Math 320-3: Midterm 2 Practice SolutionsNorthwestern University, Spring 2015

1. Give an example, with justification, of each of the following.(a) A Jordan region E ⊆ R2 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.(d) A C1 function φ : R2 → R2 such that∫

φ([0,1]×[0,1])d(x, y) 6=

∫[0,1]×[0,1]

|detDφ(u, v)| d(u, v).

(e) A continuous function on the open unit disk B1(0, 0) in R2 which is not integrable.

Solution. (a) Let E = [0, 1] × [0, 1] and E0 = E ∩ Q2. Then E is a Jordan region but E0 is notsince its boundary is all of E, which has non-zero area.

(b) The function

f(x, y) =

1 (x, y) = ( 1

n ,1m) for n,m ∈ N

0 otherwise

works. This was shown to be integrable on Homework 4 and is discontinuous at each point of theform ( 1

n ,1m), so is discontinuous infinitely often.

(c) The function described in the May 6th lecture notes, which is also in the book, works. Checkthe 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 closedunit disk and detDφ(u, v) = 4πu, so the integral on the right is∫ 1

0

∫ 1

04πu du dv = 2π,

whereas the integral on the left is the area of the unit disk, which is π. The point is that the changeof variables formula does not apply here because φ is not one-to-one, rather φ essentially traces outthe unit disk “twice”, which is why we get twice the expected area.

(e) Any unbounded function, such as f(x, y) = 11−x2−y2 , works.

2. Wade, 12.1.2c. By an interval in R2 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 everyinterval in R2 is a Jordan region.

Proof. Intervals are clearly bounded, so we need only show that they’re boundaries have Jordanmeasure zero. However, an interval in R2 equals its own boundary, so we need to show that anyinterval has Jordan measure zero. (Note that considered as a subset of R instead, the boundaryof an interval would only consist of two points.) We do the horizontal case, and leave the verticalcase 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 therectangle R = [a, b] × [c, c + 1]. For a given ε > 0, pick n ∈ N such that b−a

n < ε and let G be the

grid where we cut R up into n2 equally-sized rectangles of area b−an2 . The only small rectangles Ri

which intersect the closure of I are those along the bottom of the grid, each of which have areab−an2 , so:

V (I;G) =∑

Ri∩I 6=∅

|Ri| =n∑i=1

b− an2

=b− an

< ε.

Thus for any ε > 0 we can find a grid G such that V (I;G) < ε, showing that I has Jordan measurezero as claimed. Since I = ∂I, this shows that I is a Jordan region.

3. Wade, 12.1.7b. Let E ⊆ Rn. The dilation of E by a scalar α > 0 is the set

αE = y ∈ Rn | 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 isa rectangular box containing αE, where we use the fact from linear algebra that scalings preserveangles and thus send rectangular boxes to rectangular boxes. For a grid G on R, let αG denote thecorresponding 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 rectangularbox 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)= αn(product of terms of the form ci − di)= αn|Ri|.

Now, let ε > 0. Since E is Jordan measurable, ∂E has Jordan measure zero so there exists agrid G on R such that

V (∂E;G) =∑

Ri∩∂E 6=∅

|Ri| <ε

αn.

Then:

V (∂(αE);αG) =∑

αRi∩∂(αE)6=∅

|αRi| =∑

Ri∩∂E 6=∅

αn|Ri| = αn∑

Ri∩∂E 6=∅

|Ri| < αnε

αn= ε,

which shows that ∂(αE) has Jordan volume zero and hence that αE is a Jordan region. Since

V (αE;αG) =∑

αRi∩αE 6=∅

|αRi| =∑

Ri∩E 6=∅

αn|Ri| = αn∑

Ri∩E 6=∅

|Ri| = αnV (E;G),

we have infV (αE;αG) = αn infV (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

α(αE) = E

is Jordan measurable and that Vol(E) = Vol( 1α(αE)) = 1

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 wheref(x, y) = xy and Gm is the grid on [0, 1] × [0, 1] obtained by partitioning each side [0, 1] into 2m

subintervals of equal length 12m . Prove that

limm→∞

U(f,Gm)− L(f,Gm) = 0.

Solution. To be clear, in Gm we partition each side [0, 1] of the given square using

0 <1

2m<

2

2m<

3

2m< · · · < 2m − 1

2m< 1.

Any small rectangle Rij for i, j = 1, . . . , 2m is thus of the form

Rij =

[i− 1

2m,i

2m

]×[j − 1

2m,j

2m

]and has area 1

2m1

2m = 14m . Note that there are 4m total small rectangles in the grid Gm.

Now, on any Rij the supremum of f occurs at the upper right corner, and thus is:

f

(i

2m,j

2m

)=

ij

4m.

Thus the upper sum is:

U(f,Gm) =

2m∑i,j=1

ij

4m1

4m=

1

16m

(2m∑i=1

i

) 2m∑j=1

j

=1

16m

[2m(2m + 1)

2

]2

=(2m + 1)2

4m+1.

On any Rij the infimum of f occurs at the lower right corder, and so is given by:

f

(i− 1

2m,j − 1

2m

)=

(i− 1)(j − 1)

4m.

Thus the lower sum is (note that we are starting the indices at 0 instead to account for the i − 1and j − 1 terms):

L(f,Gm) =

2m−1∑i,j=0

ij

4m1

4m=

1

16m

(2m−1∑i=0

i

)2m−1∑j=0

j

=1

16m

[(2m − 1)2m

2

]2

=(2m − 1)2

4m+1.

We compute

U(f,Gm)− L(f,Gm) =(2m + 1)2

4m+1− (2m − 1)2

4m+1=

4 · 2m

4m+1=

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.)

5. Wade, 12.2.5. If E0 ⊆ E are Jordan regions in Rn and f : E → R is integrable on E, prove thatf is integrable on E0.

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Proof. (This problem is a little tricky when it comes to getting all the details right.) Let R be arectangular box containing E, which is thus also a rectangular box containing E0. Let ε > 0 andpick a grid G on R such that∣∣∣∣∣∣

∑Ri⊆E

Mi|Ri| −∫Ef dA

∣∣∣∣∣∣ < ε and

∣∣∣∣∣∣∑Ri⊆E

mi|Ri| −∫Ef dA

∣∣∣∣∣∣ < ε

where Mi and mi respectively denote the supremum and infimum of f over Ri. Such a grid existsby the fact that we can approximate the integral of f to whatever degree of accuracy we want usingsums involving only rectangles which are contained in the interior of E. (See Theorem 12.20 in thebook.) These inequalities imply that∑

Ri⊆E(Mi −mi)|Ri| <

(∫Ef dA+ ε

)−(∫

Ef dA− ε

)= 2ε.

Now, the rectangles Ri ⊆ E can be separated into those fully contained in E0 and those notfully contained in E0, so:∑

Ri⊆E0

(Mi −mi)|Ri| ≤∑Ri⊆E0

(Mi −mi)|Ri|+∑Ri*E0

(Mi −mi)|Ri| =∑Ri⊆E

(Mi −mi)|Ri| < 2ε.

The same inequality will hold for any grid finer than G. Using Theorem 12.20 again only nowapplied to f over E0, by making G finer if need be we get that:∣∣∣∣∣∣(U)

∫E0

f dA−∑Ri⊆E0

Mi|Ri|

∣∣∣∣∣∣ < ε and

∣∣∣∣∣∣(L)

∫E0

f dA−∑Ri⊆E0

mi|Ri|

∣∣∣∣∣∣ < ε.

These inequalities imply that

(U)

∫E0

f dA−(L)

∫E0

f dA <

∑Ri⊆E0

Mi|Ri|+ ε

− ∑Ri⊆E0

Mi|Ri| − ε

=∑Ri⊆E0

(Mi−mi)|Ri|+2ε.

Together with the previously-derived inequality we thus have

(U)

∫E0

f dA− (L)

∫E0

f dA < 4ε,

and since ε > 0 is arbitrary this implies that

(U)

∫E0

f dA = (L)

∫E0

f dA,

showing that f is integrable over E0.

6. Show that the function

f(x, y) =

1− x− y (x, y) = ( 1

2n ,1

3m ) for n,m ∈ N0 otherwise

is integrable over [0, 1]× [0, 1].

4

Solution. This essentially the same argument is that on Problem 3 of Homework 4, whose solutionwe use as a guide. In particular, note that

0 ≤ 1− x− y ≤ 1 for (x, y) of the form

(1

2n,

1

3m

).

Let ε > 0 and pick N ∈ N such that 12N+1 <

ε4 and 1

3N+1 <ε4 . Pick a rectangle Rnm of area less

than ε2N2 around each of the finitely many points ( 1

2n ,1

3m ) for n,m = 1, . . . , N . If need be we canshrink 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 havehorizontal length ε

4 , all small rectangles along the bottom side have vertical height ε4 , and the small

rectangle containing ( 12n ,

13m ) for n,m = 1, . . . , N is contained in Rnm. Any rectangle in our grid

contains points not of the form ( 12n ,

13m ), 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

ε4 , those along the bottom of height ε

4 , and those inside some Rnm containing a point of the form( 1

2n ,1

3m ) for n,m = 1, . . . , N . (On any other rectangle in our grid, sup f = 0 and so these give nocontributions to U(f,G).) On any of these rectangles we have sup f ≤ 1, so:∑

rectangles R along left

(sup f)(area R) ≤∑

rectangles along left

(area R) =ε

4∑rectangles R along bottom

(sup f)(area R) ≤∑

rectangles along bottom

(area R) =ε

4∑rectangles R inside some Rnm

(sup f)(area Rnm) ≤∑

rectangles R inside some Rnm

(area Rnm)

<∑

Rnm for n,m=1,...,N

ε

2N2=ε

2,

where in the final equality we use the fact that there are N2 rectangles among the Rnm. Thus alltogether we get:

U(f,G) <ε

4+ε

4+ε

2= ε,

so U(f,G) − L(f,G) = U(f,G) < ε for this grid and we hence conclude that f is integrable over[0, 1]× [0, 1] as desired.

7. Wade, 12.3.8. Let E be a nonempty Jordan region in R2 and let f : E → [0,∞) be integrableon E. Prove that the volume of Ω = (x, y, z) | (x, y) ∈ E and 0 ≤ z ≤ f(x, y) satisfies

Vol(Ω) =

∫∫Ef dA.

Proof. The volume of Ω is given by:

Vol(Ω) =

∫∫∫ΩdV.

Since the constant function 1 is integrable as a function of one, two, and three variables, the iteratedintegrals of the above expression all exist and so Fubini’s Theorem gives:∫∫

ΩdV =

∫∫E

(∫ f(x,y)

0dz

)dA =

∫∫Ef(x, y) dA

5

as required. (The point of this problem was really to just understand the use of Fubini’s Theoremand 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 bebounded.

(a) Prove that

(L)

∫∫Rf dA ≤ (L)

∫ b

a

((X)

∫ d

cf(x, y) dy

)dx ≤ (U)

∫ b

a

((X)

∫ d

cf(x, y) dy

)dx

for X = U or X = L. (So, omit the final inequality which is asked for, whose proof is similar to thefirst 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 fixedx, 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∫∫Rf dA =

∫ b

a

((L)

∫ d

cf(x, y) dy

)dx =

∫ b

a

((U)

∫ d

cf(x, y) dy

)dx.

(Feel free to use the additional inequality of part (a) which I omitted above. View this as a versionof 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. Takea grid G on R where we partition [a, b] as

a = x0 < x1 < x2 < · · · < xn = b

and [c, d] asc = 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

(L)

∫ yj

yj−1

mij dy ≤ (L)

∫ yj

yj−1

f(x, y) dy

using the fact that the operation of taking lower integrals preserves inequalities. The lower integralon the left is that of a constant function, so in this case the lower integral is just the ordinaryintegral (which exists), so the left side is mij∆yj . Summing over all intervals [yj−1, yj ] gives

∑j

mij∆yj ≤∑j

(L)

∫ yj

yj−1

f(x, y) dy = (L)∑j

∫ yj

yj−1

f(x, y) dy = (L)

∫ d

cf(x, y) dy.

Now taking lower integrals of both sides over [xi−1, xi] gives

(L)

∫ xi

xi−1

∑j

mij∆yj

dx ≤ (L)

∫ xi

xi−1

((L)

∫ d

cf(x, y) dy

)dx.

6

Again the lower integral on the left exists as an ordinary integral since it is the lower integral of aconstant function, so

∑j

mij∆xi∆yj =

∫ xi

xi−1

∑j

mij∆yj

dx ≤ (L)

∫ xi

xi−1

((L)

∫ d

cf(x, y) dy

)dx.

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

∑i,j

mij∆xi∆yj ≤∑i

(L)

∫ xi

xi−1

((L)

∫ d

cf(x, y) dy

)dx = (L)

∫ b

a

((L)

∫ d

cf(x, y) dy

)dx.

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

(L)

∫∫Rf dA ≤ (L)

∫ b

a

((L)

∫ d

cf(x, y) dy

)dx.

Since

(L)

∫ b

a

((L)

∫ d

cf(x, y) dy

)dx ≤ (L)

∫ b

a

((U)

∫ d

cf(x, y) dy

)dx

due to the integrand on the right being larger than or equal to that on the left, we have

(L)

∫∫Rf dA ≤ (L)

∫ b

a

((X)

∫ d

cf(x, y) dy

)dx

for either X = U or X = L, which is the first desired inequality. The remaining inequalities followfrom the fact that lower integrals are always smaller than or equal to upper integrals, but we’llleave it at that since this solutions is already getting long and tedious. Again, this problem is toohard for the midterm.

(b) If f is integrable on R, then

(L)

∫∫Rf dA = (U)

∫∫Rf dA.

Thus the inequality

(U)

∫ b

a

((X)

∫ d

cf(x, y) dy

)dx ≤ (U)

∫∫Rf dA

together with the inequalities in part (a) implies that all four expressions are equal, so∫∫Rf dA = (L)

∫ b

a

((X)

∫ d

cf(x, y) dy

)dx = (U)

∫ b

a

((X)

∫ d

cf(x, y) dy

)dx

for X = U or X = L. Hence the iterated integrals∫ b

a

((L)

∫ d

cf(x, y) dy

)dx and

∫ b

a

((U)

∫ d

cf(x, y) dy

)dx

both exist since the lower and upper integrals for each are equal. Since

(L)

∫ d

cf(x, y) dy ≤ (U)

∫ d

cf(x, y) dy,

7

we have ∫ b

a

((L)

∫ d

cf(x, y) dy

)dx ≤

∫ b

a

((U)

∫ d

cf(x, y) dy

)dx.

Combined with our previous inequalities we now have∫∫Rf dA =

∫ b

a

((L)

∫ d

cf(x, y) dy

)dx ≤

∫ b

a

((U)

∫ d

cf(x, y) dy

)dx =

∫∫Rf dA,

which finally gives∫∫Rf dA =

∫ b

a

((L)

∫ d

cf(x, y) dy

)dx =

∫ b

a

((U)

∫ d

cf(x, y) dy

)dx

as required.

9. Wade, 12.4.5cd. (c) Find ∫∫Ee(y−x)/(y+x) dA

where E is the trapezoid with vertices (1, 1), (2, 2), (2, 0), (4, 0).(d) Given

∫ 10 (1− x)f(x) dx = 5, find∫ 1

0

∫ x

0f(x− y) dy dx.

Solution. (c) Let φ : R2 → R2 be the function

φ(u, v) =

(1

2(u+ v),

1

2(u− v)

),

which is C1, one-to-one, and has Jacobian matrix:

Dφ =

(1/2 1/21/2 −1/2

)at every point. (To be clear, this function sets x = 1

2(u + v) and y = 12(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 = −vwe have φ(A) = E:

8

Thus by the change of variables formula we have:∫∫φ(A)

e(y−x)/(y+x) dA =

∫∫Aev/u|detDφ(u, v)| d(u, v) =

∫∫A

1

2ev/u d(u, v).

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

1

2ev/u d(u, v) =

1

2

∫ 4

2

∫ 0

−uev/u dv du =

1

2

∫ 4

2u(1− e−1) du = 3(1− e−1)

as the desired value.(d) Let φ : R2 → R2 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

Dφ =

(0 1−1 1

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

Thus the change of variables formula gives:∫ 1

0

∫ x

0f(x−y) dy dx =

∫∫φ(E)

f(x−y) d(x, y) =

∫∫Ef(u)| detDφ(u, v)| d(u, v) =

∫∫Ef(u) d(u, v)

since detDφ = 1. By Fubini’s Theorem the resulting integral is:∫∫Ef(u) d(u, v) =

∫ 1

0

∫ 1

uf(u) dv du =

∫ 1

0(1− u)f(u) du = 5

by the assumption that∫ 1

0 (1− x)f(x) dx = 5.

10. Wade, 12.4.7. Show that Vol is rotation invariant in R2; that is, if φ is a rotation on R2 and Eis a Jordan region in R2, then Vol(φ(E)) = Vol(E). (See Exercise 8.2.9 to remind yourself of whatrotations explicitly looks like.)

9

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

φ =

(cos θ − sin θsin 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:

Vol(φ(E)) =

∫φ(E)

dx =

∫E|detDφ(u)| du =

∫E|detφ| du.

By the explicit form of φ given above, detφ = 1 so the final integral is simply∫Edu = VolE.

Hence Vol(φ(E)) = VolE as claimed. (Thus, rotations are volume-preserving, as would be ex-pected.)

10