Problem Set #3 Math 471 Real Analysis Assignment: Chapter 3, #2, 3, 9, 10, 13

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1 Problem Set #3 Math 471 Real Analysis Assignment: Chapter 3, #2, 3, 9, 10, 13 Clayton J Lungstrum September 19, 2012

2 Exercise 32 When b =3in Exercise 1, the expansion is called the triadic or ternary expansion of x Show that the Cantor set C consists of all x such that x has some triadic expansion for which every c k is either 0 or 2 Let f(x) be the Cantor-Lebesgue function: see p 35 Show that if x 2 C and x = P 1 c k3 k, where each c k is either 0 or 2, then f(x) = P 1 ( 1 2 c k)2 k Solution Let x 2 C, the Cantor set, with ternary expansion x =0c 1 c 2 c 3 Then x 2 C n, where C n is the n th stage of construction of the Cantor set, for all n 1 Therefore, at the first stage of the construction, if x is in the leftmost interval, set c 1 in the ternary expansion to 0; if x is in the rightmost interval, set c 1 =2 Atthenextstep,theintervalx was in is divided into 2 subintervals, x is in either the leftmost subinterval of the previous interval, or it is in the rightmost subinterval of the previous interval In the former case, set c 2 =0,inthe latter, c 2 = 2 Continuing in this manner, we see that the Cantor set consists of all of those numbers whose ternary expansion consists solely of 0 s and 2 s Conversely, suppose x has a ternary expansion 1X where each c k 2 {0, 2} We will show x 2 C by induction Clearly x 2 C 0 since 0 apple x apple 1 Next, if c 1 =0,then On the other hand, if c 1 =2,then c k 3 k x apple (00222 ) 3 =(01) 3 = 1 3 x (02000 ) 3 =(02) 3 = 2 3, and hence, x 2 C 1 Now we approach the inductive step Assume x 2 C k for some k 2 N Then x is in any one of the 2 k subintervals of C k, each of length 1 Without loss of generality, say x 2 [a, b] Since we remove the interior of the 3 k middle-thirds, we have two disjoint closed intervals, apple a, a k+1 and apple b 1,b 3k+1 If c k+1 = 0, then x apple a + 1 and is therefore in the left interval If c 3 k+1 k+1 = 2, then 1 x b, and then x would be in the right interval Either way, x 2 C 3 k+1 k+1 Thus, by induction, we have shown x 2 C Combining both of these properties, we have that x 2 C if and only if its ternary expansion consists of 0 s and 2 s only For the second part of the problem, let f(x) bethecantor-lebesguefunction Then f(x) is continuous and thus f(x) =lim n!1 f(a (n) x ) By the definition of f(x), we have 1

3 f(x) =lim k!1 f k (x) andf k (a (n) x )=f k+1 (a (n) x )forallk n, where f k (x) isdefinedhereas in the textbook Thus, f(a (n) x )=f m (a (n) x )= nx fm (a (k) (k 1) x ) f m (a x ), where f m (a (0) x ) = 0 Then we have that this equals nx c k 2 2 k Thus, f(x) =lim n!1 f(a (n) x )= P 1 c k2 2 k, as desired 2

4 Exercise 33 Construct a two-dimensional Cantor set in the unit square {(x, y) :0apple x, y apple 1} as follows Subdivide the square into nine equal parts and keep only the four closed corner squares, removing the remaining region (which forms a cross) Then repeat this process in a suitably scaled version for the remaining squares, ad infinitum Show that the resulting set is perfect, has plane measure zero, and equals C C Solution Let us denote this two-dimensional Cantor set by C 2, and let C0 2 be the unit square (the first stage in the construction) while Ck 2 is the kth stage in the construction Thus, it is clear that C 2 = T 1 C2 k Note that we re removing the interior of intervals in each iteration, thus C 2 is the complement of an open set, thus is closed Now we will show that each point is a limit point To that end, let A be a corner point of the closed square in Ck 2 Then notice that A 2 C 2 since A 2 Cn 2 for all n k Using this fact, let P be a point in C 2 Then, for every k 0, there is a unique closed square containing P Let A k be a corner point di erent from P in the k th stage of construction in the same square containing P Then P is a limit point of the sequence of corner points given by {A k } 1 since P A k < p 23 k and each corner point is in C 2 as established above Thus, C 2 is closed and every point is a limit point, therefore C 2 is a perfect set It s clear that C 2 is measurable since it is the countable intersection of measurable sets (established in class) Notice that there are 4 k squares in each k th iteration, each of area 9 k Therefore, Ck 2 = 4 k 9, which tends to zero as k tends to infinity From the monotonicity of measurability, we see that C 2 =0 Our final step is to establish C 2 = C C To prove this, we ll use mathematical induction Clearly, C0 2 = C 0 C 0 Then, assume that Ck 2 = C k C k for some k 0 Now, let Q =(x, y) 2 Ck 2 If Q 2 C2 k+1, then from construction, this means that x(x) isineitherthefirstthirdor the last third of the closed square (relative to the x-axis) in which it was contained in the set Ck 2 A similar argument shows this to be true for y However, if Q/2 C2 k+1, then either the x or the y was in the middle third of its respective axis to be removed, ie, x (x) /2 C or y (y) /2 C Either way, Q/2 C k C k Hence, Ck 2 = C k C k, and it follows that!! C 2 = Ck 2 = (C C) = C C = C C k=0 3

5 Exercise 39 If {E k } 1 is a sequence of set with P 1 E k < 1, show that lim sup E k (and so also lim inf E k ) has measure zero Solution Recall that if {E k } 1 is a sequence of sets, then lim sup E k = For simplicity, let E =limsupe k Then clearly j=1! E k k=j E j=n! E k E k k=j k=n Thus, by monotonicity and countable subadditivity, we have apple 1X apple E k e E k k=n e k=n Now, since P 1 E k e < 1, for any " > 0, we can choose N 2 N such that if n N, then P 1 k=n E k e < " Thus, applying this to the above inequality, for every " > 0, we have < ", thus = 0 As we showed in class, E is then measurable and E =0 Since lim inf E k lim sup E k, by monotonicity, lim inf E k =0 4

6 Exercise 310 If E 1 and E 2 are measurable, show that E 1 [ E 2 + E 1 \ E 2 = E 1 + E 2 Solution First, assume that E 1 < 1 and E 2 < 1, otherwise the result is trivial Notice that since E 1 and E 2 are measurable, we have E 1 [ E 2 measurable, thus E 1 [ E 2 = (E 1 [ E 2 ) \ E 1 + (E 1 [ E 2 ) \ E C 1 = E 1 + E 2 \ E C 1 Similarly, notice that since E 2 is measurable, we have E 2 = E 2 \ E 1 + E 2 \ E C 1 Now, taking the left-hand side of the first identity and adding it to the right-hand side of the second identity, and similarly for the opposite sides, we have E 1 [ E 2 + E 2 \ E 1 + E 2 \ E C 1 = E 2 + E 1 + E 2 \ E C 1 From both sides we can subtract E 2 \ E C 1 and so we obtain the desired result 5

7 Exercise 313 Motivated by (37), define the inner measure of E by E i =sup F, where the supremum is taken over all closed subsets F of E Show that E i apple and if < 1, then E is measurable if and only if E i = [Use (322)] Solution Let E R n, let G be an open set containing E, andf E closed Since G is open, it is measurable, thus G = G \ F + G \ F C = F + G F Thus, F = G G F, and since Lebesgue measure is nonnegative, we clearly have F apple G Since F is a lower bound for G, it is clearly less than or equal to the greatest lower bound, ie, F apple inf G = Now we have on the other hand as an upper bound for F, thus it must be greater than or equal to the least upper bound, ie, E i =sup F apple, and so we have the desired inequality E i apple Now suppose < 1 and that E is measurable Then, given " > 0, there is a closed set F 2 M with E F e < " Since F 2 M, = E \ F e + E F e < E \ F e + " This implies E \ F e > " Now, since E i F and F is measurable, we have E i F F E {z } e = E \ F e =0 Thus, we have E i > " Since " > 0wasarbitrary,weseethat E i = Conversely, suppose < 1 and E i = Then theorem (38) implies that there exist sets of type F and G with the property that F E G, where F is F and G is G,and F = G Then E di ers only by a set of measure zero, and hence is measurable by Theorem (328) 6