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<ul><li><p>Problem Set</p><p>Problem Set #2</p><p>Math 5322, Fall 2001</p><p>December 3, 2001</p><p>ANSWERS</p></li><li><p>i</p></li><li><p>Problem 1. [Problem 18, page 32]Let A P(X) be an algebra, A the collection of countable unions of sets</p><p>in A, and A the collection of countable intersections of sets in A. Let 0 bea premeasure on A and the induced outer measure.</p><p>a. For any E X and > 0 the exists A A with E A and (A) (E) + .</p><p>Answer :</p><p>By definition,</p><p>(E) = inf</p><p>{ i=1</p><p>0(Ai) | {Ai }i=1 A, E i=1</p><p>Ai</p><p>}.</p><p>Thus, given > 0, we can find some sequence {Ai }i=1 A so that</p><p>E i=1</p><p>Ai andi=1</p><p>0(Ai) < (E) + .</p><p>Define a set A E by</p><p>A =i=1</p><p>Ai.</p><p>Since A is a countable union of sets in A, A A. Since is subadditive,</p><p>(A) i=1</p><p>(Ai) =i=1</p><p>0(Ai) < (E) + ,</p><p>since = 0 on A. This completes the proof.</p><p>b. If (E) </p></li><li><p>Then E B and B A, since it is a countable intersection of sets inA. For every n, we have</p><p>(E) (B) (An) < (E) + 1/n.</p><p>Since n is arbitrary, we must have (B) = (E).</p><p>Since E B, B is the disjoint union of E and B \ E. These three sets arein M, so we can use the additivity of on M to get</p><p>(B) = (E) + (B \ E).</p><p>All three terms are finite and (E) = (B), so simple algebra shows(B \ E) = 0.For the second part of the proof, suppose that E X and there is a setB A so that E B and (B \E) = 0. As remarked above, A Mso B is measurable. We showed in class that if F X and (F ) = 0, thenF is -measurable. Thus, in the present situation, B \ E is measurable.But E = B\(B\E) (check!). SinceM is closed under taking set differences,we conclude that E is measurable. This completes the proof.</p><p>c. If 0 is -finite, the restriction (E) </p></li><li><p>So, we proceed as follows. For each n N, we invoke part (a) to get a setAnj A such that Ej Anj and</p><p>(Anj ) < (Ej) +</p><p>1n2j</p><p>.</p><p>Since Anj and Ej are measurable and have finite measure, this gives us</p><p>(Anj \ Ej) 0, then E contains a non-measurable set. (It suffices to assumethat E [0, 1). In the notation of Section 1.1, E =</p><p>rRE Nr.)</p><p>Answer :</p><p>Briefly, if E R and m(E) > 0, we can write</p><p>E =</p><p>n=E [n, n+ 1), (disjoint union).</p><p>9</p></li><li><p>Since m(E) > 0, at least one of the sets E [n, n + 1) must have nonzeromeasure. Select one such set F = E [n, n + 1). If F contains a nonmea-surable set, so does E. The set F n is contained in [0, 1) and has the samemeasure as F . If we prove that every subset of [0, 1) of nonzero measurecontains a nonmeasurable set, then F n will contain a non-measurableset A. But then A+ n F is nonmeasurable (if A+ n was measurable, sowould be (A+ n) n = A).Thus, it will suffice to consider the case where E [0, 1) and m(E) > 0.First, observe that the process weve defined above that sends S to Sr isinvertible. Indeed, S0 = S and for t (0, 1) we define t : [0, 1) [0, 1) by</p><p>t(x) =</p><p>{x+ t, x [0, 1 t)x+ t 1, x [1 t, 1).</p><p>Then St = t(S). It is easy to check that t([0, 1 t)) = [t, 1) and t([1t, 1)) = [0, t). It is then easy to check that the inverse of t is 1t. Thus</p><p>(St)1t = S.</p><p>From the first part of the problem, we can conclude that if F is measurableand F Nr for some r then m(F ) = 0. To see this, note that if F Nrthen F1r N . As weve argued before, F1r is measurable andm(F1r) =m(F ). From the first part of the problem, m(F1r) = 0, so we concludethat m(F ) = 0.</p><p>Now suppose that E [0, 1) and m(E) > 0. The interval [0, 1) is thedisjoint union of the sets Nr for r R, so E is the disjoint union of the setsE Nr. If all of the sets E Nr were measurable, we would have</p><p>m(E) =rR</p><p>m(E Nr) =rR</p><p>0 = 0,</p><p>which contradicts the assumption that m(E) > 0. Thus at least one of thesets E Nr must be nonmeasurable and we conclude that E contains anonmeasurable set.</p><p>Before going into the problems on Page 48ff. of the book, lets recall some ofthe definitions involved.</p><p>Let (X,M) be a measurable space, soM P(X) is a -algebra. If A X,we can define a -algebra M|A P(A) by</p><p>M|A = {E A | E M} .(0.1)</p><p>If A itself is in M, we have</p><p>M|A = {E | E A, E M} .</p><p>10</p></li><li><p>If (Y,N ) is a measurable space and f : X Y , we say that f is measurableon A X if f |A : A Y (the restriction of f to A) is measurable for the-algebra M|A, which means that (f |A)1(N) = f1(N) A M|A for allN N . If A M, then f is measurable on A iff f1(N) A M for allN N . Clearly, if f is a measurable function X Y , then f is measurable onevery subset of A of X since, in this case, f1(N)A M|A for all N , becausef1(N) M.</p><p>The following proposition is pretty easy, but useful.</p><p>Proposition 0.1. Let (X,M) and (Y,N ) be measurable spaces and let f : X Y be a function. Let {E }A be a countable (finite or infinite) collection ofmeasurable subsets of X such that</p><p>X =A</p><p>E.</p><p>Then, f is measurable if and only if f is measurable on each E, i.e., f ismeasurable if and only if for each N N , f1(N) E M for all A.</p><p>Proof. If f is measurable, then f1(N) M for allN N . Since each E M,we have f1(N) E M.</p><p>Conversely, suppose that for each N N , f1(N) E is measurable forall . Then</p><p>f1(N) =A</p><p>f1(N) E</p><p>is a countable union of sets inM, so f1(N) M. Thus, f is measurable.</p><p>Problem 6. [Problem 1, page 48]Let (X,M) be a measurable space. Let f : X R and Y = f1(R). Then</p><p>f is measurable iff f1({}) M, f1({}) M and f is measurable onY .</p><p>Answer :Suppose first that f is measurable. Then f is measurable on Y , as discussedabove. The sets {} and {} are closed sets, hence Borel sets, in R, sof1({}) and f1({}) are measurable.</p><p>For the second part of proof suppose f1({}) and f1({}) are mea-surable and f is measurable on Y . The sets f1({}), f1({}) andY form a partition of X. Since f1({}) and f1({}) are measurable,Y = X \ (f1({}) f1({})) is measurable. To show f is measurableit will suffice to show it is measurable on the three sets in our partition. Ofcourse, f is measurable on Y by assumption. On the set f1({}), f isconstant and a constant function is always measurable. Similarly, f is constant,and hence measurable, on f1({}). This completes the proof.</p><p>11</p></li><li><p>Problem 7. [Problem 2, page 48]Let (X,M) be a measurable space. Suppose that f, g : X R are measur-</p><p>able.</p><p>a. fg is measurable (where 0 () = 0).</p><p>Answer :</p><p>Well consider two solutions, in slightly different spirits.</p><p>First Solution. Since f is measurable, we can partition X into the mea-surable sets F1 = f1(), F2 = f1((, 0)), F3 = f1(0), F4 =f1((0,)) and F5 = f1(). (f1(a) is the same thing as f1({ a }).) Wehave a similar partition G1 = g1(), G2 = g1((, 0)), G3 = g1(0),G4 = g1((0,)), G5 = g1() for g. Taking pairwise intersections, we geta partition of X into the 25 measurable sets (dont panic!) FiGj . To showfg is measurable it will suffice to prove that fg is measurable on each of thesets Fi Gj .Observe that fg is constant on the set F1 G1 = f1() g1().Indeed fg is constant on all of the sets F1 Gj , j = 1, . . . , 5. Similarly, fgis constant on F5 Gj , j = 1, . . . , 5 and Fi G1, i = 1, . . . 5 and Fi G5,i = 1, . . . , 5.</p><p>This leaves the nine sets Fi Gj , i, j = 2, 3, 4 to consider. But the union ofthese sets is S = f1(R) g1(R). Of course, f and g are real valued on Sand so fg is measurable on S by Proposition 2.6 (p. 45). This completes thesolution.</p><p>Second Solution We try to follow the proof of Proposition 2.6 on page 45of the book. Thus, we define F : X R R by F (x) = (f(x), g(x)). Asdiscussed in the book, this is measurable.</p><p>Next we define : R R R by (x, y) = xy (with the 0 () = 0convention). Note that is not continuous at the four points (0,),(, 0).Nonetheless, we claim that is Borel measurable. To see this, define A R R by</p><p>A = { (0,), (0,), (, 0), (, 0) } .</p><p>This is a closed subset of RR, hence a Borel subset. Thus, B = RR \Ais Borel.</p><p>Recall from our discussion in class that if Z is a metric space, and W is aBorel subset of Z,</p><p>(BZ)|W = {E W | E BZ } = BW .</p><p>Thus, the principal of the Proposition we proved above applies: If Z is theunion of two Borel sets V and W then f : Z R is Borel measurable if andonly if f is Borel measurable on V and W .</p><p>12</p></li><li><p>Thus, in order to show that is Borel measurable, it will suffice to show itis Borel measurable on A and B. On A, is constant (with value 0) andso Borel measurable. On B, is continuous, and hence Borel measurable.This proves the claim that is Borel measurable.</p><p>To complete the solution, let h = fg. Then h = F . Let B R be aBorel set. Since is Borel measurable, 1(B) is Borel in R R. Since Fis measurable, F1(1(B)) M. Thus,</p><p>h1(B) = ( F )1(B) = F1(1(B)) M,</p><p>which shows that h is measurable.</p><p>b. Fix a R and define h(x) = a if f(x) = g(x) = and h(x) = f(x)+g(x)otherwise. Then h is measurable.</p><p>Answer :</p><p>First Solution. We can partition X into the measurable sets</p><p>f1() g1()(A.1)f1() g1(R)(A.2)f1() g1()(A.3)f1(R) g1()(A.4)f1(R) g1(R)(A.5)f1(R) g1()(A.6)f1() g1()(A.7)f1() g1(R)(A.8)f1() g1()(A.9)</p><p>and it will suffice to prove that h is measurable on each of these sets. Onthe sets (A.3) and (A.7), h is constant, with value a, by our definition. Onthe sets (A.1), (A.2), (A.4), (A.6), (A.8), and (A.9), h is constant (withvalue either ). Finally, on the set f1(R) g1(R) in (A.5), both fand g are real valued and h coincides with f + g, which is measurable byProposition 2.6. This completes the solution.</p><p>Second Solution. As in the first part of the problem, define F : X RRby F (x) = (f(x), g(x)). As discussed in the book, this is measurable. LetA R be defined by</p><p>A = { (,), (,) } ,</p><p>13</p></li><li><p>which is closed, and hence Borel, in RR. Let B = RR \A, which is alsoBorel. Define : R R R by</p><p>(x, y) =</p><p>{a, (x, y) Ax+ y, (x, y) B.</p><p>This function is not continuous at the points in A, but it is nonetheless Borelmeasurable. Of course, is continuous, hence Borel measurable, on B, and is constant, hence Borel measurable, on A. Thus, is Borel measurableon R R.Since is Borel measurable, F is measurable (as discussed in the solutionof the first part of the problem), but h = F , so the proof is complete.</p><p>Problem 8. [Problem 3, page 48]Let (X,M) be a measurable space.If { fn } is a sequence of measurable functions on X, then {x | lim fn exists }</p><p>is a measurable set.</p><p>Answer :The problem is not stated very well, since it leaves some ambiguity about wherethe values of the fns are supposed to be. Going by the previous problems, Idsay we should consider functions with values in R.</p><p>Define E = {x | lim fn exists }. By Proposition 2.7, the functions g, h : X R defined by</p><p>g(x) = lim infn</p><p>fn(x)</p><p>h(x) = lim supn</p><p>fn(x)</p><p>are measurable and, of course,</p><p>E = {x X | g(x) = h(x) } ,</p><p>so it will suffice to prove that the set where these two measurable functions areequal is measurable.</p><p>For a first attempt, one could try to define (x) = h(x) g(x), claim that is measurable and observe that E = 1(0). Its not quite that easy, sincethere may be points where h(x) g(x) is undefined, e.g., if h(x) = g(x) =.</p><p>We can take care of this in the spirit of some previous solutions. We can</p><p>14</p></li><li><p>partition X into the following measurable sets</p><p>S1 = g1() h1()(A.1)S2 = g1() h1(R)(A.2)S3 = g1() h1()(A.3)S4 = g1(R) h1()(A.4)S5 = g1(R) h1(R)(A.5)S6 = g1(R) h1()(A.6)</p><p>S7 = g1() H1()(A.7)S8 = g1() h1(R)(A.8)S9 = g1() h1()(A.9)</p><p>It will suffice to show that E Sj is measurable for j = 1, . . . , 9 (since thenE is a finite union of measurable sets).</p><p>In case (A.1), we have E S1 = S1, which is measurable.In case (A.2), E S2 = , since g(x) 6= h(x) for all x S2.In case (A.3), E S3 = .In case (A.4), E S4 = .In case (A.5), both g and h are real-valued on S5, so E S5 = (h g)1(0),</p><p>which is measurable, since h g is measurable on S5 by Proposition 2.6.In case (A.6), E S6 = .In case (A.7), E S7 = .In case (A.8), E S8 = .Finally, in case (A.9), E S9 = S9, which is measurable.This completes the solution.Alternate Solution. Just for fun, heres another way to do it. Since g h,</p><p>X is the disjoint union of E = {x | g(x) = h(x) } and F = {x | g(x) < h(x) }.Since the complement of a measurable set is measurable, it will suffice to showthat F is measurable.</p><p>We claim</p><p>F =rQ{x | g(x) < r } {x | r < h(x) } .()</p><p>To see this, suppose first that p F . Then g(p) < h(p) so there is some rationals so that g(p) < s < h(p). But then</p><p>p {x | g(x) < s } {x | s < h(x) } ,</p><p>which is one of the sets in the union in ().Conversely, if p is in the union in (), then there is some r Q so that</p><p>p {x | g(x) < r } {x | r < h(x) } .</p><p>15</p></li><li><p>But then g(p) < r and r < h(p), so g(p) < h(p), i.e., p F . This completes theproof of the claim.</p><p>Of course, each of the sets</p><p>{x | g(x) < r } {x | r < h(x) } = g1([, r)) h1((r,])</p><p>is measurable, so () shows that F is a countable union of mensurable sets,hence measurable.</p><p>Problem 9. [Problem 4, page 48]Let (X,M) be a measurable space.If f : X R and f1((r,]) M for each r Q, then f is measurable.</p><p>Answer :As remarked in the book on page 45, the Borel algebra BR of R is generatedby the rays (a,] for a R. Hence, to show f is measurable, it will suffice toshow that f1((a,]) is measurable for each a R.</p><p>To do this, let a R be fixed but arbitrary. Since the rationals are dense inR, we can find a sequence of rationals { rn } that decrease to a, i.e., rn a andthe rns form a decreasing sequence.</p><p>We claim that</p><p>(a,] =n=1</p><p>(rn,].()</p><p>To see this, first suppose x (a,]. Then x > a and we can find an openinterval U around a that does not contain x. Since rn a, there is some Nsuch that rn U for n N . But then rn < x, so x (rn,], for n N . Thus,x is in the union on the right hand side of ().</p><p>Conversely, if x is in the union in (), then x (rn,] for some n. But thena < rn < x, so x (a,]. This completes the proof of the claim.</p><p>From (), we conclude that</p><p>f1((a,]) =n=1</p><p>f1((rn,]).</p><p>Since rn is rational, f1((rn,]) is measurable by our hypothesis. Thus,f1((a,]) is a countable union of measurable sets, and so is measurable. Thiscompletes the proof.</p><p>Problem 10. [Problem 9, page 48]Let f : [0, 1] [0, 1] be the Cantor function and let g(x) = f(x) + x.</p><p>16</p></li><li><p>a. g is a bijection from [0, 1] to [0, 2] and h = g1 is continuous from [0, 2] to[0, 1].</p><p>Answer :</p><p>Since the Cantor function is continuous, g is continuous.</p><p>Recall that the Cantor function is nondecreasing, i.e., if x < y then f(x) f(y). But then if x < y, g(x) = f(x) + x < f(y) + y = g(y). Thus, g isstrictly increasing and hence one-to-one. Since f(0) = 0 and f(1) = 1, wehave g(0) = 0 and g(1) = 2. If x [0, 1], 0 = g(0) < g(x) < g(1) = 2, sog maps [0, 1] into [0, 2]. By the intermediate value theorem, every point in[0, 2] is in the image of g. Thus, g is a bijection from [0, 1] to [0, 2].</p><p>The fact that h = g1 is continuous is a general fact about strictly monotonefunctions that follows from the intermediate value theorem. For complete-ness, well give a proof here, but you should probably skip it on a firstreading.</p><p>Lemma 0.2. Let f : I R be a continuous strictly increasing function,where I R is an interval. Then the image of an interval J I is aninterval of the same type (i.e., open, closed, etc.) In particular, f(I) is aninterval of the same type as I.</p><p>Proof of Lemma. Let (a, b) I be an open interval. If x (a, b) thena < x < b and f(a) < f(x) < f(b), since f is strictly increasing. Thus,f((a, b)) (f(a), f(b)). On the other hand, if y (f(a), f(b)) then there isan x (a, b) such that f(x) = y, by the intermed...</p></li></ul>

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