In this post, we present an overview of the theory of Bochner integration, a vector-valued generalization of the theory of Lebesgue integration. Specifically, we introduce an appropriate notion of measurability for functions on a measure space that take values in a Banach space and develop a basic theory of integration for these functions.
The reader is assumed to be familiar with basic real and functional analysis.
Throughout the post, we fix a measure space (Ω,Σ,μ) and a Banach space X.
1. Measurable Functions
In order to develop a theory of integration, we must first decide on a class of functions to be integrated. Since a Banach space always has plenty of bounded linear functionals, we could easily turn vector-valued functions into scalar-valued functions by letting linear functionals act on them.
Definition 1.f:Ω→X is weakly measurable if lf is measurable for each l∈X∗.
Is this a good notion of measurability? We expect, for example, to be able to approximate measurable functions with simple functions, just as in the theory of Lebesgue integration.
Definition 2.s:Ω→X is simple if it is of the form
s(t)=n=1∑Nxn1En(t),
where En∈Σ and xn∈X for all 1≤n≤N. f:Ω→X is strongly measurable if there exists a sequence of simple functions (sn)n=1∞ such that
n→∞lim∥sn(t)−f(t)∥X=0
pointwise μ-almost everywhere.
Unfortunately, not all weakly measurable functions are strongly measurable, as the following counterexample shows.
Example 3. As the image of a simple function is finite, the image of the pointwise limit of simple functions sn must necessarily be contained in the closure of the union of imsn, which is separable. Therefore, no weakly measurable function with non-separable image can be strongly measurable.
Now, if S is an uncountable set with the power set P(S) as its σ-algebra and X a Banach space, then every function f:S→X is weakly measurable, for the preimage of an arbitrary subset of X under f is a measuarble subset of S. If X is of the same cardinality as S, then we can pick a bijective f:S→X, which is then a weakly measurable map whose image is non-separable. It follows that f fails to be strongly measurable.
To construct a concrete example, we recall that the Fourier transform on T is an isometric isomorphism of L2(T) onto l2(Z). This, in particular, implies that L2(T) and l2(Z) have the same cardinality as sets. As
Parametrizing functions on T via t↦e2πit, we have a one-to-one correspondence between L∞(T) and L∞, whence
Card(L∞(T))=c.
It then follows that there is a bijection from [0,1] to L∞, whence we can produce functions [0,1]→L∞ that are weakly measurable not strongly measurable. □
The key obstruction that the above example presents is the size of its range, which we now rule out:
Definition 4.f:Ω→X is separably valued if imf is a separable subset of X, and μ-almost separably valued if there exists a null set N⊆Ω such that f(Ω∖N) is a separable subset of X.
It turns out that the lack of such a defect is enough to guarantee strong measurability.
Theorem 5 (Pettis measurability theorem). f:Ω→X is strongly measurable if and only if f is weakly measurable and μ-almost separably valued.
We shall make use of the following lemma:
Lemma 6. Suppose that X is separable. If g:Ω→X is a weakly measurable function, then t↦∥g(t)∥ is measurable.
Proof of lemma. We fix a∈R and let
E={t:∥g(t)∥≤a}
El={t:∥lg(t)∥≤a} for each l∈X∗.
If we can find a sequence (ln)n=1∞ in X∗ such that
(7)E=n=1⋂∞Eln,
then the claim will follows from the weak measurability of f.
To establish (7), we first note that E⊆El whenever ∥l∥≤1, whence
E⊆∥l∥≤1⋂El.
On the other hand, the Hahn–Banach theorem furnishes a linear functional lt∈X∗ for each t∈Ω such that ∥lt∥=1 and lt(f(t))=∥f(t)∥, so that
E⊇∥l∥≤1⋂El.
We now appeal to the following claim:
Claim. There exists a sequence (lk)k=1∞ in BX∗={l∈X∗:∥l∥≤1} such that each l∈BX∗ has a corresponding subsequence (lkN)N=1∞ that converges pointwise to l on X.
This, in particular, furnishes a sequence (ln)n=1∞ in X∗ such that
∥l∥≤1⋂El=n=1⋂∞Eln,
which establishes (7).
It thus suffices to establish the claim. Let {xn}n∈N be a dense subset of X. The mapping
l↦φN(l)=(l(x1),…,l(xn))
sends BX∗ to l2N. Since l2N is separable, there exists a sequence of linear functionals (lN,k)k=1∞ in BX∗ such that {φN(lN,j)}k∈N is dense in φN(BX∗). Therefore, each l∈BX∗ furnishes a sequence (kN)N=1∞ of indices such that
The desired result now follows from (8) and the density of {xn}n∈N in X. □
Proof of Theorem 5.(⇒) Let us first assume that f is strongly measurable. We note that simple functions are evidently weakly measurable. Let (sn)n=1∞ be a sequence of simple functions on Ω such that sn→f pointwise almost everywhere. For each l∈X∗, the continuity of l shows that
n→∞limlsn(t)=l(n→∞limsn(t))=lf(t)
μ-almost everywhere. Since each lsn is measurable, the pointwise limit lf is also measurable, whence f is weakly measurable.
We now let N be the set on which sn does not converge pointwise to f. Observe that
f(Ω∖N)=n=1⋃∞sn(Ω∖N),
the closure of the union of the images of Ω∖N under sn. Since each sn(Ω∖N) is a finite set, the union is countable. It follows that the closure is separable, whence so is f(Ω∖N).
(⇐) Suppose now that f is weakly measurable and μ-almost separably valued. We assume without loss of generality that imf is separable. Moreover, we may as well assume that X=span(imf), so that X is separable.
We fix a dense subset {xn}n∈N of X. For each x∈X, we define k(N,y) to be the smallest integer 1≤k(N,y)≤N such that
∥y−xk(N,y)∥=1≤j≤Nmin∥y−xj∥.
By the density of {xN}n∈N,
N→∞lim∥y−xk(N,y)∥=0.
We set sN′=xk(N,y) so that sN′→y as N→∞.
We now define
sN=sN′∘f
for each N∈N, so that sN→f pointwise. Note that imsN⊆{x1,…,xN}, whence
sN=n=1∑Nxn1ENn,
where ENn={t∈Ω:sN(t)=xn}. Therefore, showing that f is strongly measurable amounts to proving that each ENn is a measurable subset of Ω. To this end, we observe that
are measurable subsets of Ω, whence so is ENn. This completes the proof. □
Many properties of measurable scalar-valued functions generalize to strongly measurable functions. Below, we collect a few that follow easily from the Pettis measurability theorem.
Corollary 9. If (fn)n=1∞ is a sequence of strongly measurable X-valued functions on Ω, then its pointwise limit f is also strongly measurable.
Proof. By the Pettis measurability theorem, each fn is weakly measurable and μ-almost separably valued. We assume without loss of generality that each imfn is a separable subset of X, so that X0=⋃nimfn is a separable Banach subspace of X such that imf⊆X0. It follows that f is separably valued. To see that f is weakly measurable, it suffices to note that lf is a pointwise limit of the measurable functions lfn for each fixed l∈X∗. The strong measurability of f now follows from the Pettis measurability theorem.
Corollary 10. If f:Ω→X is strongly measurable, and if φ:X→Y is norm-continuous, then φf:Ω→Y is strongly measurable.
Proof. By the strong measurability of f, there exists a sequence of simple functions (sn)n=1∞ such that sn→f pointwise. Each φ∘sn is evidently separably valued. Furthermore, if l∈Y∗, then lφ∈X∗. sn is weakly measurable by the Pettis measurability theorem, it follows that lφsn is measurable for all l∈Y∗. Therefore, φsn is weakly measurable, and so it is strongly measurable by the Pettis measurability theorem. Now, φsn→φf by the continuity of φ, whence by Corollary 9 φf is strongly measurable.
Corollary 11. A separably valued function f:Ω→X is strongly measurable if and only if f is σ-Borel measurable, viz., f−1(B) is a measurable subset of Ω whenever B is a Borel subset of X.
Proof.(⇐) If f is Σ-Borel measurable, then lf is measurable for all l∈X∗, whence f is weakly measurable. It now follows from the Pettis measurability theorem that f is strongly measurable.
(⇒) We assume without loss of generality that X=span(imf), so that X is separable. As Borel measurability is evidently equivalent to the measurability of the preimages of open subsets of X, it suffices to consider an arbitrary open subset U of X.
Given a simple function s=∑xn1En, we note that the preimage of any subset of X under s is a finite union of E1,…,EN, which is measurable. Therefore, if (sn)n=1∞ is a sequence of simple functions that converge pointwise almost everywhere to f, and if
(12)Ur={x∈U:d(x,X∖U)>r},
then sn−1(Ur) is measurable regardless of n and r.
Fix n∈N. If x∈⋂k≥nsn−1(Ur), then sn(x)∈Ur for all n≥k, whence f(x)∈Ur. By (12), Ur⊆Ur/2, and so f(x)∈Ur/2. Therefore, f−1(Ur/2)⊇⋂k≥nsn−1(Ur) for all n, whence
Note that the value of (16) does not depend on the choice of simple functions. Indeed, if (sn)n=1∞ and (sn′)n=1∞ are two different sequences of simple functions satisfying the above three conditions, then
for all simple functions s over sets of finite measure, it now follows from (17) that
n→∞lim∫Ωsndμ=n→∞limsn′dμ,
whence (16) is well-defined.
We also remark that Bochner integrable functions are automatically strongly measuradble. In general, however, not every strongly measurable functions are Bochner integrable. If the underlying measurable space (Ω,Σ,μ) is not σ-finite, then simple functions over sets of infinite measure are not necessarily pointwise limits of simple functions over sets of finite measure. This, in particular, implies that not every strongly measurable function on a non-σ-finite measure space is a pointwise limit of simple functions over sets of finite measure. As the sum (16) makes no sense for pointwise limits of simple functions over sets of infinite measure, we rule out such cases:
Definition 19.f:Ω→X is strongly μ-measurable if there exists a sequence of simple functions over sets of finite measure converging pointwise almost everywhere to f.
Note that every Bochner integrable function is strongly μ-measurable. We have already shown that not every strongly measurable function on a non-σ-finite measure space is strongly μ-measurable. This is not the case on a σ-finite measure space:
Proposition 20. Every strongly measurable function f on a σ-finite measure space is equal almost everywhere to a strongly μ-measurable function.
Proof. If sn→f, then snχEn→f, where (En)n=1∞ is an increasing sequence of finite-measure sets whose union is the whole space. □
We now proceed to establish basic properties of the Bochner integral.
Proposition 21. A strongly μ-measurable function f:Ω→X is Bochner integrable if and only if ∫Ω∥f∥dμ<∞. In this case, we have the inequality
∫Ωfdμ≤∫Ω∥f∥dμ.
Proof. We first note that the inequality follows immediately from (18).
(⇐) If f is Bochner integrable, then we can find a sequence of simple functions (sn)n=1∞ over sets of finite measure such that sn→f pointwise almost everywhere and ∫∥sn−f∥dμ→0. We see that
∫∥f∥dμ≤∫∥f−sn∥dμ+∫∥sn∥dμ
for each n∈N, and the right-hand side is finite for all large enough n.
(⇒) If f is strongly μ-measurable and if ∫∥f∥dμ<∞, then we can find a sequence (sn′)n=1∞ of simple functions over sets of finite measure such that sn′→f pointwise almost everywhere. We set
sn=(1∥sn′∥≤2∥f∥)sn′
for each n, so that (sn)n=1∞ is a sequence of simple functions over sets of finite measure such that sn→f pointwise almost everywhere. Since ∥sn−f∥≤3∥f∥ for all n, we invoke the dominated convergence theorem to conclude that
n→∞lim∥sn−f∥dμ=0,
as was to be shown. □
In light of the above proposition, we make the following definition:
Definition 22. The Lebesgue–Bochner space of order p, 1≤p<∞, is the space of all strongly μ-measurable functions f:Ω→X with the condition
∥f∥Lp(X)=(∫Ω∥f∥p)1/p<∞,
quotiented out by the almost-everywhere equivalence relation. The Lebesgue–Bochner space of order σ is the space of all strongly measurable functions f:Ω→X with the condition
∥f∥L∞(X)=inf{r≥0:μ({t∈Ω:∥f(t)∥>r})=0}<∞,
modded out by the almost-everywhere equivalence relation. We write
Lp(X)=Lp(μ;X)=Lp(Ω,Σ,μ;X)
to denote the Lebesgue–Bochner space of order p.
Because we cannot make sense of liminf or monotone convergence of vector-valued functions, there are no analogues of Fatou’s lemma or the monotone convergence theorem. We can, on the other hand, generalize the dominated convergence theorem:
Proposition 23 (Dominated convergence theorem). Fix 1≤p<∞, and let (fn)n=1∞ be a sequence in Lp(X). If fn→f pointwise almost everywhere and if there exists a g∈Lp(Ω) such that ∥fn∥≤∣g∣ almost everywhere for all n, then f∈Lp(X), and
n→∞lim∥fn−f∥Lp(X)=0.
Proof. Since ∥fn−f∥≤2∥g∥ almost everywhere for all n∈N, the scalar-value dominated convergence theorem implies that
n→∞lim∥fn−f∥Lp(X)=n→∞lim∥∥fn−f∥X∥p=0,
as was to be shown. □
As in the scalar-valued case, the dominated convergence theorem can be used to establish the completeness of Lp(X).
Proposition 24 (Riesz–Fischer). The Lebesgue–Bochner space Lp(X) is a Banach space for all 1≤p≤∞.
Proof. It is clear that ∥⋅∥Lp(X) is a norm.
We fix a Cauchy sequence (fn)n=1∞ in Lp(X) and pick a subsequence (fnk)k=1∞ such that ∥fnk+1−fnk∥Lp(X)≤2−k for all k. Define
Therefore, sK(g)∈Lp(Ω) for all K. The scalar-valued monotone convergence theorem now implies that g is well-defined and is in Lp(Ω). As
(25)∥SK(f)(t)∥X≤∣SK(g)(t)∣≤∣g(t)∣
for each K∈N and every t∈Ω, the vector-valued dominated convergence theorem implies that f is well-defined and is in Lp(X).
It remains to show that fn→f in Lp(X) as n→∞. To this end, we first note that (25) implies
∥f(t)∥X≤∣g(t)∣
for almost every t∈Ω. Moreover, the triangle inequality implies that
∥f(t)−SK(f)(t)∥X≤∥f(t)∥X+∥SK(f)(t)∥X≤2∣g(t)∣
for almost every t∈Ω, whence we invoke the vector-valued dominated convergence theorem to conclude that fnk→f in Lp(X) as K→∞. The desired result now follows from the Cauchy hypothesis on (fn)n=1∞.
In general, most of the basic result from the theory of Lebesgue integration generalize to the theory of Bochner integration, so long as they do not involve the ordering of the real numbers.
Hölder’s inequality does not make sense in general, as a generic Banach space X lacks a multiplication operation. The inequality nevertheless generalizes quite nicely for Banach spaces with a continuous multiplication operation.
Definition 25. A Banach space X is a Banach algebra if X is a Banach space with an associative multiplication operation that satisfies the inequaltiy
∥xy∥≤∥x∥∥y∥
for all x,y∈X.
Proposition 26 (Hölder’s inequality). If X is a Banach algebra, then
∥fg∥Lr(X)≤∥f∥Lp(X)∥g∥Lq(X)
whenever 0<p,q,r≤∞, p1+q1=r1, f∈Lp(X), and g∈Lq(X).
Proof. Since X is a Banach algebra, we see that
∥fg∥Lr(X)=∥∥fg∥X∥r≤∥∥f∥X∥g∥X∥r.
The scalar-valued Hölder’s inequality implies that