will be a σ finite measure space and X will be a Banach space
which contains the values of either a function or a measure. The Banach space will be
either a real or a complex Banach space but the field of scalars does not matter and so it
is denoted by F with the understanding that F = ℂ unless otherwise stated. The theory
presented here includes the case where X = ℝ^{n} or ℂ^{n} but it does not include the
situation where f could have values in a space like [0,∞]. To begin with here is a
definition.
Definition 19.1.1A function, x : Ω → X, for X a Banach space, is a simple functionif it is of the form
n
x(s) = ∑ a X (s)
i=1 i Bi
where B_{i}∈S and μ
(Bi)
< ∞ for each i. A function x from Ω to X is saidto be strongly measurableif there exists a sequence of simple functions {x_{n}}converging pointwise to x. The function x is said to be weakly measurableif, for eachf ∈ X^{′},
f ∘x
is a scalar valued measurable function.
Earlier, a function was measurable if inverse images of open sets were measurable.
Something similar holds here. The difference is that another condition needs to
hold.
Theorem 19.1.2x is strongly measurable if and only if x^{−1}
(U)
is measurablefor all U open in X and x
(Ω )
is separable.
Proof:Suppose first x^{−1}
(U)
is measurable for all U open in X and x
(Ω )
is
separable. Let {a_{n}}_{n=1}^{∞} be the dense subset of x
(Ω)
. It follows x^{−1}
(B)
is measurable
for all B Borel because
{B : x −1(B) is measurable}
is a σ algebra containing the open sets. Let
U nk ≡ {z ∈ X : ||z − ak|| ≤ min{{||z − al||}nl=1 }.
In words, U_{k}^{m} is the set of points of X which are as close to a_{k} as they are to any of the
a_{l} for l ≤ n.
Bn ≡ x−1(Un),Dn ≡ Bn ∖ (∪k−1Bn) ,Dn ≡ Bn,
k k k k i=1 i 1 1
and
∑n
xn (s) ≡ akXDnk (s).
k=1
Thus x_{n}
(s)
is a closest approximation to x
(s)
from
{ak}
_{k=1}^{n} and so x_{n}
(s)
→ x
(s)
because {a_{n}}_{n=1}^{∞} is dense in x
(Ω)
. Furthermore, x_{n} is measurable because each D_{k}^{n} is
measurable.
Since
(Ω,S, μ)
is σ finite, there exists Ω_{n}↑ Ω with μ
(Ωn )
< ∞. Let
yn (s) ≡ XΩn (s)xn (s).
Then y_{n}
(s)
→ x
(s)
for each s because for any s, s ∈ Ω_{n} if n is large enough. Also y_{n }is a
simple function because it equals 0 off a set of finite measure.
Now suppose that x is strongly measurable. Then some sequence of simple functions,
{x_{n}}, converges pointwise to x. Then x_{n}^{−1}
(W )
is measurable for every open set W
because it is just a finite union of measurable sets. Thus, x_{n}^{−1}
(W )
is measurable for
every Borel set W. This follows by considering
{ }
W : x−n 1(W ) is measurable
and observing this is a σ algebra which contains the open sets. Since X is a metric space,
it follows that if U is an open set in X , there exists a sequence of open sets, {V_{n}} which
satisfies
-- -- ∞
V n ⊆ U,V n ⊆ Vn+1,U = ∪ n=1Vn.
Then
⋃ ⋂ --
x−1(Vm ) ⊆ x−k 1(Vm) ⊆ x− 1(Vm ).
n<∞ k≥n
This implies
⋃
x−1(U) = x−1(Vm )
m<∞
⋃ ⋃ ⋂ −1 ⋃ −1(-- ) −1
⊆ xk (Vm) ⊆ x V m ⊆ x (U) .
m< ∞n< ∞k≥n m<∞
Since
−1 ⋃ ⋃ ⋂ −1
x (U) = xk (Vm ),
m<∞ n<∞ k≥n
it follows that x^{−1}
(U)
is measurable for every open U. It remains to show x
(Ω)
is
separable. Let
D ≡ all values of the simple functions xn
which converge to x pointwise. Then D is clearly countable and dense in D, a set which
contains x
(Ω)
.
Claim: x
(Ω )
is separable.
Proof of claim:For n ∈ ℕ, let ℬ_{n}≡
{B (d,r) : 0 < r < 1, r rational, d ∈ D }
n
.
Thus ℬ_{n} is countable. Let z ∈D. Consider B
(z, 1)
n
. Then there exists d ∈ D ∩B
(z,-1)
3n
.
Now pick r ∈ ℚ ∩
(1- 1)
3n ,n
so that B
(d,r)
∈ℬ_{n}. Now z ∈ B
(d,r)
and so this shows that
x
(Ω )
⊆D⊆∪ℬ_{n} for each n. Now let ℬ_{n}^{′} denote those sets of ℬ_{n} which have nonempty
intersection with x
(Ω)
. Say ℬ_{n}^{′} =
n
{Bk}
_{n,k=1}^{∞}. By the axiom of choice, there exists
x_{k}^{n}∈ B_{k}^{n}∩ x
(Ω )
. Then if z ∈ x
(Ω )
,z is contained in some set of ℬ_{n}^{′} which also
contains a point of
n
{xk}
_{n,k=1}^{∞}. Therefore, z is at least as close as 2∕n to some point of
n
{xk}
_{n,k=1}^{∞} which shows
n
{xk}
_{n,k=1}^{∞} is a countable dense subset of x
(Ω)
. Therefore
x
(Ω )
is separable. ■
The last part also shows that a subset of a separable metric space is also separable.
Therefore, the following simple corollary is obtained.
Corollary 19.1.3If X is a separable Banach space then x is strongly measurableif and only if x^{−1}
(U)
is measurable for all U open in X.
The next lemma is interesting for its own sake. Roughly it says that if a Banach space
is separable, then the unit ball in the dual space is weak ∗ separable. This will be used to
prove Pettis’s theorem, one of the major theorems in this subject which relates weak
measurability to strong measurability.
Lemma 19.1.4If X is a separable Banach space with B^{′}the closed unit ball in X^{′},then there exists a sequence {f_{n}}_{n=1}^{∞}≡ D^{′}⊆ B^{′}with the property that for everyx ∈ X,
||x|| = sup|f (x)|
f∈D ′
If H is a dense subset of X^{′}then D^{′}may be chosen to be contained in H.
Proof:Let {a_{k}} be a countable dense set in X, and consider the mapping
ϕn : B ′ → Fn
given by
ϕ (f) ≡ (f (a) ,⋅⋅⋅,f (a )).
n 1 n
Then ϕ_{n}
(B ′)
is contained in a compact subset of F^{n} because
|f (ak)|
≤
||ak||
.
Therefore, there exists a countable dense subset of ϕ_{n}
(B ′)
,
{ϕn (fk)}
_{k=1}^{∞}.Then pick
h_{j}^{k}∈ H ∩B^{′} such that lim_{j→∞}
||||fk − hk||||
j
= 0. Then
{ϕn (hk) ,k,j}
j
must also be dense
in ϕ_{n}
(B′)
. Let D_{n}^{′} =
{hk ,k,j}
j
. Define
D ′ ≡ ∪∞ D ′.
k=1 k
Note that for each x ∈ X, there exists f_{x}∈ B^{′} such that f_{x}
(x)
=
||x||
. From the
construction,
||a || = sup{|f (a )| : f ∈ D′}
m m
because f_{am}
(am)
is the limit of numbers f
(am )
for f ∈ D_{m}^{′}⊆ D^{′}. Therefore, for x
arbitrary,
||x || ≤ ||x− a ||+ ||a || = sup{|f (a )| : f ∈ D ′}+ ||x − a ||
m m m ′ m
≤ sup{|f (am − x)+ f (x)| : f ∈ D } + ||x− am ||
≤ sup{|f (x)| : f ∈ D ′} + 2||x− am || ≤ ||x ||+ 2||x − am||.
Since a_{m} is arbitrary and the
{am}
_{m=1}^{∞} are dense, this establishes the claim of the
lemma. ■
The next theorem is one of the most important results in the subject. It is due to
Pettis and appeared in 1938.
Theorem 19.1.5If x has values in a separable Banach space X. Then x is weaklymeasurable if and only if x is strongly measurable.
Proof: It is necessary to show x^{−1}
(U )
is measurable whenever U is open. Since every
open set is a countable union of balls, it suffices to show x^{−1}
(B (a,r))
is measurable
for any ball, B
(a,r)
. Since every open ball is the countable union of closed
balls, it suffices to verify x^{−1}
which equals a countable union of measurable sets because it is assumed that f ∘ x is
measurable for all f ∈ X^{′}.
Next suppose x is strongly measurable. Then there exists a sequence of simple
functions x_{n} which converges to x pointwise. Hence for all f ∈ X^{′}, f ∘ x_{n} is measurable
and f ∘ x_{n}→ f ∘ x pointwise. Thus x is weakly measurable. ■
The same method of proof yields the following interesting corollary.
Corollary 19.1.6Let X be a separable Banach space and let ℬ
(X )
denote theσ algebra of Borel sets. Let H be a dense subset of X^{′}. Then ℬ
(X )
= σ
(H )
≡ℱ,the smallest σ algebra of subsets of X which has the property that every function,x^{∗}∈ H is measurable.
Proof: First I need to show ℱ contains open balls because then ℱ will contain the
open sets and hence the Borel sets. As noted above, it suffices to show ℱ contains
closed balls. Let D^{′} be those functionals in B^{′} defined in Lemma 19.1.4. Then
which is measurable because this is a countable intersection of measurable sets. Thus ℱ
contains open sets so σ
(H)
≡ℱ⊇ℬ
(X )
.
To show the other direction for the inclusion, note that each x^{∗} is ℬ
(X)
measurable because x^{∗−1}
(open set)
= open set. Therefore, ℬ
(X)
⊇ σ
(H )
.■
It is important to verify the limit of strongly measurable functions is itself strongly
measurable. This happens under very general conditions. Suppose X is any
separable metric space and let τ denote the open sets of X. Then it is routine to see
that
τ has a countable basis, ℬ. (19.1.1)
(19.1.1)
Whenever U ∈ℬ, there exists a sequence of open sets, {V_{m}}_{m=1}^{∞}, such that
Theorem 19.1.7Let f_{n}and fbe functions mapping Ω to X where ℱ is a σalgebra of measurable sets of Ω and (X,τ) is a topological space satisfying 19.1.1-19.1.2. Then if f_{n}is measurable, and f(ω) = lim_{n→∞}f_{n}(ω), it follows that f isalso measurable. (Pointwise limits of measurable functions are measurable.)
Proof: Let ℬbe the countable basis of 19.1.1 and let U ∈ℬ.Let {V_{m}}be the sequence
of 19.1.2. Since fis the pointwise limit of f_{n},
---
f−1(Vm) ⊆ {ω : fk(ω) ∈ Vm for all k large enough} ⊆ f−1(Vm).
It follows f^{−1}(U) ∈ℱbecause it equals the expression in the middle which is
measurable. Now let W ∈ τ. Since ℬis countable, W = ∪_{n=1}^{∞}U_{n}for some sets
U_{n}∈ℬ.Hence
f−1(W ) = ∪∞n=1f− 1(Un) ∈ ℱ.■
Note that the same conclusion would hold for any topological space with the
property that for any open set U, it has such a sequence of V_{k} attached to it as in
19.1.2.
Corollary 19.1.8x is strongly measurable if and only if x
(Ω)
is separable and xis weakly measurable.
Proof:Strong measurability clearly implies weak measurability. If x_{n}
(s)
→ x
(s)
where x_{n} is simple, then f
(xn (s))
→ f
(x(s))
for all f ∈ X^{′}. Hence f ∘ x is measurable
by Theorem 19.1.7 because it is the limit of a sequence of measurable functions. Let D
denote the set of all values of x_{n}. Then D is a separable set containing x
(Ω)
. Thus D is a
separable metric space. Therefore x
(Ω)
is separable also by the last part of the proof of
Theorem 19.1.2.
Now suppose D is a countable dense subset of x
(Ω)
and x is weakly measurable. Let
Z be the subset consisting of all finite linear combinations of Dwith the scalars coming
from the set of rational points of F. Thus, Z is countable. Letting Y =Z, Y is a
separable Banach space containing x
(Ω)
. If f ∈ Y^{′}, f can be extended to an element
of X^{′} by the Hahn Banach theorem. Therefore, x is a weakly measurable Y
valued function. Now use Theorem 19.1.5 to conclude x is strongly measurable.
■
Weakly measurable as defined above means s → x^{∗}
(x(s))
is measurable for every
x^{∗}∈ X^{′}. The next lemma ties this weak measurability to the usual version of
measurability in which a function is measurable when inverse images of open sets are
measurable.
Lemma 19.1.9Let X be a Banach space and let x :
(Ω,ℱ )
→ K ⊆ X where Kis weakly compact and X^{′}is separable. Then x is weakly measurable if and only ifx^{−1}
(U )
∈ℱ whenever U is a weakly open set.
Proof: By Corollary 15.5.9 on Page 1332, there exists a metric d, such that the
metric space topology with respect to d coincides with the weak topology. Since K is
compact, it follows that K is also separable. Hence it is completely separable and so there
exists a countable basis of open sets ℬ for the weak topology on K. It follows that if U is
any weakly open set, covered by basic sets of the form B_{A}
(x,r)
where A is a finite subset
of X^{′}, there exists a countable collection of these sets of the form B_{A}
(x,r)
which covers
U.
Suppose now that x is weakly measurable. To show x^{−1}
(U)
∈ℱ whenever U is
weakly open, it suffices to verify x^{−1}
(BA (z,r))
∈ℱ for any set, B_{A}
(z,r)
. Let
A =
∗ ∗
{x 1,⋅⋅⋅,xm}
. Then
x−1(B (z,r)) = {s ∈ Ω : ρ (x(s)− z) < r}
A { A }
≡ s ∈ Ω : max |x∗(x(s)− z)| < r
x∗∈A
= ∪mi=1{s ∈ Ω : |x∗i (x(s)− z)| < r}
= ∪mi=1{s ∈ Ω : |x∗i (x(s))− x∗i (z)| < r}
which is measurable because each x_{i}^{∗}∘ x is given to be measurable.
Next suppose x^{−1}
(U )
∈ℱ whenever U is weakly open. Then in particular this holds
when U = B_{x∗}
(z,r)
for arbitrary x^{∗}. Hence
{s ∈ Ω : x (s) ∈ Bx ∗ (z,r)} ∈ ℱ.
But this says the same as
∗ ∗
{s ∈ Ω : |x (x (s))− x (z)| < r} ∈ ℱ
Since x^{∗}
(z)
can be a completely arbitrary element of F, it follows x^{∗}∘ x is an F valued
measurable function. In other words, x is weakly measurable according to the former
definition. ■
One can also define weak ∗ measurability and prove a theorem just like the Pettis
theorem above. The next lemma is the analogue of Lemma 19.1.4.
Lemma 19.1.10Let B be the closed unit ball in X. If X^{′}is separable, there exists asequence {x_{m}}_{m=1}^{∞}≡ D ⊆ B with the property that for all y^{∗}∈ X^{′},
∗ ∗
||y || = xsu∈pD |y (x)|.
Proof:Let
{x∗k}∞k=1
be the dense subspace of X^{′}. Define ϕ_{n} : B → F^{n} by
ϕn (x) ≡ (x∗1 (x) ,⋅⋅⋅,x∗n(x)).
Then
|x∗(x)|
k
≤
||x∗||
k
and so ϕ_{n}
(B )
is contained in a compact subset of F^{n}.
Therefore, there exists a countable set, D_{n}⊆ B such that ϕ_{n}