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 )
= lim_{n→∞}f_{n}
(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 V_{m}⊆V_{m}⊆ V_{m+1} and ∪_{m}V_{m} = U. Then since V_{m} 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, 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 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 = ∪_{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 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 E_{i}∈ℱ and c_{i}∈ ℝ, the E_{i}being disjoint. Thus simple functions have finitely manyvalues and are measurable. In the next theorem, it will also be assumed that eachc_{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)
,
− 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 {s_{n}} 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 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,
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 {s_{n}} satisfies 7.1-7.2. Also each s_{n} 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 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, 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 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 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 D_{n} =
{xk}
_{k=1}^{n}. 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, A_{1} is 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
(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=1}^{n} such that for ω ∈A_{k} the best approximation to f
(ω)
from D_{n} is x_{k}. 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 x_{n}∈ 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
f_{n} is a measurable function because f_{n}^{−1}
(U )
= ∪
{Ak : xk ∈ U}
. Therefore, the conclusion
that f is measurable follows from Theorem 7.1.2 on Page 437. ■