# 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 $(\Omega, \Sigma, \mu)$ 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.$f:\Omega \to X$ isif $lf$ is measurable for each $l \in X^*$.weakly measurable

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.$s:\Omega \to X$ isif it is of the formsimple$$s(t) = \sum_{n=1}^N x_n 1_{E_n}(t),$$

where $E_n \in \Sigma$ and $x_n \in X$ for all $1 \leq n \leq N$. $f:\Omega \to X$ is

if there exists a sequence of simple functions $(s_n)_{n=1}^\infty$ such thatstrongly measurable$$\lim_{n \to \infty} \lVert s_n(t) - f(t) \rVert_X = 0$$

pointwise $\mu$-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 $s_n$ must necessarily be contained in the closure of the union of $\operatorname{im} s_n$, 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 $\mathscr{P}(S)$ as its $\sigma$-algebra and $X$ a Banach space, then every function $f:S \to X$ 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 $f:S \to X$, 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 $\mathbb{T}$ is an isometric isomorphism of $L_2(\mathbb{T})$ onto $l_2(\mathbb{Z})$. This, in particular, implies that $L_2(\mathbb{T})$ and $l_2(\mathbb{Z})$ have the same cardinality as sets. As

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

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

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

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

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

By the Schröder–Bernstein theorem, it follows that

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

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

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

It then follows that there is a bijection from $[0,1]$ to $L_\infty$, whence we can produce functions $[0,1] \to L_\infty$ 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.$f: \Omega \to X$ isif $\operatorname{im} f$ is a separable subset of $X$, andseparably valuedif there exists a null set $N \subseteq \Omega$ such that $f(\Omega \smallsetminus N)$ is a separable subset of $X$.$\mu$-almost separably valued

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

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

We shall make use of the following lemma:

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

**Proof of lemma.** We fix $a \in \mathbb{R}$ and let

- $E = \lbrace t : \lVert g(t) \rVert \leq a\rbrace$
- $E_l = \lbrace t: \lVert lg(t) \rVert \leq a\rbrace$ for each $l \in X^*$.

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

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

then the claim will follows from the weak measurability of $f$.

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

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

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

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

We now appeal to the following claim:

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

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

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

which establishes (7).

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

$$l \mapsto \varphi_N(l) = (l(x_1), \ldots, l(x_n))$$

sends $B_{X^*}$ to $l_2^N$. Since $l_2^N$ is separable, there exists a sequence of linear functionals $(l_{N,k})^\infty_{k=1}$ in $B_{X^*}$ such that $\lbrace \varphi_N(l_{N,j})\rbrace_{k \in \mathbb{N}}$ is dense in $\varphi_N(B_{X^*})$. Therefore, each $l \in B_{X^*}$ furnishes a sequence $(k_N)_{N=1}^\infty$ of indices such that

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

This, in particular, implies that

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

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

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

for all $n \in \mathbb{N}$. Since

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

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

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

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

$\mu$-almost everywhere. Since each $ls_n$ is measurable, the pointwise limit $lf$ is also measurable, whence $f$ is weakly measurable.

We now let $N$ be the set on which $s_n$ does not converge pointwise to $f$. Observe that

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

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

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

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

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

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

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

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

We now define

$$s_N = s_N' \circ f$$

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

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

where $E_N^n = \lbrace t \in \Omega : s_N(t) = x_n\rbrace$. Therefore, showing that $f$ is strongly measurable amounts to proving that each $E_N^n$ is a measurable subset of $\Omega$. To this end, we observe that

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

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

- $\phi_n(t) = \lVert f(t) - x_n\rVert$ and
- $\varphi_N(t) = \min_{1 \leq j \leq N} \phi_j(t)$,

which are measurable functions by Lemma 6. Therefore,

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

and

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

are measurable subsets of $\Omega$, whence so is $E_N^n$. 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 $(f_n)_{n=1}^\infty$ is a sequence of strongly measurable $X$-valued functions on $\Omega$, then its pointwise limit $f$ is also strongly measurable.

**Proof.** By the Pettis measurability theorem, each $f_n$ is weakly measurable and $\mu$-almost separably valued. We assume without loss of generality that each $\operatorname{im} f_n$ is a separable subset of $X$, so that $X_0 = \overline{\bigcup_n \operatorname{im} f_n}$ is a separable Banach subspace of $X$ such that $\operatorname{im} f \subseteq X_0$. 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 $lf_n$ for each fixed $l \in X^*$. The strong measurability of $f$ now follows from the Pettis measurability theorem.

Corollary 10.If $f:\Omega \to X$ is strongly measurable, and if $\varphi:X \to Y$ is norm-continuous, then $\varphi f:\Omega \to Y$ is strongly measurable.

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

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

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

$(\Rightarrow)$ We assume without loss of generality that $X = \overline{\operatorname{span}(\operatorname{im} f)}$, 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 $s = \sum x_n 1_{E_n}$, we note that the preimage of *any* subset of $X$ under $s$ is a finite union of $E_1,\ldots,E_N$, which is measurable. Therefore, if $(s_n)_{n=1}^\infty$ is a sequence of simple functions that converge pointwise almost everywhere to $f$, and if

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

then $s_n^{-1}(U_r)$ is measurable regardless of $n$ and $r$.

Fix $n \in \mathbb{N}$. If $x \in \bigcap_{k \geq n} s_n^{-1}(U_r)$, then $s_n(x) \in U_r$ for all $n \geq k$, whence $f(x) \in \overline{U_r}$. By (12), $\overline{U_r} \subseteq U_{r/2}$, and so $f(x) \in U_{r/2}$. Therefore, $f^{-1}(U_{r/2}) \supseteq \bigcap_{k \geq n} s_n^{-1}(U_r)$ for all $n$, whence

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

This implies that

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

Conversely, if $x \in f^{-1}(U_{1/m})$, then $f(x) \in U_{1/m}$. Since $U_{1/m}$ is open, there exists an $n \in \mathbb{N}$ such that $s_k(x) \in U$ for all $k \geq n$. Therefore,

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

and it follows that

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

Since each $s_n^{-1}(U_{1/m})$ is measurable, we conclude from (13) and (14) that $f^{-1}(U)$ 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.$f:\Omega \to X$ isBochner integrableif there exists a sequence $(s_n)_{n=1}^\infty$ of simple functions$$s_n = \sum_{m=1}^{M_n} x_{m,n} 1_{E_{m,n}}$$

on $\Omega$ such that

- $\mu(E_{m,n}) < \infty$ for all $m$ and $n$,
- $s_n \to f$ almost everywhere, and
- $\int_\Omega \lVert s_n - f \rVert \, d\mu \to 0$.
In this case, we define the

of $f$ to beBochner integral$$(16) \hspace{3em} \begin{align*} \int_\Omega f \, d\mu &= \lim_{n \to \infty} \int_\Omega s_n \, d\mu \\ &= \lim_{n \to \infty} \sum_{m=1}^{M_n} \mu(E_{m,n})x_{m,n}. \end{align*}$$

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

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

Since

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

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

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

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

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

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

**Proof.** If $s_n \to f$, then $s_n \chi_{E_n} \to f$, where $(E_n)_{n=1}^\infty$ 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 $\mu$-measurable function $f:\Omega \to X$ is Bochner integrable if and only if $\int_\Omega \lVert f \rVert \, d\mu < \infty$. In this case, we have the inequality$\left\lVert \int_\Omega f \, d\mu \right\rVert \leq \int_\Omega \lVert f\lVert \, d\mu.$

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

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

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

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

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

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

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

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

as was to be shown. $\square$

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

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

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

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

$$L_p(X) = L_p(\mu;X) = L_p(\Omega,\Sigma,\mu;X)$$

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 $1 \leq p < \infty$, and let $(f_n)_{n=1}^\infty$ be a sequence in $L_p(X)$. If $f_n \to f$ pointwise almost everywhere and if there exists a $g \in L_p(\Omega)$ such that $\lVert f_n\rVert \leq |g|$ almost everywhere for all $n$, then $f \in L_p(X)$, and$$\lim_{n \to \infty} \lVert f_n - f\rVert_{L_p(X)} = 0.$$

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

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

as was to be shown. $\square$

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

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

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

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

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

We claim that $f$ and $g$ are well-defined. To check this, we consider the partial sums

$$\begin{align*} S_K(f) &= f_{n_1} + \sum_{k=1}^K f_{n_{k+1}} - f_{n_k} = f_{n_{K+1}} \\ S_K(g) &= \lVert f_{n_1}\rVert_X + \sum_{k=1}^K \lVert f_{n_{k+1}} - f_{n_k}\rVert_X. \end{align*}$$

By the triangle inequality, we have the estimate

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

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

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

for each $K \in \mathbb{N}$ and every $t \in \Omega$, the vector-valued dominated convergence theorem implies that $f$ is well-defined and is in $L_p(X)$.

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

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

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

$$\lVert f(t) - S_K(f)(t)\rVert_X \leq \lVert f(t)\rVert_X + \lVert S_K(f)(t)\rVert_X \leq 2 \vert g(t) \vert$$

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

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 $X$ 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 aif $X$ is a Banach space with an associative multiplication operation that satisfies the inequaltiyBanach algebra$$\lVert xy\rVert \leq \lVert x\rVert\rVert y\rVert$$

for all $x,y \in X$.

Proposition 26(Hölder’s inequality). If $X$ is a Banach algebra, then$$\lVert fg\rVert_{L_r(X)} \leq \lVert f\rVert_{L_p(X)}\lVert g\rVert_{L_q(X)}$$

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

**Proof.** Since $X$ is a Banach algebra, we see that

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

The scalar-valued Hölder’s inequality implies that

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

whence the desired inequality follows. $\square$