Math 320-3: Midterm 2 Practice Solutions Northwestern University, Spring 2015 1. Give an example, with justification, of each of the following. 2 (a) A Jordan region E ⊆ R which contains a non-Jordan region E0 ⊆ E. (b) An integrable function on [0; 1] × [0; 1] which is discontinuous at infinitely many points. (c) A function f : [0; 1] × [0; 1] ! R whose iterated integrals exist but are not equal. 1 2 2 (d) A C function φ : R ! R such that Z Z d(x; y) 6= j det Dφ(u; v)j d(u; v): φ([0;1]×[0;1]) [0;1]×[0;1] 2 (e) A continuous function on the open unit disk B1(0; 0) in R which is not integrable. 2 Solution. (a) Let E = [0; 1] × [0; 1] and E0 = E \ Q . Then E is a Jordan region but E0 is not since its boundary is all of E, which has non-zero area. (b) The function ( 1 (x; y) = ( 1 ; 1 ) for n; m 2 f(x; y) = n m N 0 otherwise works. This was shown to be integrable on Homework 4 and is discontinuous at each point of the 1 1 form ( n ; m ), so is discontinuous infinitely often. (c) The function described in the May 6th lecture notes, which is also in the book, works. Check the lecture notes or book for the details. (d) The function φ(u; v) = (u cos 4πv; u sin 4πv) works. The image φ([0; 1] × [0; 1]) is the closed unit disk and det Dφ(u; v) = 4πu, so the integral on the right is Z 1 Z 1 4πu du dv = 2π; 0 0 whereas the integral on the left is the area of the unit disk, which is π. The point is that the change of variables formula does not apply here because φ is not one-to-one, rather φ essentially traces out the unit disk \twice", which is why we get twice the expected area. 1 (e) Any unbounded function, such as f(x; y) = 1−x2−y2 , works. 2 2. Wade, 12.1.2c. By an interval in R we mean a set of the form f(x; c) j a ≤ x ≤ bg or f(c; y) j a ≤ y ≤ bg for some a; b; c 2 R. (So an interval is just a horizontal or vertical line segment.) Prove that every 2 interval in R is a Jordan region. Proof. Intervals are clearly bounded, so we need only show that they're boundaries have Jordan 2 measure zero. However, an interval in R equals its own boundary, so we need to show that any interval has Jordan measure zero. (Note that considered as a subset of R instead, the boundary of an interval would only consist of two points.) We do the horizontal case, and leave the vertical case which is very similar to the reader. Suppose that I = f(x; c) j a ≤ x ≤ bg for some fixed a; b; c 2 R. This is contained in the b−a rectangle R = [a; b] × [c; c + 1]. For a given > 0, pick n 2 N such that n < and let G be the 2 b−a grid where we cut R up into n equally-sized rectangles of area n2 . The only small rectangles Ri which intersect the closure of I are those along the bottom of the grid, each of which have area b−a n2 , so: n X X b − a b − a V (I; G) = jR j = = < . i n2 n Ri\I6=; i=1 Thus for any > 0 we can find a grid G such that V (I; G) < , showing that I has Jordan measure zero as claimed. Since I = @I, this shows that I is a Jordan region. n 3. Wade, 12.1.7b. Let E ⊆ R . The dilation of E by a scalar α > 0 is the set n αE = fy 2 R j y = αz for some z 2 Eg: Prove that E is a Jordan region if and only if αE is a Jordan region, in which case Vol(αE) = αn Vol(E). Proof. Suppose that E is a Jordan region and let R be a rectangular box containing E. Then αR is a rectangular box containing αE, where we use the fact from linear algebra that scalings preserve angles and thus send rectangular boxes to rectangular boxes. For a grid G on R, let αG denote the corresponding grid on αG; to be clear, if ai = x0 < x1 < ··· < xn = bi is a partition of one of the side edges of R, then αai = αx0 < αx1 < ··· < αxn = αbi is the corresponding partition of one of the side edges of αR. Note that for a small Ri rectangular box in the grid G, we have: jαRij = product of the lengths of the sides of αRi = product of a bunch of terms of the form (αci − αdi) n = α (product of terms of the form ci − di) n = α jRij: Now, let > 0. Since E is Jordan measurable, @E has Jordan measure zero so there exists a grid G on R such that X V (@E; G) = jR j < : i αn Ri\@E6=; Then: X X X V (@(αE); αG) = jαR j = αnjR j = αn jR j < αn = , i i i αn αRi\@(αE)6=; Ri\@E6=; Ri\@E6=; which shows that @(αE) has Jordan volume zero and hence that αE is a Jordan region. Since X X n n X n V (αE; αG) = jαRij = α jRij = α jRij = α V (E; G); αRi\αE6=; Ri\E6=; Ri\E6=; we have inffV (αE; αG)g = αn inffV (E; G)g and hence Vol(αE) = αn Vol(E) as claimed. Conversely, if αE is Jordan measurable, then by the first part we have that 1 (αE) = E α 1 1 is Jordan measurable and that Vol(E) = Vol( α (αE)) = αn Vol(αE) as required. 2 4. Wade, 12.2.1. Compute the upper and lower sums U(f; Gm) and L(f; Gm) for m 2 N where m f(x; y) = xy and Gm is the grid on [0; 1] × [0; 1] obtained by partitioning each side [0; 1] into 2 1 subintervals of equal length 2m . Prove that lim U(f; Gm) − L(f; Gm) = 0: m!1 Solution. To be clear, in Gm we partition each side [0; 1] of the given square using 1 2 3 2m − 1 0 < < < < ··· < < 1: 2m 2m 2m 2m m Any small rectangle Rij for i; j = 1;:::; 2 is thus of the form i − 1 i j − 1 j R = ; × ; ij 2m 2m 2m 2m 1 1 1 m and has area 2m 2m = 4m . Note that there are 4 total small rectangles in the grid Gm. Now, on any Rij the supremum of f occurs at the upper right corner, and thus is: i j ij f ; = : 2m 2m 4m Thus the upper sum is: 0 1 2m 2m ! 2m 2 X ij 1 1 X X 1 2m(2m + 1) (2m + 1)2 U(f; G ) = = i j = = : m 4m 4m 16m @ A 16m 2 4m+1 i;j=1 i=1 j=1 On any Rij the infimum of f occurs at the lower right corder, and so is given by: i − 1 j − 1 (i − 1)(j − 1) f ; = : 2m 2m 4m Thus the lower sum is (note that we are starting the indices at 0 instead to account for the i − 1 and j − 1 terms): 0 1 2m−1 2m−1 ! 2m−1 2 X ij 1 1 X X 1 (2m − 1)2m (2m − 1)2 L(f; G ) = = i j = = : m 4m 4m 16m @ A 16m 2 4m+1 i;j=0 i=0 j=0 We compute (2m + 1)2 (2m − 1)2 4 · 2m 1 U(f; G ) − L(f; G ) = − = = : m m 4m+1 4m+1 4m+1 2m Thus we see that U(f; Gm)−L(f; Gm) ! 0 as m ! 1 as required. (This shows that f is integrable, which we already knew anyway since f is continuous.) n 5. Wade, 12.2.5. If E0 ⊆ E are Jordan regions in R and f : E ! R is integrable on E, prove that f is integrable on E0. 3 Proof. (This problem is a little tricky when it comes to getting all the details right.) Let R be a rectangular box containing E, which is thus also a rectangular box containing E0. Let > 0 and pick a grid G on R such that Z Z X X MijRij − f dA < and mijRij − f dA < E E Ri⊆E Ri⊆E where Mi and mi respectively denote the supremum and infimum of f over Ri. Such a grid exists by the fact that we can approximate the integral of f to whatever degree of accuracy we want using sums involving only rectangles which are contained in the interior of E. (See Theorem 12.20 in the book.) These inequalities imply that X Z Z (Mi − mi)jRij < f dA + − f dA − = 2. E E Ri⊆E Now, the rectangles Ri ⊆ E can be separated into those fully contained in E0 and those not fully contained in E0, so: X X X X (Mi − mi)jRij ≤ (Mi − mi)jRij + (Mi − mi)jRij = (Mi − mi)jRij < 2. Ri⊆E0 Ri⊆E0 Ri*E0 Ri⊆E The same inequality will hold for any grid finer than G. Using Theorem 12.20 again only now applied to f over E0, by making G finer if need be we get that: Z Z X X (U) f dA − MijRij < and (L) f dA − mijRij < .