G13MIN MEASURE AND INTEGRATION
SOLUTIONS TO QUESTION SHEET 4
1. First suppose that R is a ring of subsets of X. Then ∅ ∈ R, so we need only check (ii). But,
for A, B ∈ R, A ∩ B ∈ R, A\B and B\A ∈ R, and so A4B = (A\B) ∪ (B\A) ∈ R.
Conversely, suppose R satisfies conditions (i) and (ii) of the question. Then ∅ ∈ R, and,
for A, B ∈ R, we have A ∩ B ∈ R (given). Also A\B = A ∩ (A4B) ∈ R, and A ∪ B =
(A\B)4B ∈ R. [There are many other ways to see this.]
Note that 4 and ∩ are commutative. Among other things, this means that we only need to
check one distributive law rather than two.
The associative law for ∩ is standard. You should say this, but this is all you need to say.
For 4 the associative law is not obvious. You could work out in full what A4(B4C) and
(A4B)4C are from the definitions, but this is a long method. I accept clearly justified
shading of Venn diagrams here. Another method is to note that A4B consists of all x which
are in exactly one of A and B, so that
A4(B4C) = (A\(B4C)) ∪ ((B4C)\A)
consists of all those x which are in exactly one or exactly three of A, B and C. This is
symmetric in A, B, C so A4(B4C) = (A4B)4C.
In general, if you take the symmetric difference of finitely many sets A1 4A2 4 · · · 4An , you
get the set of those x such that the number of k with x ∈ Ak is odd. As an exercise, you can
prove this by induction, or using characteristic functions (see below).
Perhaps the best approach is to consider adding characteristic functions modulo 2. Recall
that m ≡ n (mod 2) if m − n is divisible by 2. For convenience, we will simply write m ≡ n
below.
It is standard that this is an equivalence relation on Z, and that if m ≡ m0 and n ≡ n0 then
m + n ≡ m0 + n0 and mn ≡ m0 n0 .
It is clear that two sets are equal if and only if their characteristic functions agree mod 2.
Also, it is easy to see that, for A, B ⊆ X, χA∩B = χA χB , and, for x ∈ X,
χA4B (x) ≡ χA (x) + χB (x) .
Thus, for A, B, C ∈ R and x ∈ X, we have
χ(A4B)4C (x) ≡ χA4B (x) + χC (x) ≡ χA (x) + χB (x) + χC (x)
≡ χA (x) + χB4C (x) ≡ χA4(B4C) (x)
It follows that (A4B)4C = A4(B4C).
The distributive law may be proved using by direct calculation as follows:
A ∩ (B4C) = A ∩ ((B\C) ∪ (C\B))
= (A ∩ (B\C)) ∪ (A ∩ (C\B))
= ((A ∩ B)\(A ∩ C)) ∪ ((A ∩ C)\(A ∩ B))
= (A ∩ B)4(A ∩ C).
Alternatively, you can use Venn diagrams, or the following characteristic function proof. We
have, for all x ∈ X,
χA∩(B4C) (x) = χA (x)χB4C (x) ≡ χA (x)(χB (x) + χC (x))
= χA (x)χB (x) + χA (x)χC (x) = χA∩B (x) + χA∩C (x) ≡ χ(A∩B)4(A∩C) (x) .
1
Thus A ∩ (B4C) = (A ∩ B)4(A ∩ C).
A more sophisticated approach is to regard {0, 1} as the finite field F2 , so that {0, 1}X is the
set of functions from X to F2 . This is easily seen to be a ring with pointwise addition and
multiplication, and the obvious bijection from {0, 1}X to P(X) may be used to transfer these
operations over to P(X). The resulting operations on P(X) are then symmetric difference
and intersection, and these make P(X) into a ring isomorphic to FX
2 . Since associativity and
distributivity hold on the whole of P(X), they also hold on R.
2. There are many examples, the most standard being counting measure on N, µ, and An =
{n, n + 1, n + 2, . . . }, so that
∞
\
An = ∅,
n=1
but
N
\
µ(∅) = 0 6= lim µ
N →∞
!
An
= lim µ(AN ) = ∞.
N →∞
n=1
3. In this question we make heavy use of the (easy to check) facts that
f −1 (A\B) = f −1 (A)\f −1 (B),


[
[
f −1 
Aγ  =
f −1 (Aγ ).
γ∈Γ
Note that
γ∈Γ

f

[
Aγ  =
γ∈Γ
[
f (Aγ )
γ∈Γ
is true, but that, in general, f (A\B) is not f (A)\f (B). We have f : X → Y a function, F1
a σ-field on X, F2 a σ-field on Y . Set
F3 = {f −1 (E) : E ∈ F2 }.
Then
(i) ∅ = f −1 (∅) and X = f −1 (Y ), so ∅ and X are in F3 .
(ii) Let A, B ∈ F3 . Then ∃ E, F ∈ F2 with f −1 (E) = A, f −1 (F ) = B. But then
A\B = f −1 (E)\f −1 (F ) = f −1 (E\F ).
Since E\F ∈ F2 , we have A\B ∈ F3 .
(iii) Let A1 , A2 , · · · ∈ F3 . Then ∃ E1 , E2 , . . . in F2 with An = f −1 (En ) (n ∈ N). But then
!
∞
∞
∞
[
[
[
An =
f −1 (En ) = f −1
En ,
n=1
and since
∞
S
n=1
En ∈ F2 , we have
n=1
∞
S
n=1
An ∈ F3 . Thus F3 is a σ-field on X.
n=1
Now set F4 = {E ⊆ Y : f −1 (E) ∈ F1 }. Then
2
(i) f −1 (∅) = ∅ and f −1 (Y ) = X, and so ∅ and Y are in F4 .
(ii) If A, B ∈ F4 , then
f −1 (A) ∈ F1 , f −1 (B) ∈ F1 ,
f −1 (A\B) = f −1 (A)\f −1 (B),
and so f −1 (A\B) ∈ F1 , i.e. A\B ∈ F4 .
(iii) If A1 , A2 , A3 · · · ∈ F4 , then
f −1 (An ) ∈ F1
Thus
f
−1
∞
[
!
An
and so
f −1 (An ) ∈ F1 ,
n=1
n=1
∞
S
=
∞
[
(n ∈ N).
An ∈ F 4 .
n=1
Thus F4 is a σ-field on Y .
In general, {f (E) : E ∈ F1 } need not be a σ-field on Y . In particular, whenever f is not
surjective, we find Y 6∈ {f (E) : E ∈ F1 }, so you cannot have a σ-field on Y .
It is always best to give a specific counterexample, e.g. X = Y = N, f (n) = 2n, F1 =
F2 = P(N). Then {f (E) : E ∈ F1 } = P(2N), which is not a σ-field on N. (Of course, it IS a
σ-field on 2N.)
There are several other ways that such a collection can fail to be a σ-field on Y . However,
it is NOT enough to say, for example, that f (A ∩ B) is not always equal to f (A) ∩ f (B).
Although this is true, it only shows that one attempted proof that the collection is a σ-field
would be invalid, and this is not enough to show that the result is actually false. There are
many possible invalid attempted proofs of correct results!
4. Set
S1 = {(a, b) : a, b ∈ R}
S2 = {[a, b] : a, b ∈ R}
S3 = {(−∞, a] : a ∈ R}.
These collections generate σ-fields on R, FR (S1 ), FR (S2 ), FR (S3 ).
Note. All the sets listed are Borel sets, since (a, b) is open, while [a, b] and (−∞, a] are closed
subsets of R. Thus B is a σ-field containing S1 , S2 and S3 , so that
FR (Sk ) ⊆ B
(k = 1, 2, 3).
To obtain equality:
(i) Every open subset of R is a countable union of open intervals, and so FR (S1 ) includes
all open sets. Thus FR (S1 ) contains the σ-field generated by the open sets, giving us
FR (S1 ) ⊇ B,
and so FR (S1 ) = B.
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(ii) If a < b, then
∞ [
b−a
b−a
,
(a, b) =
a+
,b −
n
n
n=3
so that FR (S2 ) ⊇ FR (S1 ) = B, giving us FR (S2 ) = B.
(iii) If a < b, then
(a, b] = (−∞, b]\(−∞, a],
so that our standard semi-ring P of half open intervals must be contained in FR (S3 ).
Thus FR (P ) ⊆ FR (S3 ). But it is standard that FR (P ) = B, and so we have B ⊆ FR (S3 ),
giving us equality.
5. (a) Since ∅ = ∅ ∩ E and E = Y ∩ E, we see that ∅ and E are in FE . Next, if B ∈ FE , then
there exists A ∈ F with A ∩ E = B. But then E \ B = E \ A = (Y \ A) ∩ E. Since Y \ A ∈ F,
we have E \ B ∈ FE . Finally, if B1 , B2 , S
. . . are sets in FE , S
then there are
S sets A1 , A2 , . . . in
F
with
A
∩
E
=
B
(n
∈
N).
But
then
A
∈
F,
and
B
=
n
n
n
n
n∈N
n∈N
n∈N An ∩ E . Thus
S
B
∈
F
,
as
required.
E
n∈N n
(b) Given that E ∈ F, then since an intersection of two sets in F is in F,
{A ∩ E : A ∈ F} ⊆ {A ∈ F : A ⊆ E } .
However, if A is in the set on the right hand side, then A ∈ F and A = A ∩ E, so that A is
also in the set on the left hand side. The result follows.
6. Set F = {A ∪ B : A ∈ FX (C), B ∈ { ∅, Y \ X} } . Following the hint, we begin by showing
that F is a σ-field on Y . Certainly ∅ = ∅ ∪ ∅ ∈ F and Y = X ∪ (Y \ X) ∈ F. Now suppose
that E ∈ F. Then E = A ∪ B for some A ∈ FX (C) and B ∈ { ∅, Y \ X} . But then we see,
because A ⊆ X and B ⊆ Y \ X, that Y \ E = (X \ A) ∪ ((Y \ X) \ B). Since X \ A ∈ FX (C)
and (Y \ X) \ B is one of ∅ or Y \ X, we have Y \ E ∈ F.
Finally, suppose that En ∈ F (n ∈ N). then, for all n ∈ N, there are An ∈ FX (C) and
Bn ∈ { ∅, Y \ X} such that En = An ∪ Bn . We then have
!
!
[
[
[
En =
An ∪
Bn .
n∈N
n∈N
S
But n∈N An ∈ FX (C) and it is easy to see that
F, as required.
n∈N
S
n∈N Bn
∈ { ∅, Y \X} , so we have
S
n∈N En
∈
We now know that F is a σ-field on Y . Clearly C ⊆ F, and so FY (C) ⊆ F.
The reverse inequality requires us to use the fact that X ∈ FY (C) (given in the question).
Without this assumption, the result is false. (You may see this in the case where X 6= ∅,
Y 6= X and C = {∅}.) The proof also involves a rather subtle point. Since C generates FX (C)
on X and C ⊆ FY (C), you would expect that FY (C) contains the σ-field on X generated by
C, i.e. FX (C). This is, however, not obvious, as we do not yet know how the σ-fields generated
on X and on Y are related: that is what this question is about! This unrigorous approach
can be made more formal using transfinite induction, but I recommend the following proof
(following the hint).
Set F 0 = FY (C). Using the notation of Question 5, with E = X, we have a σ-field on X
0 = {A ∩ X : A ∈ F 0 }. Since C ⊆ F 0 , and all sets in C are subsets of X, we also have
FX
0 . Since F 0 is a σ-field on X (rather than Y ), we have F (C) ⊆ F 0 . But, because
C ⊆ FX
X
X
X
0 ⊆ F 0 , and so F (C) ⊆ F (C), as expected.
we are given that X ∈ FY (C), we have FX
X
Y
4
Since the question gives us that X ∈ F 0 , we also have Y \ X ∈ F 0 .
Finally, let E ∈ F. Then E = A ∪ B, where A ∈ FX (C) and B is ∅ or Y \ X. Thus A and B
are in F 0 , and hence E = A ∪ B is also in F 0 , as required.
7. This question can be solved directly in a similar way to Question 6. However we can also use
Question 6 to help solve this question. Here is how it works. Set C = {(−∞, a] : a ∈ R } .
We know, by Question 4, that FR (C) = B. Set Y = R ∪ {−∞} and set G = FY (C). To
applySQuestion 6, with X = R, we first need to check that R ∈ FY (C). This is OK, because
R = n∈N (−∞, n]. Now, by Question 6,
G = FY (C) = {A ∪ B : A ∈ B, B ⊆ {−∞} } .
(1)
Note that saying that B ⊆ {−∞} is another way of saying that B is either ∅ or {−∞}. Note
also that {−∞} ∈ G.
Set C 0 = {[−∞, a] : a ∈ RT} . We claim that G = FY (C 0 ). Clearly C 0 ⊆ G, and so FY (C 0 ) ⊆ G.
However, since {−∞} = n∈N [−∞, −n], we have {−∞} ∈ FY (C 0 ). From this it follows that
C ⊆ FY (C 0 ), and hence G ⊆ FY (C 0 ), and equality follows.
S
We now wish to apply Question 6 again. First note that Y = n∈N [−∞, n] ∈ FR̄ (C 0 ). Thus,
by Question 6 again, noting that R̄ \ Y = {+∞}, we have
FR̄ (C 0 ) = {A0 ∪ B 0 : A0 ∈ G, B 0 ⊆ {+∞} } .
(2)
From (1) and (2) it follows easily that (with F as defined in the question) FR̄ (C 0 ) = F. In
other words, F is a σ-field on R̄, and F is generated by the collection {[−∞, a] : a ∈ R } , as
required.
8. Let F 0 be the σ-field on R̄ generated by the collection of open subsets of R̄ (as defined in the
notes). (Thus F 0 is the set of Borel subsets of R̄ in the topological sense.) What are
the open subsets of R̄? From the notes we know that the open sets in R̄ are all countable
unions of sets of the form [−∞, a), (a, b) or (b, ∞], for a, b ∈ R.
Certainly all sets of these three types are in the σ-field F, hence so are all the open subsets
of R̄. It follows that F 0 ⊆ F. On the other hand, F is generated (as a σ-field on R̄) by the
collection {[−∞, a] : a ∈ R}, and these sets are all closed in R̄ (you should check this!), and
hence are in F 0 (they are the complements in R̄ of open subsets of R̄). Thus F ⊆ F 0 and so
we have equality.
Recall: the Riemann lower integral of a bounded function f over an interval is the
supremum of all the Riemann lower sums for f , while the Riemann upper integral is
the infimum of all the Riemann upper sums for f . The Riemann lower integral is always
less than or equal to the Riemann upper integral, and we say that the function is Riemann
integrable if the Riemann lower integral is equal to the Riemann upper integral.
9. This is standard, and there are many ways to do it. First note that, since E is only a finite
set, it is clear that the Riemann lower integral of χE from 0 to 1 is 0. Thus it is enough to
show that the infimum of the Riemann upper sums for χE is ≤ 0. Let n be the number of
points in E.
Let ε > 0, and partition [0, 1] up into N equal intervals, where N is chosen such that
The partition is x0 < x1 < · · · < xN , x0 = 0, x1 = n1 , . . . , xN = 1.
2n
N
< ε.
Then, since E has only n points, and each point is in at most two of the intervals [xi , xi+1 ],
it follows that at most 2n of these intervals contain points of E, and so
N
−1
X
(xi+1 − xi ) sup{χE (t) : t ∈ [xi , xi+1 ]} ≤
i=0
5
1
× 2n < ε.
N
Thus the infimum of the Riemann upper sums for χE is < ε. Since ε > 0 was arbitrary, the
result follows.
10. We may take E = Q ∩ [0, 1]. For then we see that all the Riemann lower sums for χE are
≤ 0, and so it is easy to see that the Riemann lower integral of χE is 0. But, similarly, all
the Riemann upper sums for χE are ≥ 1, and the Riemann upper integral from 0 to 1 of χE
is 1. Thus χE is NOT Riemann integrable on [0, 1].
11. We may enumerate the countable set Q ∩ [0, 1]: say Q ∩ [0, 1] = {q1 , q2 , q3 , . . . }. Set En =
{q1 , q2 , . . . , qn } (n ∈ N). Then each function χEn is Riemann integrable on [0, 1], but
χEn → χE
pointwise,
where E = Q ∩ [0, 1]. By Question 10, χE is not Riemann integrable on [0, 1].
12. Here we use the facts that
χE∩F
= χE χF
(3)
χE\F
= χE − χE∩F
(4)
χE∪F
= χE + χF − χE∩F
(5)
and χ∅ , χ[0,1] are constantly 0, 1 respectively (on [0, 1]).
Thus ∅ ∈ F, and, if E, F ∈ F, then χE , χF are Riemann integrable; so, by (3), (4), (5) we
have (since a product, sum or difference of two Riemann integrable functions is still Riemann
integrable) χE∩F , χE\F and χE∪F are all Riemann integrable.
Thus E ∩ F , E\F and E ∪ F are all in F, and so F is a ring. Finally, since χ[0,1] is the
constant function 1, χ[0,1] is Riemann integrable, and [0, 1] ∈ F. Thus F is a field on [0, 1].
F is not a σ-field, because Q ∩ [0, 1] is not in F, but Q ∩ [0, 1] is countable, and so, again
writing Q ∩ [0, 1] = {q1 , q2 , q3 , . . . }
Q ∩ [0, 1] =
∞
[
{qn },
n=1
a countable union of sets in F. Thus F is not a σ-field.
13.
(i) Yes, χC is Riemann integrable, with Riemann integral
Z 1
χC (x) dx = 0.
0
To see this, first note that the lower integral
Z 1
χC (x) dx
0
is non-negative. However, C is contained in each of a sequence of sets Xn , where Xn
consists of a disjoint union of 2n intervals, each of length 3−n . Thus, for all n ∈ N, the
R1
upper integral of χC , 0 χC (x) dx, satisfies
Z 1
Z 1
χC (x) dx ≤
χXn (x) dx
0
0
2n
=
.
3n
Letting n → ∞, we deduce that the upper integral of χC (x) is ≤ 0, and so χC is Riemann
integrable, with integral 0.
6
(ii) This is tricky, but one way is to modify the construction of the cantor set. Usually we
delete the middle third of each interval in Xn to obtain Xn+1 . Suppose you start with
X0 = [0, 1], delete the middle 41 to obtain X1 = [0, 38 ] ∪ [ 58 , 1], and form Xn+1 from Xn
1
by deleting the middle (n+2)
2 th. Then, because
∞ Y
n=0
1
1−
(n + 2)2
=
1
> 0,
2
(exercise!)
T
we can set E = ∞
n=0 Xn , and then E contains no intervals of positive length, so the
lower integral of χE over [0, 1] is 0. But the upper integral is ≥ 12 , because, essentially
the length of E is 12 .
Exercise: find an argument (possibly using compactness) to prove this last statement
without using Lebesgue measure.
Since the lower and upper integrals are different, E will do (it is closed). Note that
Q ∩ [0, 1] will not do, because it is not closed. Also (unfortunately) it is not enough
to take, for example, E = {0} ∪ { 21 , 13 , 14 , 15 , . . . } because, for this E, χE is Riemann
integrable.
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