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 7.1.1Let
(Ω,ℱ )
be a measurable space, one for which ℱ is a σalgebra contained in P
(Ω)
. Let f : Ω → X where X is a metric space. Then f ismeasurable means that f−1
(U)
∈ℱ whenever 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 7.1.2Let
{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 alsothat f
(ω )
= limn→∞fn
(ω)
for all ω. Then f is also ameasurable function.
Proof:It is required to show f−1
(U)
is measurable for all U open. Let
{ }
( C) 1-
Vm ≡ x ∈ U : dist x,U > m .
Thus, since dist is continuous,
{ }
V ⊆ x ∈ U : dist(x,UC) ≥ 1-
m m
and Vm⊆Vm⊆ Vm+1 and ∪mVm = U. Then since Vm is open,
f− 1(Vm ) = ∪ ∞n=1 ∩∞k=n f−k1(Vm )
and so
−1 ∞ −1
f (U ) = ∪m=1f (Vm)
= ∪∞m=1 ∪∞n=1 ∩∞k=nf−k1(Vm )
⊆ ∪∞ f −1(Vm) = f−1(U )
m=1
which shows f−1
(U )
is measurable. ■
An important example of a metric space is of course ℝ, ℂ, Fn, 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 7.1.3The open balls in (−∞,∞] consist of sets of the form
(a,b)
for a,breal numbers and (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
(− ∞ )
≡−
π
2
,
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 7.1.4Let f : Ω → (−∞,∞] where ℱ is a σ algebra of subsets of Ω. Here
(−∞,∞] is the metric space 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,
−1
f ((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 < ∞,
f−1((a,b)) = f−1((− ∞, b)) ∩f− 1((a,∞ ]) ∈ ℱ.
Finally, if the last condition holds,
f−1 ([d,∞ ]) = (∪ ∞k=1f −1((− k+ d,d)))C ∈ ℱ
and so the third condition holds. Therefore, all five conditions are equivalent.
Since (−∞,∞] is a separable metric space, it follows from Theorem 2.4.2 that every open
set U is a countable union of open intervals U = ∪kIk where Ik is of the form
(a,b)
or (a,∞]
and, as just shown if any of the equivalent conditions holds, then f−1
(U )
= ∪kf−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 7.1.5Let E ∈ℱ for ℱ a σ algebra. Then
{
X (ω ) ≡ 1 if ω ∈ E
E 0 if ω ∕∈ E
This is called the indicator function of the set E. Let s :
(Ω,ℱ)
→ ℝ. Then s is a simplefunction if it is of the form
∑n
s(ω) = ciXEi (ω )
i=1
where Ei∈ℱ and ci∈ ℝ, the Eibeing disjoint. Thus simple functions have finitely manyvalues and are measurable. In the next theorem, it will also be assumed that eachci≥ 0.
Each simple function is measurable. This is easily seen as follows. First of all, you can
assume the ci are distinct because if not, you could just replace those Ei which
correspond to a single value with their union. Then if you have any open interval
(a,b)
,
− 1
s ((a,b)) = ∪ {Ei : ci ∈ (a,b)}
and this is measurable because it is the finite union of measurable sets.
Theorem 7.1.6Let f ≥ 0 be measurable. Then there exists a sequence ofnonnegative simple functions {sn} satisfying
0 ≤ sn(ω) (7.1)
(7.1)
⋅⋅⋅sn(ω) ≤ sn+1(ω)⋅⋅⋅
f(ω) = lim s (ω) for all ω ∈ Ω. (7.2)
n→∞ n
(7.2)
If f is bounded, the convergence is actually uniform. Conversely, if f is nonnegative and isthe pointwise limit of such simple functions, then f is measurable.
Proof: Letting I ≡
{ω : f (ω ) = ∞ }
, define
∑2n k
tn(ω) = nXf−1([kn,k+n1))(ω)+ 2nXI(ω).
k=0
Then tn(ω) ≤ f(ω) for all ω and limn→∞tn(ω) = f(ω) for all ω. This is because tn
(ω)
= 2n
for ω ∈ I and if f
(ω)
∈ [0,
n
2n+1
), then
1
0 ≤ f (ω)− tn(ω) ≤-. (7.3)
n
(7.3)
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 {sn} satisfies 7.1-7.2. Also each sn has finitely many values and is
measurable. To see this, note that
s−1((a,∞]) = ∪n t−1((a,∞ ]) ∈ ℱ
n k=1 k
To verify the last claim, note that in this case the term 2nXI(ω) is not present and for n
large enough, 2n∕n is larger than all values of f. Therefore, for all n large enough, 7.3 holds
for all ω. Thus the convergence is uniform.
The last claim follows right away from Theorem 7.1.2. ■
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 Ek are disjoint measurable sets and the xk 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 xk on the set Ek.
Theorem 7.1.7Let
(Ω, ℱ)
be a measurable space and let f : Ω → X where
(X, d)
is a separable metric space. Then f is a measurable function if and onlyif there exists a sequence of simple functions,
{fn}
such that for each ω ∈ Ω andn ∈ ℕ,
d (fn (ω ),f (ω)) ≥ d(fn+1(ω),f (ω )) (7.4)
(7.4)
and
nli→m∞ d(fn(ω),f (ω)) = 0. (7.5)
(7.5)
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 Dn =
{xk}
k=1n. Let
{ }
A1 ≡ ω : d(x1,f (ω)) = min d(xk,f (ω))
k≤n
= ∩n {ω : d(x ,f (ω))− d(x ,f (ω)) ≥ 0}
k=1 k 1
That is, A1 is those ω such that f
(ω)
is approximated best out of Dn by x1. Why is this a
measurable set? It is because ω → d
(xk,f (ω ))
− d
(x1,f (ω ))
is a real valued measurable
function, being the composition of a continuous function, y → d
(xk,y)
− d
(x1,y)
and a
measurable function, ω → f
(ω)
. Next let
{ }
A ≡ ω∈∕A : d(x ,f (ω)) = min d(x ,f (ω))
2 1 2 k≤n k
and continue in this manner obtaining disjoint measurable sets,
{Ak}
k=1n such that for ω ∈Ak the best approximation to f
(ω)
from Dn is xk. Then
∑n
fn(ω) ≡ xkXAk (ω).
k=1
Note
min d (xk,f (ω)) ≤ mind (xk,f (ω ))
k≤n+1 k≤n
and so this verifies 7.4. It remains to verify 7.5.
Let ε > 0 be given and pick ω ∈ Ω. Then there exists xn∈ D such that d
(xn,f (ω))
< ε. It
follows from the construction that d
(fn (ω ),f (ω))
≤ d
(xn,f (ω))
< ε. This proves the first
half.
Now suppose the existence of the sequence of simple functions as described above. Each
fn is a measurable function because fn−1
(U )
= ∪
{Ak : xk ∈ U}
. Therefore, the conclusion
that f is measurable follows from Theorem 7.1.2 on Page 437. ■