MA676 Assignment #2
Morgan Schreffler
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Prove the following properties of elementary measure.
Exercise 1 (a) If E, F are elementary sets and E ⊂ F , then m(E) ≤ m(F ).
N
[
Proof. Since E is elementary, E =
Bi for disjoint boxes Bi . Further, since F is elemeni=1
tary, we know F − E 6= ∅ is elementary as well and can be written F − E =
M
[
Bj for
j=N +1
disjoint boxes Bj and M > N . Finally, we know F = E ∪ (F − E) =
are disjoint because E and F − E are. Knowing this, it follows that
m(E) ≤ m(E) + m(F − E) =
N
X
|Bi | +
i=1
M
X
|Bj | =
j=N +1
M
X
M
[
Bk , where the Bk ’s
k=1
|Bk | = m(F ),
k=1
thus completing the proof.
Exercise 1 (b) If E, F are elementary, then m(E ∪ F ) ≤ m(E) + m(F ).
Proof. First, we know that A = E ∩ F , C = E − (E ∩ F ), and D = F − (E ∩ F ) are
all elementary, and we know E = A ∪ C and F = A ∪ D. Now, for disjoint boxes Bn and
L
M
N
[
[
[
integers 0 < L < M < N , we have A =
Bi , C =
Bj , and D =
Bk . Hence,
i=1
j=L+1
k=M +1
m(E ∪ F ) = m(A ∪ C ∪ D)
!
N
[
=m
Bi
i=1
=
=
N
X
i=1
L
X
|Bi |
|Bi | +
i=1
≤
L
X
i=1
M
X
j=L+1
|Bi | +
N
X
|Bj | +
M
X
|Bk |
k=M +1
!
|Bj |
+
L
X
i=1
j=L+1
|Bi | +
N
X
k=M +1
= m(A ∪ C) + m(A ∪ D) = m(E) + m(F ),
which completes the proof.
1
!
|Bk |
(1)
MA676 Assignment #2
Morgan Schreffler
Exercise 1 (c) If E1 , E2 , . . . , EN are disjoint elementary sets, then
m(E1 ∪ E2 ∪ . . . ∪ EN ) =
N
X
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m(Ei ).
i=1
Proof. We can adapt the proof of Exercise 1(b) to show that if E and F are elementary
and disjoint then m(E ∪ F ) = m(E) + m(F ). The proof of Exercise 1(c) follows immediately
by induction.
Note that ∅ ⊂ Rd is elementary, since ∅ = (0, 0) × (0, 0) × . . . × (0, 0), and m(∅) = 0d = 0.
Adapting the same notation from Exercise 1(b), when E and F are disjoint, the set A = ∅,
L
X
so let L = 1 and B1 be the empty box described above. Then
|Bi | = 0, and equality
i=1
follows in line (1) of the previous proof.
Exercise 1 (d) If E is an elementary set, x ∈ Rd , and E + x denotes the set
{z ∈ Rd | z = y + x, y ∈ E},
then m(E + x) = m(E).
Proof. It suffices to show that m(B + x) = |B + x| = |B| = m(B) for a box B, since E is
a finite union of (disjoint) boxes. Note also that the type of interval in each dimension of
B has no impact on the volume of the box, so without loss of generality, we will use closed
intervals.
For an interval [a, b] ⊂ R, [a, b]+x = [a+x, b+x]. Similarly, for a box [a1 , b1 ]×[a2 , b2 ] ⊂ R2
and a point x = (x1 , x2 ) ∈ R2 , [a1 , b1 ] × [a2 , b2 ] + x = [a1 + x1 , b1 + x1 ] × [a2 + x2 , b2 + x2 ].
We can extend this process inductively for any box and point in Rd .
Let B = [a1 , b1 ]×[a2 , b2 ]×. . .×[ad , bd ] ⊂ Rd with volume |B| = (b1 −a1 )(b2 −a2 ) . . . (bd −ad ),
and let x = (x1 , x2 , . . . , xd ) ∈ Rd . Then
B + x = [a1 + x1 , b1 + x1 ] × [a2 + x2 , b2 + x2 ] × . . . × [ad + xd , bd + xd ],
and
|B + x| = ((b1 + x1 ) − (a1 + x1 ))((b2 + x2 ) − (a2 + x2 )) . . . ((bd + xd ) − (ad + xd ))
= (b1 − a1 )(b2 − a2 ) . . . (bd − ad )
= |B|,
and we are done.
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MA676 Assignment #2
Morgan Schreffler
d
Exercise 2 For bounded E ⊂ R , prove that the following are equivalent:
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(a) E is Jordan measurable
(b) For each ε > 0 there are elementary sets A and B so that A ⊂ E ⊂ B and m(B −A) < ε.
(c) For each ε > 0 there is an elementary set A such that m∗,J (A∆E) < ε, where A∆E
denotes (A − E) ∪ (E − A).
Proof.
(a)⇒(b) Let E be Jordan measurable and let ε > 0 be given. Then there is an elementary
set A ⊂ E such that m(A) > m(E)− 2ε and an elementary set B ⊃ E such that m(B) <
m(E)+ 2ε . Combining these inequalities, we have m(B)+m(E)− 2ε < m(A)+m(E)+ 2ε or
equivalently, m(B)−m(A) < ε. Now, we know A and B−A are disjoint and elementary,
and we know A ⊂ B, so we know m(B) = m(A ∪ (B − A)) = m(A) + m(B − A). In
particular, this tells us that m(B) − m(A) = m(B − A). Hence, m(B − A) < ε, which
was the desired result.
(b)⇒(c) Let ε > 0 be given and let E ⊂ Rd be such that there exist elementary sets
A, B so that A ⊂ E ⊂ B and m(B − A) < ε. As we established previously, this
gives us m(B) − m(A) < ε, and since B and A can easily be shown to be Jordan
measurable, we have m∗,J (B) − m∗,J (A) < ε. Now, surely m∗,J (E) ≥ m∗,J (A), so
m∗,J (B) − m∗,J (E) < ε. By our previous argument, we now have m∗,J (B − E) < ε.
But E − B = ∅ by our construction of B, so B − E = (B − E) ∪ (E − B) = B∆E, and
we have an elementary set B such that m∗,J (B∆E) < ε.
(c)⇒(a) Let ε > 0, and suppose there is an elementary set A such that m∗,J (A∆E) < ε.
Since it is clear that m∗,J (E) ≥ m∗,J (E) for any bounded set E ⊂ Rd , it suffices to
show that m∗,J (E) ≤ m∗,J (E).
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