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= | det Dφ(u, v)| 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 ∈ 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 {(x, c) | a ≤ x ≤ b} or {(c, y) | a ≤ y ≤ b} for some a, b, c ∈ 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 = {(x, c) | a ≤ x ≤ b} for some ﬁxed a, b, c ∈ R. This is contained in the b−a rectangle R = [a, b] × [c, c + 1]. For a given > 0, pick n ∈ 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) = |R | = = < . 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 = {y ∈ R | y = αz for some z ∈ E}. 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:

|αRi| = 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 |Ri|. 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) = |R | < . i αn Ri∩∂E6=∅ Then: X X X V (∂(αE); αG) = |αR | = αn|R | = αn |R | < 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) = |αRi| = α |Ri| = α |Ri| = α V (E; G),

αRi∩αE6=∅ Ri∩E6=∅ Ri∩E6=∅ we have inf{V (αE; αG)} = αn inf{V (E; 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 ∈ 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→∞

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:   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   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):   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   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 → ∞ 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 Mi|Ri| − f dA < and mi|Ri| − 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)|Ri| < 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)|Ri| ≤ (Mi − mi)|Ri| + (Mi − mi)|Ri| = (Mi − mi)|Ri| < 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 − Mi|Ri| < and (L) f dA − mi|Ri| < . E0 E0 Ri⊆E0 Ri⊆E0 These inequalities imply that     Z Z X X X (U) f dA−(L) f dA <  Mi|Ri| + − Mi|Ri| −  = (Mi−mi)|Ri|+2. E0 E0 Ri⊆E0 Ri⊆E0 Ri⊆E0

Together with the previously-derived inequality we thus have Z Z (U) f dA − (L) f dA < 4, E0 E0 and since > 0 is arbitrary this implies that Z Z (U) f dA = (L) f dA, E0 E0 showing that f is integrable over E0.

6. Show that the function ( 1 − x − y (x, y) = ( 1 , 1 ) for n, m ∈ f(x, y) = 2n 3m N 0 otherwise is integrable over [0, 1] × [0, 1].

4 Solution. This essentially the same argument is that on Problem 3 of Homework 4, whose solution we use as a guide. In particular, note that 1 1 0 ≤ 1 − x − y ≤ 1 for (x, y) of the form , . 2n 3m

1 1 Let > 0 and pick N ∈ N such that 2N+1 < 4 and 3N+1 < 4 . Pick a rectangle Rnm of area less 1 1 than 2N 2 around each of the ﬁnitely many points ( 2n , 3m ) for n, m = 1,...,N. If need be we can shrink each Rnm so that they are all disjoint. Take a ﬁne enough grid G on [0, 1] × [0, 1] so that the small rectangles along the left side have horizontal length 4 , all small rectangles along the bottom side have vertical height 4 , and the small 1 1 rectangle containing ( 2n , 3m ) for n, m = 1,...,N is contained in Rnm. Any rectangle in our grid 1 1 contains points not of the form ( 2n , 3m ), so f attains the value zero on any rectangle and hence inf f = 0 on any rectangle. Thus L(f, G) = 0. There are three types of rectangles contributing to U(f, G): those along the left side of length 4 , those along the bottom of height 4 , and those inside some Rnm containing a point of the form 1 1 ( 2n , 3m ) for n, m = 1,...,N. (On any other rectangle in our grid, sup f = 0 and so these give no contributions to U(f, G).) On any of these rectangles we have sup f ≤ 1, so: X X (sup f)(area R) ≤ (area R) = 4 rectangles R along left rectangles along left X X (sup f)(area R) ≤ (area R) = 4 rectangles R along bottom rectangles along bottom X X (sup f)(area Rnm) ≤ (area Rnm)

rectangles R inside some Rnm rectangles R inside some Rnm X < = , 2N 2 2 Rnm for n,m=1,...,N

2 where in the ﬁnal equality we use the fact that there are N rectangles among the Rnm. Thus all together we get: U(f, G) < + + = , 4 4 2 so U(f, G) − L(f, G) = U(f, G) < for this grid and we hence conclude that f is integrable over [0, 1] × [0, 1] as desired.

2 7. Wade, 12.3.8. Let E be a nonempty Jordan region in R and let f : E → [0, ∞) be integrable on E. Prove that the volume of Ω = {(x, y, z) | (x, y) ∈ E and 0 ≤ z ≤ f(x, y)} satisﬁes ZZ Vol(Ω) = f dA. E Proof. The volume of Ω is given by: ZZZ Vol(Ω) = dV. Ω Since the constant function 1 is integrable as a function of one, two, and three variables, the iterated integrals of the above expression all exist and so Fubini’s Theorem gives:

ZZ ZZ Z f(x,y) ! ZZ dV = dz dA = f(x, y) dA Ω E 0 E

5 as required. (The point of this problem was really to just understand the use of Fubini’s Theorem and to set up the inner bounds on z correctly.)

8. Wade, 12.3.9ab. Let R = [a, b] × [c, d] be a two-dimensional rectangle and f : R → R be bounded. (a) Prove that

ZZ Z b Z d Z b Z d (L) f dA ≤ (L) (X) f(x, y) dy dx ≤ (U) (X) f(x, y) dy dx R a c a c for X = U or X = L. (So, omit the final inequality which is asked for, whose proof is similar to the first inequality asked for. This is too hard for the midterm, but good to think about nonetheless. This is similar to Lemma 12.30, only here we are not assuming that f(x, ·) is integrable for a fixed x, which is why we’re having to use upper and lower integrals for the “inner” integrals.) (b) Prove that if f is integrable on R, then

ZZ Z b Z d Z b Z d f dA = (L) f(x, y) dy dx = (U) f(x, y) dy dx. R a c a c (Feel free to use the additional inequality of part (a) which I omitted above. View this as a version of Fubini’s Theorem which holds without assuming the single-variable function f(x, ·) is integrable.)

Proof. (a) We mimic the proof of Fubini’s Theorem we gave in class and in the lecture notes. Take a grid G on R where we partition [a, b] as

a = x0 < x1 < x2 < ··· < xn = b and [c, d] as c = y0 < y1 < y2 < ··· < yn = b.

Let mij denote the inﬁmum of f over the rectangle Rij = [xi−1, xi] × [yj−1, yj]. Fix x ∈ [xi−1, xi]. Then mij ≤ f(x, y) for all y ∈ [yj−1, yj] so Z yj Z yj (L) mij dy ≤ (L) f(x, y) dy yj−1 yj−1 using the fact that the operation of taking lower integrals preserves inequalities. The lower integral on the left is that of a constant function, so in this case the lower integral is just the ordinary integral (which exists), so the left side is mij∆yj. Summing over all intervals [yj−1, yj] gives

X X Z yj X Z yj Z d mij∆yj ≤ (L) f(x, y) dy = (L) f(x, y) dy = (L) f(x, y) dy. j j yj−1 j yj−1 c

Now taking lower integrals of both sides over [xi−1, xi] gives   Z xi X Z xi Z d (L)  mij∆yj dx ≤ (L) (L) f(x, y) dy dx. xi−1 j xi−1 c

6 Again the lower integral on the left exists as an ordinary integral since it is the lower integral of a constant function, so   X Z xi X Z xi Z d mij∆xi∆yj =  mij∆yj dx ≤ (L) (L) f(x, y) dy dx. j xi−1 j xi−1 c

Summing over all intervals [xi−1, xi] gives

X X Z xi Z d Z b Z d mij∆xi∆yj ≤ (L) (L) f(x, y) dy dx = (L) (L) f(x, y) dy dx. i,j i xi−1 c a c RR Thus the supremum of the terms on the left, which is (L) R f dA, is less than or equal to the term on the right, so ZZ Z b Z d (L) f dA ≤ (L) (L) f(x, y) dy dx. R a c Since Z b Z d Z b Z d (L) (L) f(x, y) dy dx ≤ (L) (U) f(x, y) dy dx a c a c due to the integrand on the right being larger than or equal to that on the left, we have

ZZ Z b Z d (L) f dA ≤ (L) (X) f(x, y) dy dx R a c for either X = U or X = L, which is the ﬁrst desired inequality. The remaining inequalities follow from the fact that lower integrals are always smaller than or equal to upper integrals, but we’ll leave it at that since this solutions is already getting long and tedious. Again, this problem is too hard for the midterm. (b) If f is integrable on R, then ZZ ZZ (L) f dA = (U) f dA. R R Thus the inequality Z b Z d ZZ (U) (X) f(x, y) dy dx ≤ (U) f dA a c R together with the inequalities in part (a) implies that all four expressions are equal, so

ZZ Z b Z d Z b Z d f dA = (L) (X) f(x, y) dy dx = (U) (X) f(x, y) dy dx R a c a c for X = U or X = L. Hence the iterated integrals

Z b Z d Z b Z d (L) f(x, y) dy dx and (U) f(x, y) dy dx a c a c both exist since the lower and upper integrals for each are equal. Since

Z d Z d (L) f(x, y) dy ≤ (U) f(x, y) dy, c c

7 we have Z b Z d Z b Z d (L) f(x, y) dy dx ≤ (U) f(x, y) dy dx. a c a c Combined with our previous inequalities we now have

ZZ Z b Z d Z b Z d ZZ f dA = (L) f(x, y) dy dx ≤ (U) f(x, y) dy dx = f dA, R a c a c R which ﬁnally gives

ZZ Z b Z d Z b Z d f dA = (L) f(x, y) dy dx = (U) f(x, y) dy dx R a c a c as required.

9. Wade, 12.4.5cd. (c) Find ZZ e(y−x)/(y+x) dA E where E is the trapezoid with vertices (1, 1), (2, 2), (2, 0), (4, 0). R 1 (d) Given 0 (1 − x)f(x) dx = 5, ﬁnd Z 1 Z x f(x − y) dy dx. 0 0

2 2 Solution. (c) Let φ : R → R be the function 1 1 φ(u, v) = (u + v), (u − v) , 2 2 which is C1, one-to-one, and has Jacobian matrix:

1/2 1/2 Dφ = 1/2 −1/2

1 1 at every point. (To be clear, this function sets x = 2 (u + v) and y = 2 (u − v), so u = y + x and v = y − x.) For the region A in the uv-plane bounded by the line u = 2, u = 4, v = 0, and u = −v we have φ(A) = E:

8 Thus by the change of variables formula we have: ZZ ZZ ZZ 1 e(y−x)/(y+x) dA = ev/u| det Dφ(u, v)| d(u, v) = ev/u d(u, v). φ(A) A A 2

Since the integrand is continuous on A, Fubini’s Theorem gives

ZZ 1 1 Z 4 Z 0 1 Z 4 ev/u d(u, v) = ev/u dv du = u(1 − e−1) du = 3(1 − e−1) A 2 2 2 −u 2 2 as the desired value. 2 2 (d) Let φ : R → R be the function

φ(u, v) = (v, v − u), so x = v and y = v − u. This is C1 and one-to-one and has Jacobian matrix

0 1 Dφ = −1 1 everywhere. For the region E in the uv-plane bounded by u = 0, u = v, and v = 1 we have that φ(E) is described by the bounds on the given double integral:

Thus the change of variables formula gives:

Z 1 Z x ZZ ZZ ZZ f(x − y) dy dx = f(x − y) d(x, y) = f(u)| det Dφ(u, v)| d(u, v) = f(u) d(u, v) 0 0 φ(E) E E since det Dφ = 1. By Fubini’s Theorem the resulting integral is:

ZZ Z 1 Z 1 Z 1 f(u) d(u, v) = f(u) dv du = (1 − u)f(u) du = 5 E 0 u 0 R 1 by the assumption that 0 (1 − x)f(x) dx = 5.

2 2 10. Wade, 12.4.7. Show that Vol is rotation invariant in R ; that is, if φ is a rotation on R and E 2 is a Jordan region in R , then Vol(φ(E)) = Vol(E). (See Exercise 8.2.9 to remind yourself of what rotations explicitly looks like.)

9 Proof. Let φ be the rotation given by the 2 × 2 matrix

cos θ − sin θ φ = sin θ cos θ for some θ ∈ [0, 2π). Since φ is a linear transformation, it is continuously diﬀerentiable and Dφ = φ, which is invertible everywhere. Thus the change of variables formula applies to give: Z Z Z Vol(φ(E)) = dx = | det Dφ(u)| du = | det φ| du. φ(E) E E

By the explicit form of φ given above, det φ = 1 so the ﬁnal integral is simply Z du = Vol E. E Hence Vol(φ(E)) = Vol E as claimed. (Thus, rotations are volume-preserving, as would be expected.)