The main content of this section is the pointwise approximation of a measurable map taking values in a finite-dimensional Banach space by a suitable sequence of simple maps. As we will see, in one dimension, this reduces to the classical result that every non-negative measurable function is the pointwise limit of an increasing sequence of non-negative simple functions (see for instance Theorem 11.6 in [2]).
We start with a measurable space (Ω,F ) and a metric space (E, %). The Borel
σ-field of E is denoted by B. In our setting, a map f : Ω → E is F -measurable if f−1(B) ∈F for all B ∈ B. Since the set of all B ∈ B such that f−1(B) ∈ F is a
σ-field in E and the topology of E generates B, it follows that f is F -measurable
if and only if f−1(O) ∈ F for each open set O in E. Let us consider two results on measurability, stated as Lemma 8.1.9 and Proposition 8.1.10 in [5]. First, f is F -measurable if and only if
ϕ ◦ f isF -measurable for all ϕ ∈ C(E, R+).
Since every R+-valued continuous function on E is Borel measurable, we only have
to check the if direction. To this end, let O be an open set in E. Then Lemma A.13 yields that ϕ : E → R+ defined by ϕ(z) := dist(z, Oc) is Lipschitz continuous and
satisfies O = {z ∈ E | ϕ(z) > 0} = ϕ−1((0, ∞)). Hence, f−1(O) = f−1(ϕ−1((0, ∞))) = (ϕ ◦ f )−1((0, ∞)) ∈F , as desired. The second result deals with pointwise limits of sequences of measurable maps.
A.18 Lemma. Let (fn)n∈N be a sequence of E-valued F -measurable maps on Ω
that converges pointwise to some map f : Ω → E, then f is F -measurable.
Proof. By the preceding discussion, it is enough to show that ϕ ◦ f is F -measurable
for each ϕ ∈ C(E, R+). As fnisF -measurable, so is ϕ◦fnfor all n ∈ N. Continuity
of ϕ entails that the sequence (ϕ ◦ fn)n∈N of R+-valued F -measurable functions on
Ω converges pointwise to ϕ ◦ f . Hence, ϕ ◦ f is F -measurable.
From here on, let E be a finite-dimensional linear space and ρ be induced by a complete norm k · k on E. That is, %(z, z0) = kz − z0k for each z, z0 ∈ E. We
set k := dim(E), then there is an isomorphism φ : E → Rk, which is necessarily
bimeausurable in the sense that φ and its inverse φ−1 are Borel measurable.
A.19 Lemma. A map f : Ω → E is F -measurable if and only if the i-th coordinate
function φi◦ f of the map φ ◦ f : Ω → Rk is F -measurable for all i ∈ {1, . . . , k}.
Proof. The only if direction is valid, since the composition of two measurable maps
is measurable. For if it suffices to show that φ ◦ f is F -measurable, because from
f = φ−1◦ (φ ◦ f ) theF -measurability of f follows. We note that (φ ◦ f )−1(B) = (φ1◦ f )−1(B1) ∩ · · · ∩ (φk◦ f )−1(Bk) ∈F
for each B ∈ B(Rk) of the form B = B
1 × · · · × Bk for some B1, . . . , Bk ∈ B(R).
Since the set of all B ∈ B(Rk) with (φ ◦ f )−1(B) ∈ F is a σ-field in Rk and
×k
We recall that a map f : Ω → E is F -simple if it is F -measurable and takes finitely many values. In this case, there exists m ∈ N such that z1, . . . , zm ∈ E are
the pairwise distinct values of f . The sets A1 := {f = z1}, . . . , Am := {f = zm}
belong to F and form a decomposition of Ω. That means, A1, . . . , Am are pairwise
disjoint andSm i=1Ai = Ω. Furthermore, f = m X i=1 zi1Ai. (A.4)
Conversely, assume that m ∈ N, z1, . . . , zm ∈ E, and A1, . . . , Am ∈F are such that
above representation holds. Then f is F -measurable and takes at most m pairwise distinct values. Hence, f is F -simple if and only if (A.4) is valid for some m ∈ N,
z1, . . . , zm ∈ E, and A1, . . . , Am ∈ F . Every representation (A.4) for f in which
the sets A1, . . . , Am form a decomposition of Ω is called normal. As we wish to
approximate every E-valued F -measurable map on Ω by an appropriate sequence of E-valued F -simple maps on Ω, we introduce set partitions.
A.20 Definition. Let C, D ∈ B be non-empty with C ⊂ D and (Cn)n∈N be an
increasing sequence in B with C1 6= ∅.
(i) A set partition of C is a countable system T of bounded and pairwise disjoint Borel sets in C with S
B∈TB = C. If C is compact, then we also require that
T has finitely many elements.
(ii) Let T be a set partition of C. Then |T| := supB∈Tdiam(B) is called the mesh
of T.
(iii) Let S and T be two set partitions of C and D, respectively. We say that T refines S if for each B ∈ S there are n ∈ N and pairwise distinct sets
B1, . . . , Bn ∈ T such thatSni=1Bi = B.
(iv) A refining sequence of set partitions of (Cn)n∈N is a sequence (Tn)n∈N, where
Tn is a set partition of Cn for all n ∈ N, such that Tn+1 refines Tn for each
n ∈ N and limn↑∞|Tn| = 0.
We justify the existence of set partitions and refining sequences of set partitions.
A.21 Lemma. Let C ∈ B be non-empty and δ > 0, then there is a set partition
T of C with |T| ≤ δ that has finitely many elements if C is bounded. Moreover,
for each increasing sequence (Cn)n∈N in B with C1 6= ∅ there is a refining sequence
(Tn)n∈N of set partitions of (Cn)n∈N.
Proof. Let first assume that C is bounded. SinceC is compact and {Bδ/2(z) | z ∈ C}
is an open covering of C, there are n ∈ N and pairwise distinct z1, . . . , zn ∈ C such
that Sn
i=1Bδ/2(zi) includes C while m
[
j=1
for each m ∈ {1, . . . , n − 1} and every i1, . . . , im ∈ {1, . . . , n} with i1 < · · · < im.
We let B1 := Bδ/2(z1) ∩ C and define recursively Bi := (Bδ/2(zi) ∩ C)\Bi−1 for all
i ∈ {2, . . . , n}. Then B1, . . . , Bnare non-empty and pairwise disjoint Borel sets in C
with Sn
i=1Bi =Sni=1(Bδ/2(zi) ∩ C) = C. Thus, T := {B1, . . . , Bn} is a set partition
of C and |T| ≤ supi∈{1,...,n}diam(Bδ/2(zi)) = δ.
Now, let C be unbounded. We define a sequence (νn)n∈N in N recursively by
ν0 := 0 and νn := min{m ∈ N | m > νn−1, m − 1 ≤ kzk < m for some z ∈ C} for
each n ∈ N, then (νn)n∈N is strictly increasing. Moreover, we set
Cn:= {z ∈ C | νn− 1 ≤ kzk < νn} for all n ∈ N.
Then (Cn)n∈N is a sequence of non-empty, bounded, and pairwise disjoint Borel sets
in C with S
n∈NCn = C. By what we have shown, for each n ∈ N there exists a set
partition Tn of Cn with finitely many elements such that |Tn| ≤ δ. Consequently,
T :=
[
n∈N
Tn
is a set partition of C. Clearly, T must be countable. In addition, for each A, B ∈ T there are m, n ∈ N such that A ∈ Tm and B ∈ Tn. If m = n, then A ∩ B = ∅,
as Tn is a set partition of Cn. Otherwise, it follows from A ⊂ Cm, B ⊂ Cn, and
Cm∩ Cn = ∅ that A ∩ B = ∅. Moreover, [ A∈T A = [ n∈N [ B∈Tn B = [ n∈N Cn= C.
In the same manner, |T| = supn∈NsupB∈Tndiam(B) = supn∈N|Tn| ≤ δ. Hence, the
first assertion is proven.
We turn to the second claim. Let (Cn)n∈N be an increasing sequence in B with
C1 6= ∅. We use the first assertion to construct a sequence of set partitions (Tn)n∈N
of (Cn)n∈N recursively as follows:
(i) Let T1 be a set partition of C1 with |T1| ≤ 1.
(ii) For n ∈ N suppose that Tn is a set partition of Cn with |Tn| ≤ 1/n. For each
B ∈ Tn we choose a set partition TB of B with |TB| ≤ 1/(n + 1) and let Sn+1
be a set partition of Cn+1\Cn with |Sn+1| ≤ 1/(n + 1) provided Cn+1\Cn 6= ∅,
otherwise let Sn+1 := ∅. Finally, we set Tn+1 :=
S
B∈TnTB
∪ Sn+1.
This yields the correct result. To see this, let n ∈ N, then Tn+1 contains only
bounded and pairwise disjoint Borel sets in Cn+1, by construction. This system
must be countable, and has finitely many elements whenever Cn+1 is bounded. We
also observe that
[ B∈Tn+1 = [ B∈Tn [ A∈TB A∪Cn+1\Cn) = Cn+1.
Finally, let B ∈ Tn and write TB = {A1, . . . , Am} for some m ∈ N and some
bounded and pairwise disjoint Borel sets A1, . . . , Am in B, then B =Smi=1Ai. Since
A1, . . . , Am ∈ Tn+1, this shows that Tn+1 refines Tn. Hence, we set |∅| := 0, then
A.22 Example. Let E = Rkand k · k be the Euclidean norm | · |, thenB = B(Rk).
We set Cn := [−n, n)k for each n ∈ N, which gives an increasing sequence (Cn)n∈N
of bounded sets in B(Rk) with S
n∈NCn = Rk, and for each compact set K in Rk
there is n ∈ N with K ⊂ Cn. We readily see that
max
z∈Cn
|z| =√kn = min
z∈Cn+1\Cn◦
|z| for all n ∈ N.
Furthermore, for each n ∈ N we define Tn to be the system of all sets B ⊂ Cn
that can be written in the form B = 2−n([i1, i1 + 1) × · · · × [ik, ik+ 1)) for some
i1, . . . , ik ∈ {−n2n, −n2n+ 1, . . . , n2n− 1}, then Tn is a set partition of Cn with
diam(B) =√k2−n for each B ∈ Tn.
The definition entails that Tn ⊂ Tn+1 for all n ∈ N. Hence, (Tn)n∈N is a refining
sequence of set partitions of (Cn)n∈N.
Set partitions allow for a local uniform approximation of the identity map on a closed set by a suitable sequence.
A.23 Proposition. Let D be a non-empty closed set in E. Then there is a sequence
(ϕn)n∈N of E-valued B-simple maps on E with ϕn(D) ⊂ D for all n ∈ N that
converges locally uniformly to the identity map E → E, z 7→ z on D such that
kϕn(z)k ≤ kϕn+1(z)k ≤ kϕ(z)k for all n ∈ N and each z ∈ D.
Proof. We choose an increasing sequence (Cn)n∈N of bounded Borel sets in D with
C1 6= ∅ satisfying the following two properties:
(i) S
n∈NCn= D and for each compact set K in D there is n ∈ N with K ⊂ Cn.
(ii) maxz∈Cnkzk ≤ minz∈C
n+1\Cn◦kzk for all n ∈ N.
For instance, we could choose c > 0 such that kzk ≤ c for at least one z ∈ D and let Cn = {z ∈ D | kzk ≤ cn} for each n ∈ N. By Lemma A.21, there is a
refining sequence of set partitions (Tn)n∈N of (Cn)n∈N. For each n ∈ N we choose
zn ∈ Cn such that kznk = maxz∈Cnkzk and for each B ∈ Tn we let zB ∈ B satisfy
kzBk = minz∈Bkzk. Then ϕn: E → E defined by
ϕn(z) :=
X
B∈Tn
zB1B(z) + zn1Cc n(z)
is B-simple and fulfills ϕn(D) ⊂ D. In fact, each set B ∈ Tn is Borel and Cnc ∈B,
which implies that ϕn is B-simple. Next, let z ∈ D. If z ∈ Cn, then there is a
unique B ∈ Tn with z ∈ B, which gives ϕn(z) = zB ∈ B. If instead z ∈ Cnc, then
ϕn(z) = zn ∈ Cn. Hence, from the fact that B ⊂ Cn ⊂ D for all B ∈ Tn we get
Now, let K be a compact set in D. Then (i) gives n0 ∈ N such that K ⊂ Cn for
all n ∈ N with n ≥ n0. We choose such an n ∈ N, then for each z ∈ K there is a
unique set B ∈ Tn with z ∈ B, which entails that
kϕn(z) − zk = kzB− zk ≤ diam(B) = diam(B) ≤ |Tn|.
So, limn↑∞supz∈Kkϕn(z)−zk = 0. As E is locally compact, this shows that (ϕn)n∈N
converges locally uniformly to the identity map E → E, z 7→ z on D.
Let us prove that kϕn(z)k ≤ kϕn+1(z)k for all n ∈ N and each z ∈ D. Assume
initially that z ∈ Cn, then there is a unique B ∈ Tn with z ∈ B. Since Tn+1 refines
Tn, there are a unique m ∈ N and unique pairwise distinct sets B1, . . . , Bm ∈ Tn+1
with B =Sm
i=1Bi. We choose the unique i ∈ {1, . . . , m} such that z ∈ Bi, then
kϕn(z)k = kzBk = min z0∈Bkz 0k ≤ min z0∈B i kz0k = kzBik = kϕn+1(z)k.
For z ∈ Cn+1\Cn there is a unique B ∈ Tn+1 with B ⊂ Cn+1\Cn and z ∈ B, because
Tn+1\Tn is a set partition of Cn+1\Cn. Thus, (ii) yields that
kϕn(z)k = kznk = max z0∈C n kz0k ≤ min Cn+1\C◦n kz0k ≤ min z0∈Bkz 0k = kz Bk = kϕn+1(z)k.
In the last case z /∈ Cn+1, we have that kϕn(z)k = kznk ≤ kzn+1k = kϕn+1(z)k.
Hence, the claim is established.
This gives the main result of this section.
A.24 Corollary. Let D be a non-empty closed set in E. Then to eachF -measurable
map f : Ω → D there is a sequence (fn)n∈N of D-valued F -simple maps on Ω that
converges pointwise to f such that
kfn(ω)k ≤ kfn+1(ω)k ≤ kf (ω)k for all n ∈ N and every ω ∈ Ω.
Moreover, if f is bounded, then the convergence is uniform.
Proof. Proposition A.23 yields a sequence (ϕn)n∈N of E-valued B-simple maps on
E with ϕn(D) ⊂ D for all n ∈ N that converges locally uniformly to the identity
map E → E, z 7→ z on D such that
kϕn(z)k ≤ kϕn+1(z)k ≤ kϕ(z)k for all n ∈ N and each z ∈ D.
We set fn:= ϕn◦ f for all n ∈ N. Then fn is a D-valuedF -simple map on Ω, since
f (Ω) ⊂ D. We also see that limn↑∞fn(ω) = limn↑∞ϕn(f (ω)) = f (ω) and
kfn(ω)k = kϕn(f (ω))k ≤ kϕn+1(f (ω))k = kfn+1(ω)k
for all n ∈ N and each ω ∈ Ω. To justify the second claim, let f be bounded. Then there is c ≥ 0 with kf (ω)k ≤ c for each ω ∈ Ω. Since K := {z ∈ D| kzk ≤ c} is a compact set in D and
sup
ω∈Ω
kfn(ω) − f (ω)k ≤ sup z∈K
kϕn(z) − zk
for each n ∈ N, we obtain that limn↑∞supω∈Ωkfn(ω) − f (ω)k = 0, which completes
The pointwise approximation of measurable maps by simple maps leads us to a classical statement (cf. Theorem 18 in [8, Section 1.2]).
A.25 Corollary. Let (Ω0,F0) be another measurable space and h : Ω → Ω0 be a
F -F0-measurable map. Then a F -measurable map f : Ω → E is measurable with
respect to σ(h) if and only if there is a F0-measurable map g : Ω0 → E such that
f (ω) = g(h(ω)) for all ω ∈ Ω.
Proof. As the composition of two measurable maps is measurable, the condition is
clearly sufficient. To prove its necessity, we first assume that f is F -simple. Then there are m ∈ N and pairwise distinct z1, . . . , zm ∈ E with f (Ω) = {z1, . . . , zm}.
The sets A1 := {f = z1}, . . . , Am := {f = zm} belong to F , form a decomposition
of Ω, and satisfy f = m X i=1 zi1Ai.
Because σ(h) = {h−1(A0) | A0 ∈ F0} ⊂ F and f is σ(h)-measurable, there must
exist A01, . . . , A0m ∈ F0 such that A
i = {h ∈ A0i} for all i ∈ {1, . . . , m}. Hence, the
map g := m X i=1 zi1A0i
is F0-simple and satisfies f (ω) = g(h(ω)) for each ω ∈ Ω, which follows directly from 1A0
i(h(ω)) = 1Ai(ω) for all i ∈ {1, . . . , m}. We turn to the general case. The
preceding corollary provides a sequence (fn)n∈N of E-valued σ(h)-simple maps on Ω
that converges pointwise to f and fulfills kfn(ω)k ≤ kfn+1(ω)k ≤ kf (ω)k for each
n ∈ N and all ω ∈ Ω. By what we have shown, for each n ∈ N there is a F0-simple map gn : Ω0 → E such that
fn(ω) = gn(h(ω)) for all ω ∈ Ω.
Let A0 be the set of all ω0 ∈ Ω0 for which the sequence (g
n(ω0))n∈N converges, then
A0 ∈F0 and h(Ω) ⊂ A0. For this reason, we can conclude that g : Ω0 → E defined
by g(ω0) := limn↑∞gn(ω0), if ω0 ∈ A0, and g(ω0) := 0, if ω0 ∈ A/ 0, is the demanded
map.