hypotheses are satis ed with A 0 except that fj are not nonnegativei Again the conclusion is false Many solutions introduced a bit of notation and then simply claimed the conclusion without offering any analysis at all In particular if you didn t use the monotonicity hypothesis fj S fj1 and the hypothesis that fj 2 0 then you can t possibly have writtenwritten a correct proofl A more sophisticated fallacy to which several victims fell prey is to write fj out in the form fj 221 Cj XBM where 01 are nonnegative scalars and the sets BMc are pairwise disjoint for each xed j as k varies and are measurable Often this was done with de cient notation which didn t involve a double subscript and didn t acknowledge that the number Nj of terms in the jth sum could depend on j and could even tend to in nity with j 7 a potential source of trouble in any proof It then follows directly that either 01 0 in which case this term may be dropped or Bj k C Al The more sophisticatd error was then to assert that as j A 00 mink 01 tends to l in other words fj A XA uniformly as a function of z E E as j A 00 This does not follow from the hypotheses For example we could have with E R1 0 for all I S lj and 1 for z gt lj so that A 0 The hypotheses are satis ed yet the conclusion is not uniformi Worse the number of terms Nj could tend to in nity allowing worsi cations of this basic example 5 Please see text for a solution Some very good solutions of this longish problem were sub mitted Other students proved beyond doubt that they had studied the proof but produced only disconnected fragments rather than a coherent argumenti t7s not possible to nd an exact cover of any set by rectanglesi For instance its impossible for a Cantor set of measure zero in any exact cover by intervals each interval would be forced to have length zero hence would contain a single point so such a cover would cover only countably many points hence would miss most of the Cantor set What we proved is that this can be done for any open set it can be covered by a nonoverlapping family of cubes lying inside it the sum of whose measures equals the total measurei Therefore a correct proof begins by approximating A from the outside by an open set whose measure is nearly zeroi Its true and immediate that for any set S of diameter 6 is contained in a set of diameter S 36 However some asserted implicitly or explicitly that for any rectangle R of diameter 6 is contained in a rectangle whose outer measure is S ClRl where C is a certain nite constant whose exact value is not important here This is false lf zy 7 12 for all zy in some neighborhood of the origin then f transforms a rectangle R 0 c X 0 6 to a set which is shaped for small 6 like a very skinny banana draw it to see Let 5 remain xed and imagine what happens as 6 A 0 I you draw any rectangle containing fR you nd that its volume is bounded below by a positive constant which does not tend to zero with 6

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