MAT4300 - Mandatory Assignment
Sindre Froyn
(a)
We have A1 , . . . , An ∈ A. We use (i),(ii) and (iii) as the axioms of algebras.
(iii)
A1 , A2 ∈ A =⇒ A1 ∪ A2 ∈ A
We define the set B2 := A1 ∪ A2 , which we just showed was in A.
(iii)
B2 , A3 ∈ A =⇒ B2 ∪ A3 ∈ A =⇒ A1 ∪ A2 ∪ A3 ∈ A
We define B3 := A1 ∪A2 ∪A3 and as we have just shown, B3 ∈ A. We assume
this is true for Bk := A1 ∪ . . . ∪ Ak , so Bk ∈ A. We check the next step.
(iii)
Bk , Ak+1 ∈ A =⇒ Bk ∪ Ak+1 ∈ A =⇒ A1 ∪ . . . ∪ Ak ∪ Ak+1 ∈ A
By induction, and repeated use of property (iii) for algebras, we have verified
that the algebra A is closed under a countable and finite number of unions,
so
(a)
A1 , . . . , An ∈ A =⇒ A1 ∪ A2 ∪ . . . ∪ An ∈ A.
(b)
We assume A, B ∈ A.
(ii)
(iii)
A, B ∈ A =⇒ Ac , B c ∈ A =⇒ Ac ∪ B c ∈ A
We apply DeMorgans identity.
(ii)
Ac ∪ B c ∈ A ⇐⇒ (A ∩ B)c ∈ A =⇒ A ∩ B ∈ A
We have now shown that for A, B ∈ A, we have A ∩ B ∈ A. We use this in
the next part.
(ii)
(b)
A, B ∈ A =⇒ A, B c ∈ A =⇒ A ∩ B c ∈ A ⇐⇒ A \ B ∈ A
(c)
We are going to determine if the set B is an algebra. To do this we must
verify the three axioms of algebras.
∅ ⊂ N, ∅ is finite =⇒ ∅ ∈ B
1
(i)
We assume we have a set B ∈ B, and show that this implies B c ∈ B. Since
B ∈ B, we have B or B c finite, which we can use directly as properties for
the set B c (since (B c )c = B).
B ∈ B =⇒ B ⊂ N =⇒ B c = N \ B ⊂ N =⇒ B c ∈ B
(ii)
For the last property we assume A, B ∈ B. Since A ⊂ N and B ⊂ N we
obviously have A ∪ B ⊂ N, since the “worst case scenario” is N ⊂ N. We
check if (A ∪ B) or (A ∪ B)c are finite, and using DeMorgan’s law on the
complemented union we can rewrite it as Ac ∩B c . There are four possibilities:
so if A and B are finite, then A ∪ B is finite. In any other case (A ∪ B)c is
finite, since the intersection Ac ∩ B c restricts us to a finite set.
A, B ∈ B =⇒ A ∪ B ∈ B
(iii)
We have checked the three axioms of algebras for B and verified that it
is an algebra.
(d)
The σ-algebra generated by B is the smallest σ-algebra containing it, so
B ⊂ σ(B). This means the first two properties we verified for the algebra
B also apply for σ(B) for all sets B ∈ B. The σ-algebra allows countable
unions, which the algebra did not, and these unions provide new sets. For
sets A1 , A2 , A3 , . . . there are two possibilities. First if all are finite: then their
union is also finite.
[
A1 , A2 , . . . ∈ σ(B) =⇒
Ai ∈ σ(B)
i∈N
If at least one of the sets Ak is not finite, then the complement to the union
is finite.
[ c
\
[
Ai ∈ σ(B) =⇒
Ai ∈ σ(B) =⇒
Aci ∈ σ(B)
i∈N
i∈N
i∈N
As described in (c), the one, or more, finite set(s) Ack restricts this intersection
to a finite set, so (∪i∈N Ai )c is finite. All the sets in the σ-algebra generated
2
by B satisfy the same conditions as the sets in the underlying algebra: either
it is finite, or it has a finite complement.
σ(B) = B ⊂ N B or B c finite
(e)
We assume A1 , A2 , . . . , An ∈ A are all mutually disjoint. We use (i) and (ii)
as the axioms of the finitely additive measure. We define B2 = A1 ∪ A2 .
(ii)
µ(B2 ) = µ A1 ∪ A2 = µ(A1 ) + µ(A2 )
We define B3 := A1 ∪ A2 ∪ A3 = B2 ∪ A3 .
(ii)
µ(B3 ) = µ(B2 ∪ A3 ) = µ(B2 ) + µ(A3 ) = µ(A1 ) + µ(A2 ) + µ(A3 )
P
We assume this is true for Bk , so µ(Bk ) = µ(A1 ∪ . . . ∪ Ak ) = ki µ(Ak ). We
check the next step, Bk+1 = Bk ∪ Ak+1 .
(ii)
µ(Bk+1 ) = µ(Bk ∪ Ak+1 ) = µ(Bk ) + µ(Ak+1 ) =
k+1
X
µ(Ai )
i=1
By induction, this is true for a countable, finite amount of sets, so
µ
n
[
i=1
Ai =
n
X
µ(Ai ).
i=1
(f)
To check that ν : B 7→ [0, ∞] is a finitely additive measure, we must verify
the two defining properties. Property (i) is verified directly.
ν(∅) = 0, since ∅ is finite.
To check property (ii) we must consider different scenarios. We recall that A
and B are disjoint. When A and B both are finite, then so is A ∪ B.
ν(A ∪ B) = 0 = ν(A) + ν(B)
When Ac is finite and B is finite, (A ∪ B)c = Ac ∩ B c is finite. By symmetry
the same argument applies for the reverse case.
ν(A ∪ B) = 1 = 1 + 0 = ν(A) + ν(B)
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For the last scenario, Ac and B c finite, it is impossible to choose disjoint
sets A and B. To illustrate this, we can set Ac = {1, . . . , 10} in which case
A = {11, 12, . . .}, and now we cannot choose a finite B c such that B is disjoint
from A, so the function ν does not apply to this case.
Both properties are met, and ν is a finitely additive measure.
(g)
To extend the measure ν to the σ-algebra σ(B) we need to use Caratheodory’s
theorem. To do this we must verify if the algebra B is a semi-ring, and check
if ν satisfies the conditions of a pre-measure on B.
To verify that B is a semi-ring, we must check the three defining properties.
Obviously, ∅ ∈ B. As we showed in (b), the intersection of two sets in an
algebra, is again in the algebra: S, T ∈ B ⇒ S ∩ T ∈ B. For the last property,
we must check that for two sets S, T ∈ B, there are finitely many disjoint
sets S1 , . . . , Sn such that
n
[
S\T =
Sj .
j=1
However, as we showed in (b), if S, T ∈ B then S \ T ∈ B, since B is an
algebra, so this difference is always a single set in B.
With all the three properties of semi-rings met, B is a semi-ring.
We verify that ν satisfies the two necessary properties of pre-measures on the
semi-ring. The first is ν(∅) = 0 which is satisfied since ν(∅) = 0 by (f).
For the second property, we assume P
(Sj )j∈N ⊂ B are disjoint, set
S = ∪j∈N Sj ∈ BPand check that
ν(S)
=
j∈N ν(Sj ). If Sj is finite for all
P
j, then the sum j∈N ν(Sj ) = j∈N 0 = 0, but on the left side we combine
all natural numbers and get an infinite set, so ν(S) = 1. This property is
not met for the finitely additive measure ν, so we cannot use Caratheodory’s
theorem.
In conclusion we cannot extend ν to be a measure on the σ-algebra σ(B).
(h)
Due to laziness, procrastination, excessive workload and confusion I have
chosen not to answer this question. My apologies!
(i)
Pn
Let f and
g
be
step
functions,
f
≤
g,
with
representations
f
=
i=1 ai 1Ai
Pm
and g = j=1 bj 1Aj , where we have chosen the standard representation over
the same partition {Ai } of X. Using the representation of f and g:
n
X
i=1
ai 1Ai ≤
n
X
i=1
bi 1Ai =⇒ 0 ≤
n
X
i=1
4
bi 1Ai −
n
X
i=1
ai 1Ai
n
X
=
(bi − ai )1Ai
i=1
We have a step function representation of a function that is greater or equal
to 0. Using the preliminary integral, we replace the indicator function 1Ai
with the non-negative measure µ(Ai ), and we preserve the inequality. It is
permissible with a negative measure, since we know µ(X) < ∞.
0≤
n
X
(bi − ai )µ(Ai) =
i=1
n
X
bi µ(Ai ) −
i=1
n
X
ai µ(Ai ) ≤
i=1
n
X
n
X
ai µ(Ai ) =⇒
i=1
bi µ(Ai ) =⇒ I(f ) ≤ I(g)
i=1
(j)
We have a non-negative, simple function f .
Z
f dµ = inf I(h) h a simple function, f ≤ h = I(f )
Z
f dµ = sup I(h) h a simple function, 0 ≤ h ≤ f = I(f )
In both cases we get I(f ) since f itself is among the simple functions h in
the sets. Since the upper and lower integral are the same, f is integrable.
Z
Z
Z
f dµ = f dµ = f dµ = I(f )
(k)
We have a function h : N 7→ R, given by h(n) = n1 . This function has a simple
function representation, and thus we have an expression for the preliminary
integral.
n
n
X
X
ai ν(Ai )
ai 1Ai =⇒ I(h) = lim
h = lim
n→∞
n→∞
i=1
i=1
1
1
The sets {A1 } = 1, {A2 } = 2, etc and a1 = = 1, a2 = 12 and so on. Since h
has a simple function representation, both the upper and lower integral will
be I(h) as in part (j).
Z
Z
Z
n
X
hdν = hdν = hdν = I(h) = lim
ai ν(Ai )
n→∞
i=1
All the sets Ai are finite so we get ν(Ai ) = 0 for all i.
Z
n
X
ai ν(Ai ) = 0.
hdν = I(h) = lim
n→∞
5
i=1
(l)
For the measure space (Y, C, λ) we have λ(Y ) 6= 0. By definition of measures
λ: Y 7→ [0, ∞], but the inequality means that there is at least one set A ⊂ Y
such that λ(A) > 0. A strictly positive, measurable function f : Y 7→ R
means f = 0 on a null set: λ({f = 0}) = 0. By definition of the integral of
measurable functions:
Z
Z
Z
Z
+
−
f dλ = f dλ − f dλ = f + dλ
We only consider the positive part since f is strictly positive and f − = 0. By
definition of the integral of positive functions:
Z
f dλ = sup I(h) h a simple, strictly positive function, h ≤ f
We can choose some h from this set, which can be a crude approximation of
f , but still strictly positive and over the sets Ai ⊂ Y .
h=
n
X
ai 1Ai
i=1
For this simple function we know that there exists at least one set Ak such
that λ(Ak ) > 0, and since this set is not a null set h is positive on Ak so
ak > 0. To sum up we know there exists som k ∈ N such that ak λ(Ak ) > 0.
With this we can prove the desired inequality.
Z
n
X
0 < ak λ(Ak ) ≤
ai λ(Ai ) = I(h) ≤ f dλ
i=1
We have assumed that we integrate over the entire set Y . Even if the measure
of Y isn’t 0 doesn’t mean there aren’t subsets with measure 0. We can
construct a simple example to illustrate this.
Y ∗ = [1], [2], (3, 4)
We define a strictly positive function f ∗ on this set, where g is some strictly
positive function.

x=1
 1
∗
f (x) =
1
x=2

g(x) x ∈ (3, 4)
The Lebesgue measure λ∗ gives us λ∗ (Y ∗ ) 6= 0, the function f ∗ : Y ∗ 7→ R is
strictly positive, but the integral over a subset of Y ∗ is zero.
Z
f ∗ dλ∗ = 0.
[1]∪[2]
6