Bochner integration

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)(\Omega, \Sigma, \mu) and a Banach space XX.

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:ΩXf:\Omega \to X is weakly measurable if lflf is measurable for each lXl \in 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:ΩXs:\Omega \to X is simple if it is of the form

s(t)=n=1Nxn1En(t),s(t) = \sum_{n=1}^N x_n 1_{E_n}(t),

where EnΣE_n \in \Sigma and xnXx_n \in X for all 1nN1 \leq n \leq N. f:ΩXf:\Omega \to X is strongly measurable if there exists a sequence of simple functions (sn)n=1(s_n)_{n=1}^\infty such that

limnsn(t)f(t)X=0\lim_{n \to \infty} \lVert s_n(t) - f(t) \rVert_X = 0

pointwise μ\mu-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 sns_n must necessarily be contained in the closure of the union of imsn\operatorname{im} s_n, which is separable. Therefore, no weakly measurable function with non-separable image can be strongly measurable.

Now, if SS is an uncountable set with the power set P(S)\mathscr{P}(S) as its σ\sigma-algebra and XX a Banach space, then every function f:SXf:S \to X is weakly measurable, for the preimage of an arbitrary subset of XX under ff is a measuarble subset of SS. If XX is of the same cardinality as SS, then we can pick a bijective f:SXf:S \to X, which is then a weakly measurable map whose image is non-separable. It follows that ff fails to be strongly measurable.

To construct a concrete example, we recall that the Fourier transform on T\mathbb{T} is an isometric isomorphism of L2(T)L_2(\mathbb{T}) onto l2(Z)l_2(\mathbb{Z}). This, in particular, implies that L2(T)L_2(\mathbb{T}) and l2(Z)l_2(\mathbb{Z}) have the same cardinality as sets. As

Card(l2(Z))=Card(CZ)=c,\operatorname{Card}(l_2(\mathbb{Z})) = \operatorname{Card}(\mathbb{C}^{\mathbb{Z}}) = \mathfrak{c},

the space L2(T)L_2(\mathbb{T}) also has the cardinality of the continuum c\mathfrak{c}. The set inclusion L(T)L2(T)L_\infty(\mathbb{T}) \subseteq L_2(\mathbb{T}) implies that

Card(L(T))Card(L2(T))=c,\operatorname{Card}(L_\infty(\mathbb{T})) \leq \operatorname{Card}(L_2(\mathbb{T})) = \mathfrak{c},

and the injection tt1Tt \mapsto t 1_{\operatorname{T}} from R\mathbb{R} to L(T)L_\infty(\mathbb{T}) implies that

c=Card(T)Card(L(T)).\mathfrak{c} = \operatorname{Card}(\operatorname{T}) \leq \operatorname{Card}(L_\infty(\mathbb{T})).

By the Schröder–Bernstein theorem, it follows that

Card(L(T))=c.\operatorname{Card}(L_\infty(\operatorname{T})) = \mathfrak{c}.

Parametrizing functions on T\operatorname{T} via te2πitt \mapsto e^{2\pi i t}, we have a one-to-one correspondence between L(T)L_\infty(\mathbb{T}) and LL_\infty, whence

Card(L(T))=c.\operatorname{Card}(L_\infty(\mathbb{T})) = \mathfrak{c}.

It then follows that there is a bijection from [0,1][0,1] to LL_\infty, whence we can produce functions [0,1]L[0,1] \to L_\infty that are weakly measurable not strongly measurable. \square

The key obstruction that the above example presents is the size of its range, which we now rule out:

Definition 4. f:ΩXf: \Omega \to X is separably valued if imf\operatorname{im} f is a separable subset of XX, and μ\mu-almost separably valued if there exists a null set NΩN \subseteq \Omega such that f(ΩN)f(\Omega \smallsetminus N) is a separable subset of XX.

It turns out that the lack of such a defect is enough to guarantee strong measurability.

Theorem 5 (Pettis measurability theorem). f:ΩXf:\Omega \to X is strongly measurable if and only if ff is weakly measurable and μ\mu-almost separably valued.

We shall make use of the following lemma:

Lemma 6. Suppose that XX is separable. If g:ΩXg:\Omega \to X is a weakly measurable function, then tg(t)t \mapsto \lVert g(t) \rVert is measurable.

Proof of lemma. We fix aRa \in \mathbb{R} and let

If we can find a sequence (ln)n=1(l_n)_{n=1}^\infty in XX^* such that

(7)E=n=1Eln,(7) \hspace{3em} E = \bigcap_{n=1}^\infty E_{l_n},

then the claim will follows from the weak measurability of ff.

To establish (7), we first note that EElE \subseteq E_l whenever l1\lVert l \rVert \leq 1, whence

El1El.E \subseteq \bigcap_{\lVert l \rVert \leq 1} E_l.

On the other hand, the Hahn–Banach theorem furnishes a linear functional ltXl_t \in X^* for each tΩt \in \Omega such that lt=1\lVert l_t\rVert = 1 and lt(f(t))=f(t)l_t(f(t)) = \lVert f(t) \rVert, so that

El1El.E \supseteq \bigcap_{\lVert l \rVert \leq 1} E_l.

We now appeal to the following claim:

Claim. There exists a sequence (lk)k=1(l_k)^\infty_{k=1} in BX={lX:l1}B_{X^{*}} = \lbrace l \in X^{*} : \lVert l\rVert \leq 1\rbrace such that each lBXl \in B_{X^*} has a corresponding subsequence (lkN)N=1(l_{k_N})_{N=1}^\infty that converges pointwise to ll on XX.

This, in particular, furnishes a sequence (ln)n=1(l_n)_{n=1}^\infty in XX^* such that

l1El=n=1Eln,\bigcap_{\lVert l\rVert \leq 1} E_l = \bigcap^\infty_{n=1} E_{l_n},

which establishes (7).

It thus suffices to establish the claim. Let {xn}nN\lbrace x_n\rbrace_{n \in \mathbb{N}} be a dense subset of XX. The mapping

lφN(l)=(l(x1),,l(xn))l \mapsto \varphi_N(l) = (l(x_1), \ldots, l(x_n))

sends BXB_{X^*} to l2Nl_2^N. Since l2Nl_2^N is separable, there exists a sequence of linear functionals (lN,k)k=1(l_{N,k})^\infty_{k=1} in BXB_{X^*} such that {φN(lN,j)}kN\lbrace \varphi_N(l_{N,j})\rbrace_{k \in \mathbb{N}} is dense in φN(BX)\varphi_N(B_{X^*}). Therefore, each lBXl \in B_{X^*} furnishes a sequence (kN)N=1(k_N)_{N=1}^\infty of indices such that

φ(lN,kN)φ(l)l2N<1N.\lVert \varphi(l_{N,k_N}) - \varphi(l) \rVert_{l_2^N} < \frac{1}{N}.

This, in particular, implies that

lN,kN(xn)l(xn)<1N\vert l_{N, k_N}(x_n) - l(x_n) \vert < \frac{1}{N}

for all 1nN1 \leq n \leq N, whence

(8)limNlN,kN(xn)=l(xn)(8) \hspace{3em} \lim_{N \to \infty} l_{N,k_N}(x_n) = l(x_n)

for all nNn \in \mathbb{N}. Since

lN,kN(x)l(x)lN,kN(x)lN,kN(xn)+lN,kN(xn)l(xn)+l(xn)l(x).\begin{align*} \vert l_{N,k_N}(x) - l(x) \vert \leq & \vert l_{N,k_N}(x) - l_{N,k_N}(x_n) \vert \\ & + \vert l_{N,k_N}(x_n) - l(x_n) \vert \\ & + \vert l(x_n) - l(x) \vert. \end{align*}

The desired result now follows from (8) and the density of {xn}nN\lbrace x_n\rbrace_{n \in \mathbb{N}} in XX. \square

Proof of Theorem 5. ()(\Rightarrow) Let us first assume that ff is strongly measurable. We note that simple functions are evidently weakly measurable. Let (sn)n=1(s_n)_{n=1}^\infty be a sequence of simple functions on Ω\Omega such that snfs_n \to f pointwise almost everywhere. For each lXl \in X^*, the continuity of ll shows that

limnlsn(t)=l(limnsn(t))=lf(t)\lim_{n \to \infty} ls_n(t) = l\left(\lim_{n \to \infty} s_n(t)\right) = lf(t)

μ\mu-almost everywhere. Since each lsnls_n is measurable, the pointwise limit lflf is also measurable, whence ff is weakly measurable.

We now let NN be the set on which sns_n does not converge pointwise to ff. Observe that

f(ΩN)=n=1sn(ΩN),f(\Omega \smallsetminus N) = \overline{\bigcup_{n=1}^\infty s_n(\Omega \smallsetminus N)},

the closure of the union of the images of ΩN\Omega \smallsetminus N under sns_n. Since each sn(ΩN)s_n(\Omega \smallsetminus N) is a finite set, the union is countable. It follows that the closure is separable, whence so is f(ΩN)f(\Omega \smallsetminus N).

()(\Leftarrow) Suppose now that ff is weakly measurable and μ\mu-almost separably valued. We assume without loss of generality that imf\operatorname{im} f is separable. Moreover, we may as well assume that X=span(imf)X = \overline{\operatorname{span}(\operatorname{im} f)}, so that XX is separable.

We fix a dense subset {xn}nN\lbrace x_n\rbrace_{n \in \mathbb{N}} of XX. For each xXx \in X, we define k(N,y)k(N, y) to be the smallest integer 1k(N,y)N1 \leq k(N,y) \leq N such that

yxk(N,y)=min1jNyxj.\lVert y - x_{k(N,y)}\rVert = \min_{1 \leq j \leq N} \lVert y - x_j \rVert.

By the density of {xN}nN\lbrace x_N\rbrace_{n \in \mathbb{N}},

limNyxk(N,y)=0.\lim_{N \to \infty} \lVert y - x_{k(N,y)}\rVert = 0.

We set sN=xk(N,y)s_N' = x_{k(N,y)} so that sNys_N' \to y as NN \to \infty.

We now define

sN=sNfs_N = s_N' \circ f

for each NNN \in \mathbb{N}, so that sNfs_N \to f pointwise. Note that imsN{x1,,xN}\operatorname{im} s_N \subseteq \lbrace x_1,\ldots,x_N\rbrace, whence

sN=n=1Nxn1ENn,s_N = \sum_{n=1}^N x_n1_{E_N^n},

where ENn={tΩ:sN(t)=xn}E_N^n = \lbrace t \in \Omega : s_N(t) = x_n\rbrace. Therefore, showing that ff is strongly measurable amounts to proving that each ENnE_N^n is a measurable subset of Ω\Omega. To this end, we observe that

ENn={tΩ:xn=xk(N,f(t))}={tΩ:f(t)xn=min1jNf(t)xj}(p=1n1{tΩ:f(t)xp>min1jNf(t)xj}),\begin{align*} & E_N^n \\ =& \lbrace t \in \Omega : x_n = x_k(N,f(t))\rbrace \\ =& \left\lbrace t \in \Omega : \lVert f(t) - x_n\rVert = \min_{1 \leq j \leq N} \lVert f(t) - x_j\rVert\right\rbrace \\ &\cap \left( \bigcap_{p=1}^{n-1} \left\lbrace t \in \Omega : \lVert f(t) - x_p \rVert > \min_{1 \leq j \leq N} \lVert f(t) - x_j\rVert \right\rbrace \right), \end{align*}

as xn=xk(N,f(t))x_n = x_{k(N, f(t))} implies that nn is the first index among 1,,N1, \ldots, N to minimize the norm. We now let

which are measurable functions by Lemma 6. Therefore,

(ϕnφN)1({0})={tΩ:f(t)xn=min1jNf(t)xj}(\phi_n - \varphi_N)^{-1}(\lbrace 0\rbrace) = \left\lbrace t \in \Omega : \lVert f(t) - x_n\rVert = \min_{1 \leq j \leq N} \lVert f(t) - x_j\rVert \right\rbrace

and

(ϕnφN)1((0,))={tΩ:f(t)xn>min1jNf(t)xj}(\phi_n - \varphi_N)^{-1}((0, \infty)) = \left\lbrace t \in \Omega : \lVert f(t) - x_n\rVert > \min_{1 \leq j \leq N} \lVert f(t) - x_j\rVert \right\rbrace

are measurable subsets of Ω\Omega, whence so is ENnE_N^n. This completes the proof. \square

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(f_n)_{n=1}^\infty is a sequence of strongly measurable XX-valued functions on Ω\Omega, then its pointwise limit ff is also strongly measurable.

Proof. By the Pettis measurability theorem, each fnf_n is weakly measurable and μ\mu-almost separably valued. We assume without loss of generality that each imfn\operatorname{im} f_n is a separable subset of XX, so that X0=nimfnX_0 = \overline{\bigcup_n \operatorname{im} f_n} is a separable Banach subspace of XX such that imfX0\operatorname{im} f \subseteq X_0. It follows that ff is separably valued. To see that ff is weakly measurable, it suffices to note that lflf is a pointwise limit of the measurable functions lfnlf_n for each fixed lXl \in X^*. The strong measurability of ff now follows from the Pettis measurability theorem.

Corollary 10. If f:ΩXf:\Omega \to X is strongly measurable, and if φ:XY\varphi:X \to Y is norm-continuous, then φf:ΩY\varphi f:\Omega \to Y is strongly measurable.

Proof. By the strong measurability of ff, there exists a sequence of simple functions (sn)n=1(s_n)_{n=1}^\infty such that snfs_n \to f pointwise. Each φsn\varphi \circ s_n is evidently separably valued. Furthermore, if lYl \in Y^*, then lφXl \varphi \in X^*. sns_n is weakly measurable by the Pettis measurability theorem, it follows that lφsnl \varphi s_n is measurable for all lYl \in Y^*. Therefore, φsn\varphi s_n is weakly measurable, and so it is strongly measurable by the Pettis measurability theorem. Now, φsnφf\varphi s_n \to \varphi f by the continuity of φ\varphi, whence by Corollary 9 φf\varphi f is strongly measurable.

Corollary 11. A separably valued function f:ΩXf:\Omega \to X is strongly measurable if and only if ff is σ\sigma-Borel measurable, viz., f1(B)f^{-1}(B) is a measurable subset of Ω\Omega whenever BB is a Borel subset of XX.

Proof. ()(\Leftarrow) If ff is Σ\Sigma-Borel measurable, then lflf is measurable for all lXl \in X^*, whence ff is weakly measurable. It now follows from the Pettis measurability theorem that ff is strongly measurable.

()(\Rightarrow) We assume without loss of generality that X=span(imf)X = \overline{\operatorname{span}(\operatorname{im} f)}, so that XX is separable. As Borel measurability is evidently equivalent to the measurability of the preimages of open subsets of XX, it suffices to consider an arbitrary open subset UU of XX.

Given a simple function s=xn1Ens = \sum x_n 1_{E_n}, we note that the preimage of any subset of XX under ss is a finite union of E1,,ENE_1,\ldots,E_N, which is measurable. Therefore, if (sn)n=1(s_n)_{n=1}^\infty is a sequence of simple functions that converge pointwise almost everywhere to ff, and if

(12)Ur={xU:d(x,XU)>r},(12) \hspace{3em} U_r = \lbrace x \in U : d(x, X \smallsetminus U) > r\rbrace,

then sn1(Ur)s_n^{-1}(U_r) is measurable regardless of nn and rr.

Fix nNn \in \mathbb{N}. If xknsn1(Ur)x \in \bigcap_{k \geq n} s_n^{-1}(U_r), then sn(x)Urs_n(x) \in U_r for all nkn \geq k, whence f(x)Urf(x) \in \overline{U_r}. By (12), UrUr/2\overline{U_r} \subseteq U_{r/2}, and so f(x)Ur/2f(x) \in U_{r/2}. Therefore, f1(Ur/2)knsn1(Ur)f^{-1}(U_{r/2}) \supseteq \bigcap_{k \geq n} s_n^{-1}(U_r) for all nn, whence

f1(Ur/2)n1knsn1(Ur).f^{-1}(U_{r/2}) \supseteq \bigcup_{n \geq 1} \bigcap_{k \geq n} s_n^{-1}(U_r).

This implies that

(13)f1(U)=m1f1(U1/2m)m1n1knsn1(U1/m).(13) \hspace{3em} f^{-1}(U) = \bigcup_{m \geq 1} f^{-1}(U_{1/2m}) \supseteq \bigcup_{m \geq 1} \bigcup_{n \geq 1} \bigcap_{k \geq n} s_n^{-1}(U_{1/m}).

Conversely, if xf1(U1/m)x \in f^{-1}(U_{1/m}), then f(x)U1/mf(x) \in U_{1/m}. Since U1/mU_{1/m} is open, there exists an nNn \in \mathbb{N} such that sk(x)Us_k(x) \in U for all knk \geq n. Therefore,

f1(U1/m)n1knsn1(U1/m),f^{-1}(U_{1/m}) \subseteq \bigcup_{n \geq 1} \bigcap_{k \geq n} s_n^{-1}(U_{1/m}),

and it follows that

(14)f1(U)=m1f1(U1/m)m1n1knsn1(U1/m).(14) \hspace{3em} f^{-1}(U) = \bigcup_{m \geq 1} f^{-1}(U_{1/m}) \subseteq \bigcup_{m \geq 1} \bigcup_{n \geq 1} \bigcap_{k \geq n} s_n^{-1}(U_{1/m}).

Since each sn1(U1/m)s_n^{-1}(U_{1/m}) is measurable, we conclude from (13) and (14) that f1(U)f^{-1}(U) is measurable. This completes the proof. \square

2. The Bochner integral

Having studied the basic properties of measurable Banach-valued functions, we are now in a position to study the Bochner integral.

Definition 15. f:ΩXf:\Omega \to X is Bochner integrable if there exists a sequence (sn)n=1(s_n)_{n=1}^\infty of simple functions

sn=m=1Mnxm,n1Em,ns_n = \sum_{m=1}^{M_n} x_{m,n} 1_{E_{m,n}}

on Ω\Omega such that

  1. μ(Em,n)<\mu(E_{m,n}) < \infty for all mm and nn,
  2. snfs_n \to f almost everywhere, and
  3. Ωsnfdμ0\int_\Omega \lVert s_n - f \rVert \, d\mu \to 0.

In this case, we define the Bochner integral of ff to be

(16)Ωfdμ=limnΩsndμ=limnm=1Mnμ(Em,n)xm,n.(16) \hspace{3em} \begin{align*} \int_\Omega f \, d\mu &= \lim_{n \to \infty} \int_\Omega s_n \, d\mu \\ &= \lim_{n \to \infty} \sum_{m=1}^{M_n} \mu(E_{m,n})x_{m,n}. \end{align*}

Note that the value of (16) does not depend on the choice of simple functions. Indeed, if (sn)n=1(s_n)^\infty_{n=1} and (sn)n=1(s_n')^\infty_{n=1} are two different sequences of simple functions satisfying the above three conditions, then

(17)limnΩsnsnlimnΩsnfdμ+Ωfsndμ=0.(17) \hspace{3em} \begin{align*} & \lim_{n \to \infty} \int_\Omega \lVert s_n - s_n'\rVert \\ &\leq \lim_{n \to \infty} \int_\Omega \lVert s_n - f\rVert \, d\mu + \int_\Omega \lVert f - s_n'\rVert \, d\mu \\ &= 0. \end{align*}

Since

(18)Ωsdμ=m=1Mμ(Em)xmm=1Mμ(Em)xm=Ωsdμ(18) \hspace{3em} \begin{align*} \left\lVert \int_\Omega s \, d\mu \right\rVert &= \left\lVert \sum_{m=1}^M \mu(E_m) x_m \right\rVert \\ &\leq \sum_{m=1}^M \mu(E_m) \lVert x_m\rVert \\ &= \int_\Omega \lVert s\rVert \, d\mu \end{align*}

for all simple functions ss over sets of finite measure, it now follows from (17) that

limnΩsndμ=limnsndμ,\lim_{n \to \infty} \int_\Omega s_n \, d\mu = \lim_{n \to \infty} s_n' \, d\mu,

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)(\Omega, \Sigma, \mu) is not σ\sigma-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\sigma-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:ΩXf:\Omega \to X is strongly μ\mu-measurable if there exists a sequence of simple functions over sets of finite measure converging pointwise almost everywhere to ff.

Note that every Bochner integrable function is strongly μ\mu-measurable. We have already shown that not every strongly measurable function on a non-σ\sigma-finite measure space is strongly μ\mu-measurable. This is not the case on a σ\sigma-finite measure space:

Proposition 20. Every strongly measurable function ff on a σ\sigma-finite measure space is equal almost everywhere to a strongly μ\mu-measurable function.

Proof. If snfs_n \to f, then snχEnfs_n \chi_{E_n} \to f, where (En)n=1(E_n)_{n=1}^\infty is an increasing sequence of finite-measure sets whose union is the whole space. \square

We now proceed to establish basic properties of the Bochner integral.

Proposition 21. A strongly μ\mu-measurable function f:ΩXf:\Omega \to X is Bochner integrable if and only if Ωfdμ<\int_\Omega \lVert f \rVert \, d\mu < \infty. In this case, we have the inequality

ΩfdμΩfdμ.\left\lVert \int_\Omega f \, d\mu \right\rVert \leq \int_\Omega \lVert f\lVert \, d\mu.

Proof. We first note that the inequality follows immediately from (18).

()(\Leftarrow) If ff is Bochner integrable, then we can find a sequence of simple functions (sn)n=1(s_n)_{n=1}^\infty over sets of finite measure such that snfs_n \to f pointwise almost everywhere and snfdμ0\int \lVert s_n - f\rVert \, d\mu \to 0. We see that

fdμfsndμ+sndμ\int \lVert f\rVert \, d\mu \leq \int \lVert f-s_n\rVert \, d\mu + \int\lVert s_n\rVert\, d\mu

for each nNn \in \mathbb{N}, and the right-hand side is finite for all large enough nn.

()(\Rightarrow) If ff is strongly μ\mu-measurable and if fdμ<\int \lVert f \rVert \, d\mu < \infty, then we can find a sequence (sn)n=1(s_n')_{n=1}^\infty of simple functions over sets of finite measure such that snfs_n' \to f pointwise almost everywhere. We set

sn=(1sn2f)sns_n = \left(1_{\lVert s_n'\rVert \leq 2\lVert f\rVert}\right)s_n'

for each nn, so that (sn)n=1(s_n)_{n=1}^\infty is a sequence of simple functions over sets of finite measure such that snfs_n \to f pointwise almost everywhere. Since snf3f\lVert s_n - f\rVert \leq 3\lVert f\rVert for all nn, we invoke the dominated convergence theorem to conclude that

limnsnfdμ=0,\lim_{n \to \infty} \lVert s_n - f\rVert \, d\mu = 0,

as was to be shown. \square

In light of the above proposition, we make the following definition:

Definition 22. The Lebesgue–Bochner space of order pp, 1p<1 \leq p < \infty, is the space of all strongly μ\mu-measurable functions f:ΩXf:\Omega \to X with the condition

fLp(X)=(Ωfp)1/p<,\lVert f\rVert_{L_p(X)} = \left( \int_\Omega \lVert f\rVert^p \right)^{1/p} < \infty,

quotiented out by the almost-everywhere equivalence relation. The Lebesgue–Bochner space of order σ\sigma is the space of all strongly measurable functions f:ΩXf: \Omega \to X with the condition

fL(X)=inf{r0:μ({tΩ:f(t)>r})=0}<,\lVert f\rVert_{L_\infty(X)} = \inf\lbrace r \geq 0 : \mu(\lbrace t \in \Omega : \lVert f(t)\rVert > r\rbrace) = 0\rbrace < \infty,

modded out by the almost-everywhere equivalence relation. We write

Lp(X)=Lp(μ;X)=Lp(Ω,Σ,μ;X)L_p(X) = L_p(\mu;X) = L_p(\Omega,\Sigma,\mu;X)

to denote the Lebesgue–Bochner space of order pp.

Because we cannot make sense of lim inf\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 1p<1 \leq p < \infty, and let (fn)n=1(f_n)_{n=1}^\infty be a sequence in Lp(X)L_p(X). If fnff_n \to f pointwise almost everywhere and if there exists a gLp(Ω)g \in L_p(\Omega) such that fng\lVert f_n\rVert \leq |g| almost everywhere for all nn, then fLp(X)f \in L_p(X), and

limnfnfLp(X)=0.\lim_{n \to \infty} \lVert f_n - f\rVert_{L_p(X)} = 0.

Proof. Since fnf2g\lVert f_n - f\rVert \leq 2\lVert g \rVert almost everywhere for all nNn \in \mathbb{N}, the scalar-value dominated convergence theorem implies that

limnfnfLp(X)=limnfnfXp=0,\lim_{n \to \infty} \lVert f_n - f\rVert_{L_p(X)} = \lim_{n \to \infty} \left\lVert \lVert f_n - f\rVert_X \right\rVert_p = 0,

as was to be shown. \square

As in the scalar-valued case, the dominated convergence theorem can be used to establish the completeness of Lp(X)L_p(X).

Proposition 24 (Riesz–Fischer). The Lebesgue–Bochner space Lp(X)L_p(X) is a Banach space for all 1p1 \leq p \leq \infty.

Proof. It is clear that Lp(X)\lVert\cdot\rVert_{L_p(X)} is a norm.

We fix a Cauchy sequence (fn)n=1(f_n)^\infty_{n=1} in Lp(X)L_p(X) and pick a subsequence (fnk)k=1(f_{n_k})^\infty_{k=1} such that fnk+1fnkLp(X)2k\lVert f_{n_{k+1}} - f_{n_k}\rVert_{L_p(X)} \leq 2^{-k} for all kk. Define

f=fn1+k=1fnk+1fnkg+fn1X+k=1fnk+1fnkX.\begin{align*} f &= f_{n_1} + \sum_{k=1}^\infty f_{n_{k+1}} - f_{n_k} \\ g &+ \lVert f_{n_1}\rVert_X + \sum_{k=1}^\infty \lVert f_{n_{k+1}} - f_{n_k}\rVert_X. \end{align*}

We claim that ff and gg are well-defined. To check this, we consider the partial sums

SK(f)=fn1+k=1Kfnk+1fnk=fnK+1SK(g)=fn1X+k=1Kfnk+1fnkX.\begin{align*} S_K(f) &= f_{n_1} + \sum_{k=1}^K f_{n_{k+1}} - f_{n_k} = f_{n_{K+1}} \\ S_K(g) &= \lVert f_{n_1}\rVert_X + \sum_{k=1}^K \lVert f_{n_{k+1}} - f_{n_k}\rVert_X. \end{align*}

By the triangle inequality, we have the estimate

SK(g)pfn1Lp(X)+k=1fnk+1fnkLp(X)fn1Lp(X)+k=1K2k.\begin{align*} \lVert S_K(g)\rVert_p &\leq \lVert f_{n_1}\rVert_{L_p(X)} + \sum_{k=1}^\lVert f_{n_{k+1}} - f_{n_k}\rVert_{L_p(X)} \\ &\leq \lVert f_{n_1}\rVert_{L_p(X)} + \sum_{k=1}^K 2^{-k}. \end{align*}

Therefore, sK(g)Lp(Ω)s_K(g) \in L_p(\Omega) for all KK. The scalar-valued monotone convergence theorem now implies that gg is well-defined and is in Lp(Ω)L_p(\Omega). As

(25)SK(f)(t)XSK(g)(t)g(t)(25) \hspace{3em} \lVert S_K(f)(t)\rVert_X \leq \vert S_K(g)(t) \vert \leq \vert g(t) \vert

for each KNK \in \mathbb{N} and every tΩt \in \Omega, the vector-valued dominated convergence theorem implies that ff is well-defined and is in Lp(X)L_p(X).

It remains to show that fnff_n \to f in Lp(X)L_p(X) as nn \to \infty. To this end, we first note that (25) implies

f(t)Xg(t)\lVert f(t)\rVert_X \leq \vert g(t) \vert

for almost every tΩt \in \Omega. Moreover, the triangle inequality implies that

f(t)SK(f)(t)Xf(t)X+SK(f)(t)X2g(t)\lVert f(t) - S_K(f)(t)\rVert_X \leq \lVert f(t)\rVert_X + \lVert S_K(f)(t)\rVert_X \leq 2 \vert g(t) \vert

for almost every tΩt \in \Omega, whence we invoke the vector-valued dominated convergence theorem to conclude that fnkff_{n_k} \to f in Lp(X)L_p(X) as KK \to \infty. The desired result now follows from the Cauchy hypothesis on (fn)n=1(f_n)_{n=1}^\infty.

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 XX lacks a multiplication operation. The inequality nevertheless generalizes quite nicely for Banach spaces with a continuous multiplication operation.

Definition 25. A Banach space XX is a Banach algebra if XX is a Banach space with an associative multiplication operation that satisfies the inequaltiy

xyxy\lVert xy\rVert \leq \lVert x\rVert\rVert y\rVert

for all x,yXx,y \in X.

Proposition 26 (Hölder’s inequality). If XX is a Banach algebra, then

fgLr(X)fLp(X)gLq(X)\lVert fg\rVert_{L_r(X)} \leq \lVert f\rVert_{L_p(X)}\lVert g\rVert_{L_q(X)}

whenever 0<p,q,r0 < p, q, r \leq \infty, 1p+1q=1r\frac{1}{p} + \frac{1}{q} = \frac{1}{r}, fLp(X)f \in L_p(X), and gLq(X)g \in L_q(X).

Proof. Since XX is a Banach algebra, we see that

fgLr(X)=fgXrfXgXr.\lVert fg \rVert_{L_r(X)} = \lVert \lVert fg \rVert_{X} \rVert_r \leq \lVert \lVert f \rVert_{X} \lVert g \rVert_{X} \rVert_{r}.

The scalar-valued Hölder’s inequality implies that

fXgXrfXpgXq=fLp(X)gLq(X),\lVert \lVert f\rVert_X \lVert g\rVert_X \rVert_r \leq \lVert \lVert f \rVert_X \rVert_p \lVert \lVert g \rVert_X \rVert_q = \lVert f \rVert_{L_p(X)} \lVert g \rVert_{L_q(X)},

whence the desired inequality follows. \square