Recall that a σ algebra is a collection of subsets of a set Ω which includes ∅,Ω, and is closed with respect to
countable unions and complements.
Definition 19.1.1Let
(Ω,ℱ )
be a measurable space, one for which ℱ is a σ algebra contained inP
(Ω)
. Let f : Ω → X where X is a metric space. Then f is measurable means that f^{−1}
(U )
∈ℱwhenever U is open.
It is important to have a theorem about pointwise limits of measurable functions, those with the
property that inverse images of open sets are measurable. The following is a fairly general such theorem
which holds in the situations to be considered in these notes.
Theorem 19.1.2Let
{fn}
be a sequence of measurable functions mapping Ω to
(X,d)
where
(X,d)
is a metric space and
(Ω,ℱ )
is a measureable space. Suppose also that f
(ω)
= lim_{n→∞}f_{n}
(ω)
forall ω. Then f is also ameasurable function.
Proof:It is required to show f^{−1}
(U)
is measurable for all U open. Let
{ ( ) }
Vm ≡ x ∈ U : dist x,U C > 1 .
m
Thus, since dist is continuous,
{ ( ) 1}
Vm ⊆ x ∈ U : dist x,UC ≥--
m
and V_{m}⊆V_{m}⊆ V_{m+1} and ∪_{m}V_{m} = U. Then since V_{m} is open,
f −1(V ) = ∪ ∞ ∩∞ f−1(V )
m n=1 k=n k m
and so
− 1 ∞ −1
f (U ) = ∪m∞=1f ∞ (Vm)∞ −1
= ∪m=1 ∪n=1(-∩k=)nfk (Vm )
⊆ ∪∞m=1f −1 Vm = f−1 (U )
which shows f^{−1}
(U )
is measurable. ■
An important example of a metric space is of course ℝ, ℂ, F^{n}, where F is either ℝ or ℂ and
so forth. However, it is also very convenient to consider the metric space (−∞,∞], the real
line with ∞ tacked on at the end. This can be considered as a metric space in a very simple
way.
ρ(x,y) = |arctan(x)− arctan (y)|
with the understanding that arctan
(∞ )
≡ π∕2. It is easy to show that this metric restricted to ℝ gives the
same open sets on ℝ as the usual metric given by d
(x,y)
=
|x− y|
but in addition, allows the inclusion of
that ideal point out at the end of the real line denoted as ∞. This is considered mainly because it makes
the development of the theory easier. The open sets in (−∞,∞] are described in the following
lemma.
Lemma 19.1.3The open balls in (−∞,∞] consist of sets of the form
(a,b)
for a,b real numbersand (a,∞]. This is a separable metric space.
Proof: If the center of the ball is a real number, then the ball will result in an interval
(a,b)
where a,b
are real numbers. If the center of the ball is ∞, then the ball results in something of the form (a,∞]. It is
obvious that this is a separable metric space with the countable dense set being ℚ since every ball contains
a rational number. ■
If you kept both −∞ and ∞ with the obvious generalization that arctan
(− ∞)
≡−
π2
, then the
resulting metric space would be a complete separable metric space. However, it is not convenient to include
−∞, so this won’t be done. The reason is that it will be desired to make sense of things like
f + g.
Then for functions which have values in (−∞,∞] we have the following extremely useful description of
what it means for a function to be measurable.
Lemma 19.1.4Let f : Ω → (−∞,∞] where ℱ is a σ algebra of subsets of Ω. Here (−∞,∞] is the metricspace just described with the metric given by
ρ(x,y) = |arctan(x)− arctan (y)|.
Then the following are equivalent.
f−1((d,∞]) ∈ ℱfor all finite d,
f− 1((− ∞, d)) ∈ ℱfor all finite d,
f− 1([d,∞ ]) ∈ ℱfor all finite d,
f− 1((− ∞, d]) ∈ ℱfor all finite d,
f−1 ((a,b)) ∈ ℱ for all a < b,− ∞ < a < b < ∞.
Any of these equivalent conditions is equivalent to the function being measurable.
Proof: First note that the first and the third are equivalent. To see this, observe
−1 ∞ − 1
f ([d,∞ ]) = ∩n=1f ((d− 1∕n,∞ ]),
and so if the first condition holds, then so does the third.
−1 ∞ −1
f ((d,∞]) = ∪n=1f ([d+ 1∕n,∞ ]),
and so if the third condition holds, so does the first.
Similarly, the second and fourth conditions are equivalent. Now
−1 −1 C
f ((− ∞, d]) = (f ((d,∞ ]))
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 < ∞,
−1 −1 − 1
f ((a,b)) = f ((− ∞, b)) ∩f ((a,∞ ]) ∈ ℱ.
Finally, if the last condition holds,
− 1 ( ∞ −1 )C
f ([d,∞ ]) = ∪k=1f ((− k +d,d)) ∈ ℱ
and so the third condition holds. Therefore, all five conditions are equivalent.
Since (−∞,∞] is a separable metric space, it follows from Theorem 11.1.29 that every open set U is a
countable union of open intervals U = ∪_{k}I_{k} where I_{k} is of the form
(a,b)
or (a,∞] and, as just shown if
any of the equivalent conditions holds, then f^{−1}
(U )
= ∪_{k}f^{−1}
(Ik)
∈ℱ. Conversely, if f^{−1}
(U )
∈ℱ for any
open set U ∈ (−∞,∞], then f^{−1}
((a,b))
∈ℱ which is one of the equivalent conditions and so all the
equivalent conditions hold. ■
There is a fundamental theorem about the relationship of simple functions to measurable functions
given in the next theorem.
Definition 19.1.5Let E ∈ℱ for ℱ a σ algebra. Then
{
1 if ω ∈ E
XE (ω ) ≡ 0 if ω ∕∈ E
This is called the indicator function of the set E. Let s :
(Ω, ℱ)
→ ℝ. Then s is a simple function if it is ofthe form
∑n
s (ω ) = ciXEi (ω)
i=1
where E_{i}∈ℱ and c_{i}∈ ℝ, the E_{i}being disjoint. Thus simple functions have finitely many values and aremeasurable. In the next theorem, it will also be assumed that each c_{i}≥ 0.
Each simple function is measurable. This is easily seen as follows. First of all, you can assume the c_{i} are
distinct because if not, you could just replace those E_{i} which correspond to a single value with their union.
Then if you have any open interval
(a,b)
,
s−1((a,b)) = ∪ {E : c ∈ (a,b)}
i i
and this is measurable because it is the finite union of measurable sets.
Theorem 19.1.6Let f ≥ 0 be measurable. Then there exists a sequence of nonnegative simple functions{s_{n}} satisfying
0 ≤ sn(ω) (19.1)
(19.1)
⋅⋅⋅sn(ω) ≤ sn+1(ω)⋅⋅⋅
f (ω ) = nli→m∞ sn(ω) for all ω ∈ Ω. (19.2)
(19.2)
If f is bounded, the convergence is actually uniform. Conversely, if f is nonnegative and is the pointwiselimit of such simple functions, then f is measurable.
Proof: Letting I ≡
{ω : f (ω ) = ∞ }
, define
2n
t (ω) = ∑ kX k+1 (ω)+ 2nX (ω ).
n k=0 n f−1([nk,n-)) I
Then t_{n}(ω) ≤ f(ω) for all ω and lim_{n→∞}t_{n}(ω) = f(ω) for all ω. This is because t_{n}
(ω)
= 2^{n} for ω ∈ I and
if f
(ω)
∈ [0,
2n+1
n
), then
0 ≤ f (ω)− t (ω) ≤ 1. (19.3)
n n
(19.3)
Thus whenever ω
∕∈
I, the above inequality will hold for all n large enough. Let
s1 = t1,s2 = max (t1,t2),s3 = max (t1,t2,t3),⋅⋅⋅.
Then the sequence {s_{n}} satisfies 19.1-19.2. Also each s_{n} has finitely many values and is measurable. To see
this, note that
−1 n −1
sn ((a,∞ ]) = ∪ k=1tk ((a,∞ ]) ∈ ℱ
To verify the last claim, note that in this case the term 2^{n}X_{I}(ω) is not present and for n large enough,
2^{n}∕n is larger than all values of f. Therefore, for all n large enough, 19.3 holds for all ω. Thus the
convergence is uniform.
The last claim follows right away from Theorem 19.1.2. ■
Although it is not needed here, there is a more general theorem which applies to measurable functions
which have values in a separable metric space. In this context, a simple function is one which is of the
form
∑m
xkXEk (ω)
k=1
where the E_{k} are disjoint measurable sets and the x_{k} are in X. I am abusing notation somewhat by using a
sum. You can’t add in a general metric space. The symbol means the function has value x_{k} on the set
E_{k}.
Theorem 19.1.7Let
(Ω, ℱ)
be a measurable space and let f : Ω → X where
(X, d)
is a separable metricspace. Then f is a measurable function if and only if there exists a sequence of simple functions,
{fn }
suchthat for each ω ∈ Ω and n ∈ ℕ,
d(fn(ω),f (ω)) ≥ d (fn+1 (ω ),f (ω)) (19.4)
(19.4)
and
nl→im∞ d(fn(ω),f (ω)) = 0. (19.5)
(19.5)
Proof: Let D =
{xk}
_{k=1}^{∞} be a countable dense subset of X. First suppose f is measurable. Then
since in a metric space every open set is the countable intersection of closed sets, it follows
f^{−1}
(closed set)
∈ℱ. Now let D_{n} =
{xk}
_{k=1}^{n}. Let
{ }
A ≡ ω : d(x ,f (ω)) = min d(x ,f (ω))
1 1 k≤n k
That is, A_{1} are those ω such that f
(ω )
is approximated best out of D_{n} by x_{1}. Why is this a
measurable set? It is because ω → d
(x,f (ω))
is a real valued measurable function, being the
composition of a continuous function, y → d