This chapter is on the basics of measure theory and integration. A measure is a real
valued mapping from some subset of the power set of a given set which has values in
[0,∞ ]
. Many apparently different things can be considered as measures and also there is
an integral defined. By discussing this in terms of axioms and in a very abstract setting,
many different topics can be considered in terms of one general theory. For example, it
will turn out that sums are included as an integral of this sort. So is the usual
integral as well as things which are often thought of as being in between sums and
integrals.
Let Ω be a set and let ℱ be a collection of subsets of Ω satisfying
∅ ∈ ℱ,Ω ∈ ℱ, (9.1.1)
(9.1.1)
E ∈ ℱ implies EC ≡ Ω ∖ E ∈ ℱ,
If {En}∞n=1 ⊆ ℱ, then ∪∞n=1 En ∈ ℱ. (9.1.2)
(9.1.2)
Definition 9.1.1A collection of subsets of a set, Ω, satisfying Formulas9.1.1-9.1.2is called a σ algebra.
As an example, let Ω be any set and let ℱ = P(Ω), the set of all subsets of Ω (power
set). This obviously satisfies Formulas 9.1.1-9.1.2.
Lemma 9.1.2Let C be a set whose elements are σ algebras of subsets of Ω. Then∩C is a σ algebra also.
Be sure to verify this lemma. It follows immediately from the above definitions but it
is important for you to check the details.
Example 9.1.3Let τ denote the collection of all open sets in ℝ^{n}and let σ
(τ)
≡intersection of all σ algebras that contain τ. σ
(τ)
is called the σ algebra of Borelsets. In general, for a collection of sets, Σ, σ
(Σ )
is the smallest σ algebra whichcontains Σ.
This is a very important σ algebra and it will be referred to frequently as the Borel
sets. Attempts to describe a typical Borel set are more trouble than they are
worth and it is not easy to do so. Rather, one uses the definition just given in
the example. Note, however, that all countable intersections of open sets and
countable unions of closed sets are Borel sets. Such sets are called G_{δ}and F_{σ}respectively.
Definition 9.1.4Let ℱ be a σ algebra of sets of Ω and let μ : ℱ→ [0,∞]. μ is calleda measure if
∞ ∞
μ (⋃ Ei) = ∑ μ (Ei) (9.1.3)
i=1 i=1
(9.1.3)
whenever the E_{i}are disjoint sets of ℱ.The triple, (Ω,ℱ,μ) is called a measure spaceandthe elements of ℱ are called the measurable sets.
(Ω,ℱ, μ)
is a finite measure spacewhenμ
(Ω )
< ∞.
Note that the above definition immediately implies that if E_{i}∈ℱ and the sets E_{i} are
not necessarily disjoint,
∞⋃ ∞∑
μ( Ei) ≤ μ(Ei).
i=1 i=1
To see this, let F_{1}≡ E_{1}, F_{2}≡ E_{2}∖ E_{1},
⋅⋅⋅
,F_{n}≡ E_{n}∖∪_{i=1}^{n−1}E_{i}, then the sets F_{i} are
disjoint sets in ℱ and
∞ ∞ ∞ ∞
μ(⋃ E ) = μ(⋃ F ) = ∑ μ (F ) ≤ ∑ μ(E )
i=1 i i=1 i i=1 i i=1 i
because of the fact that each E_{i}⊇ F_{i} and so
μ (Ei) = μ(Fi)+ μ (Ei ∖Fi)
which implies μ
(Ei)
≥ μ
(Fi)
.
The following theorem is the basis for most of what is done in the theory of measure
and integration. It is a very simple result which follows directly from the above
definition.
Theorem 9.1.5Let {E_{m}}_{m=1}^{∞}be a sequence of measurable sets in a measure space
(Ω,ℱ,μ). Then if
⋅⋅⋅
E_{n}⊆ E_{n+1}⊆ E_{n+2}⊆
⋅⋅⋅
,
μ(∪∞ Ei) = lim μ (En ) (9.1.4)
i=1 n→ ∞
(9.1.4)
and if
⋅⋅⋅
E_{n}⊇ E_{n+1}⊇ E_{n+2}⊇
⋅⋅⋅
and μ(E_{1}) < ∞, then
∞
μ(∩i=1Ei ) = lni→m∞ μ(En). (9.1.5)
(9.1.5)
Stated more succinctly, E_{k}↑ E implies μ
(Ek )
↑ μ
(E)
and E_{k}↓ E with μ
(E1 )
< ∞implies μ
(Ek)
↓ μ
(E )
.
Proof: First note that ∩_{i=1}^{∞}E_{i} = (∪_{i=1}^{∞}E_{i}^{C})^{C}∈ℱso ∩_{i=1}^{∞}E_{i}is measurable. Also
note that for A and B sets of ℱ, A∖B ≡
( )
AC ∪ B
^{C}∈ℱ. To show 9.1.4, note that 9.1.4
is obviously true if μ(E_{k}) = ∞ for any k. Therefore, assume μ(E_{k}) < ∞ for all k.
Thus
μ(Ek+1 ∖Ek )+ μ(Ek) = μ(Ek+1)
and so
μ(Ek+1 ∖Ek) = μ(Ek+1)− μ(Ek).
Also,
∞⋃ ⋃∞
Ek = E1 ∪ (Ek+1 ∖ Ek)
k=1 k=1
and the sets in the above union are disjoint. Hence by 9.1.3,
n
= μ(E1) −nl→im∞ μ(∩i=1Ei) = μ(E1)− nli→m∞ μ(En ),
Hence, subtracting μ
(E1)
from both sides,
lim μ(En) = μ(∩∞i=1Ei).
n→∞
This proves the theorem.
It is convenient to allow functions to take the value +∞. You should think of +∞,
usually referred to as ∞ as something out at the right end of the real line and its only
importance is the notion of sequences converging to it. x_{n}→∞ exactly when for all
l ∈ ℝ, there exists N such that if n ≥ N, then
xn > l.
This is what it means for a sequence to converge to ∞. Don’t think of ∞ as a number. It
is just a convenient symbol which allows the consideration of some limit operations more
simply. Similar considerations apply to −∞ but this value is not of very great interest. In
fact the set of most interest is the complex numbers or some vector space. Therefore, this
topic is not considered.
Lemma 9.1.6Let f : Ω → (−∞,∞] where ℱ is a σ algebra of subsets of Ω. Then thefollowing are equivalent.
− 1
f ((d,∞ ]) ∈ ℱfor all finite d,
− 1
f ((− ∞,d)) ∈ ℱfor all finite d,
− 1
f ([d,∞ ]) ∈ ℱfor all finite d,
f −1((− ∞,d]) ∈ ℱfor all finite d,
f− 1((a,b)) ∈ ℱ for all a < b,− ∞ < a < b < ∞.
Proof: First note that the first and the third are equivalent. To see this,
observe
f−1([d,∞ ]) = ∩∞ f −1((d− 1∕n,∞ ]),
n=1
and so if the first condition holds, then so does the third.
f−1((d,∞ ]) = ∪ ∞n=1f−1([d+ 1∕n,∞ ]),
and so if the third condition holds, so does the first.
Similarly, the second and fourth conditions are equivalent. Now
f−1((− ∞, d]) = (f−1((d,∞ ]))C
so the first and fourth conditions are equivalent. Thus the first four conditions are
equivalent and if any of them hold, then for −∞ < a < b < ∞,
f−1((a,b)) = f−1((− ∞, b))∩f −1((a,∞ ]) ∈ ℱ.
Finally, if the last condition holds,
f −1([d,∞ ]) = (∪∞ f−1((− k + d,d)))C ∈ ℱ
k=1
and so the third condition holds. Therefore, all five conditions are equivalent. This proves
the lemma.
This lemma allows for the following definition of a measurable function having values
in (−∞,∞].
Definition 9.1.7Let (Ω,ℱ,μ) be a measure space and let f : Ω → (−∞,∞].Then f is said to be measurableif any of the equivalent conditions of Lemma 9.1.6hold. When the σ algebra, ℱ equals the Borel σ algebra, ℬ, the function is calledBorel measurable.More generally, if f : Ω → X where X is a topological space, fis said to be measurable if f^{−1}
(U)
∈ℱ whenever U is open.
Theorem 9.1.8Let f_{n}and fbe functions mapping Ω to (−∞,∞] where ℱis a σ algebra of measurable sets of Ω. Then if f_{n}is measurable, and f(ω) =
lim_{n→∞}f_{n}(ω), it follows that f is also measurable. (Pointwise limits of measurablefunctions are measurable.)
Proof: First it is shown f^{−1}
((a,b))
∈ℱ. Let V_{m}≡
( )
a+ m1,b− m1
and
V_{m} =
[ ]
a+ m1,b− m1
. Then for all m, V_{m}⊆
(a,b)
and
∞ ∞ --
(a,b) = ∪m=1Vm = ∪ m=1V m.
Note that V_{m}≠∅ for all m large enough. Since fis the pointwise limit of f_{n},
--
f−1(Vm) ⊆ {ω : fk(ω) ∈ Vm for all k large enough} ⊆ f−1(Vm ).
You should note that the expression in the middle is of the form
∪∞ ∩ ∞ f−1(Vm).
n=1 k=n k
Therefore,
f−1((a,b)) = ∪∞ f−1(V ) ⊆ ∪ ∞ ∪∞ ∩∞ f −1(V )
m=1 m m=1 n=1 k=n k m
--
⊆ ∪∞m=1f −1(Vm ) = f−1((a,b)).
It follows f^{−1}(
(a,b)
) ∈ℱbecause it equals the expression in the middle which is
measurable. This shows f is measurable.
The following theorem considers the case of functions which have values in a metric
space. Its proof is similar to the proof of the above.
Theorem 9.1.9Let
{fn}
be a sequence of measurable functions mapping Ω to
(X, d)
where
(X,d)
is a metric space and
(Ω,ℱ )
is a measure space. Suppose alsothat f
(ω )
= lim_{n→∞}f_{n}
(ω)
for all ω. Then f is also a measurable function.
Proof:It is required to show f^{−1}
(U)
is measurable for all U open. Let
{ }
V ≡ x ∈ U : dist(x,UC) > 1 .
m m
Thus
{ }
( C ) -1
Vm ⊆ x ∈ U : dist x,U ≥ m
and V_{m}⊆V_{m}⊆ V_{m+1} and ∪_{m}V_{m} = U. Then since V_{m} is open,
f−1(Vm) = ∪∞ ∩ ∞ f−1 (Vm )
n=1 k=n k
and so
f− 1(U) = ∪∞m=1f −1(Vm )
= ∪∞ ∪∞ ∩∞ f− 1(Vm )
m∞=1 n−=11(--k=)n k−1
⊆ ∪m=1f Vm = f (U )
which shows f^{−1}
(U )
is measurable. This proves the theorem.
Now here is a simple observation.
Observation 9.1.10Let f : Ω → X where X is some topological space. Suppose
m∑
f (ω) = xkXAk (ω )
k=1
where each x_{k}∈ X and the A_{k}are disjoint measurable sets. (Such functions are oftenreferred to as simple functions.) The sum means the function has value x_{k}on set A_{k}.Then f is measurable.
Proof: Letting U be open, f^{−1}
(U )
= ∪
{Ak : xk ∈ U }
, a finite union of measurable
sets.
There is also a very interesting theorem due to Kuratowski [?] which is presented
next.
To summarize the proof, you get an increasing sequence of 2^{−n} nets C_{n} and you
obtain a corresponding sequence of simple functions
{sn}
such that
{sn (ω )}
_{n=1}^{∞} is a
Cauchy sequence, and the maximum value of x → ψ