4. (10 pts) Prove or nd a counterexample: For any subsets A 1 ;A ;::: of R n, the Hausdor dimension dim ([ 1 i=1 A i) = sup dim A i : i Proof : Let s

Size: px

Start display at page:

Download "4. (10 pts) Prove or nd a counterexample: For any subsets A 1 ;A ;::: of R n, the Hausdor dimension dim ([ 1 i=1 A i) = sup dim A i : i Proof : Let s"

Sabina Garrett
5 years ago
Views:

1 Math 46 - Final Exam Solutions, Spring (10 pts) State Rademacher's Theorem and the Area and Co-area Formulas. Suppose that f : R n! R m is Lipschitz and that A is an H n measurable subset of R n. Rademacher's Thm. Then f is dierentiable H n almost everywhere. Area Formula. If n m, then A [[Df(x)]] dh n x = f (A) H 0 (A \ f,1 fyg) dh n y : Co-area Formula. If n m, then A [[Df(x)]] dh n x = f (A) H n,m (A \ f,1 fyg) dh m y :. (10 pts) Suppose A is an H 1 measurable subset of R with H 1 (A) < 1. True or False (No proofs necessary.) (a) H 1 [A \ B(a; r)] lim r#0 r = 1 for H 1 almost all a A. FALSE (See the example from the 1st week.) (b) H 1 (A) = inffh 1 (U) : U is an open neighborhood of Ag : FALSE Nonempty open sets have innite H 1 measure. (c) H 1 (A) = supfh 1 (K) : K is a compact subset of Ag :T RU E 3. (10 pts) Find the approximate Hausdor outer measures H 1 1(S 1 ) and H 1 p (S 1 ) where S 1 is the unit circle in the plane. H 1 1(S 1 )=and H 1 p (S 1 )=4 p.

2 4. (10 pts) Prove or nd a counterexample: For any subsets A 1 ;A ;::: of R n, the Hausdor dimension dim ([ 1 i=1 A i) = sup dim A i : i Proof : Let s = sup i dim A i. Then, for t>s, H t ([ 1 i=1 A i) i=1 H t (A i ) = 0 because t > dim A i for all i. Also, for r < s, there is a j with r < dim A j s so that H r ([ 1 i=1 A i) H r (A j ) = 1 : Thus, dim ([ 1 i=1 A i)=s. 5. (10 pts) Let S 0 be the unit square [0; 1] [0; 1] in the plane. Dividing S 0 into 9 congruent squares, let S 1 be obtained from S 0 by removing the middle third square: S 1 = S 0 n, [ 1 3 ; 3 ] [1 3 ; 3 ] : Similarly, let S be obtained from S 1 by omitting the middle third squares from each of the 8 remaining squares. Continuing, we let S i+1 be obtained from S 1 by omitting the middle third squares from the remaining squares of S i. What is the Hausdor dimension of S = \ 1 i=1 S i? Does S have nite Hausdor measure in this dimension? This is a self-similar set consisting of 8 pieces each of which can be expanded by a factor of 3 to give a congruent copy of the original set. Its dimension is t = log 8. log 3 For each i, we may cover S by 8 i squares of diameter p 3,i so that H t p 3,i(S) 8 i t ( p 3,i ) t = t ( p ) t (8 3,t ) i = t ( p ) t < 1 : Thus dim S t. The fact that H t (S) > 0, hence dim S t follows as in our discussion of the Cantor set. 6. (10 pts) A function f : R n! R is approximately continuous at a point a R n if, for every positive there is a positive so that, fx B(a; r) : jf(x), f(a)j >g, B(a; r) >

3 whenever 0 <r<. (a) Prove or nd a counterexample: If a is a Lebesgue point of f, then f is approximately continuous at a. Proof : At a Lebesgue point a of f we may choose, for >0, a positive so that, for 0 <r<, [(B(a; r)],1 jf(x), f(a)j dx ; hence, B(a;r), fx B(a; r) : jf(x), f(a)j >g, B(a; r) : (b) Prove or nd a counterexample: If f is approximately continuous at a, then a is a Lebesgue point of f. For a counterexample one can take f = i=1 1 jb i, a i+1 j [ai+1;bi] where positive numbers a 1 > b 1 > a > b > ::: are chosen inductively so that! 1 as i!1. ai,bi ai 7. (10 pts) Suppose A is a closed subset of R n of Lebesgue measure zero, and f(x) = dist (x; A) inffjx, aj : a Ag. (a) Show that f is dierentiable almost everywhere. For x; y R n the triangular inequality implies that dist (x; A) jx, yj + dist (y; A) and dist (y; A) jx, yj + dist (x; A) ; so that jf(x), f(y)j jx, yj : Thus f is Lipschitz and so dierentiable almost eveywhere by Rademacher's theorem. (b) Show that for any g L 1 (R n ) 1 g(x) dx = 0 f,1 ftg g(y) dh n,1 y : At apoint x of dierentiability off we showed that [[Df(x)]] = 1, and so the above formula follows from the co-area change of variables formula.

4 8. (10 pts) Suppose that K is a compact subsets of R n with H n,1 (K) =0. Prove that any points in R n n K may beconnected by a path in R n n K. For distinct points a; b R n n K, choose rst 0 << 1 ja, bj so that the two closed balls B (a) and B (b) do not intersect K. Note that the retraction map f : R n n B (a) ; f(x) = a + x, a jx, aj ; is Lipschitz with Lip (f) 1. Thus H n,1, f(k) 1 k Hn,1 (K) = 0 : Since also (b) is a nonempty open region (a), we may chose a (a) n (b). Then the half-line f,1 f!g misses K and contains an interval joining the two (a) (b). Joining the endpoints of this interval radially to a and b respectively, gives the desired path in R n n K joining a and b. 9. (10 points) Suppose and are Borel measures on R n with 0 < (R n ) < 1 and 0 <(R n ) < 1. Show that fx R n : lim sup r#0 (B r (x)) (B r (x)) = 1g = 0 : Proof : The above set is the decreasing intersection of the sets E i = fx R n : lim sup r#0 (B r (x)) (B r (x)) >ig : Thus, if the statement is false, then (E i ) >for some positive number and all i suciently large. For each point x E i, we can choose a positive r x so that (B r x(x)) (B r x(x)) >i: Now, for each i, wemay repeat the proof of the Vitali Covering Theorem to choose from these balls a sequence of balls B r1 (a 1 ); B r (a );::: covering E i so that the shrunken balls B r1 =5(a 1 ); B r =5(a );::: are disjoint. The \essentially largest rst" aspect of the proof of Vitali gives us a bound c n (depending only on n) on the number of balls B r j (a j ) which may intersect B rk (a k ) with j k. Thus, although

5 the original balls B r1 (a 1 ); B r (a );::: are not disjoint, any given point lies in at most c n of them. It follows that c n (R n ) j=1 contradicting the niteness of (R n ). (B r j (a j )) i j=1 (B r j (a j )) i(e i ) i! 1 as i!1; 10. (10 pts) P(a) Prove or nd a counterexample: If I is a (possibly uncountable) family of open rectangles in R and A I is any set with I A I Clos I for each I I, then [ II A I is Lebesgue measurable in R. Proof : Replacing I by [ II S I where S I = fopen squares Ig and letting A S = A I \ S whenever I I; S S I ; we may assume that I itself consists of squares which are ne about each point of [ II I. Thus by Vitali's theorem, we may choose a countable disjointed subfamily C of fi : I Igwhich covers almost all of [ II I. Since [ II A I contains the open set [ II Int I and since, [ II A I n [ II Int I [ II A I is measurable. HAPPY SUMMER!, [ II I n[c +, [ = ;

Problem set 1, Real Analysis I, Spring, 2015.

Problem set 1, Real Analysis I, Spring, 015. (1) Let f n : D R be a sequence of functions with domain D R n. Recall that f n f uniformly if and only if for all ɛ > 0, there is an N = N(ɛ) so that if n