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/Part 4: Measure Theory and Integration/Chapter 20: The Bochner Integral

20.2 The Bochner Integral

Definition 20.2.1 (Bochner Integral). Let $(X, \cm, \mu)$ be a measure space and $E$ be a Banach space over $K \in \RC$, then there exists a unique $I \in L(L^{1}(X; E); E)$ such that:

  1. For any $x \in E$ and $A \in \cm$ with $\mu(A) < \infty$, $I(x \cdot \one_{A}) = x \cdot \mu(A)$.

  2. For all $f \in L^{1}(X; E)$, $\norm{If}_{E} \le \int \norm{f}_{E} d\mu$.

For any $f \in L^{1}(X; E)$, $If = \int f d\mu$ is the Bochner integral of $f$.

Proof. (1): For any $\phi \in \Sigma(X; E) \cap L^{1}(X; E)$, let

\[I\phi = \sum_{y \in \phi(X)}^{n} y \cdot \mu\bracs{\phi = y}\]

For any $\lambda \in K$, if $\lambda \ne 0$, then the mapping $x \mapsto \lambda x$ is a bijection, so

\[I(\lambda \phi) = \sum_{\lambda y \in \lambda \phi(X)}^{n} (\lambda y) \cdot \mu\bracs{\lambda \phi = \lambda y}= \lambda \sum_{y \in \phi(X)}y \cdot \mu\bracs{\phi = y}= \lambda I\phi\]

If $\lambda = 0$, then $I(\lambda \phi) = I(0) = 0$.

Let $\phi, \psi \in \Sigma(X; E) \cap L^{1}(X; E)$, then

\begin{align*}I\phi + I\psi&= \sum_{y \in \phi(X)}y \cdot \mu\bracs{\phi = y}+ \sum_{z \in \psi(X)}z \cdot \mu\bracs{\psi = z}\\&= \sum_{y \in \phi(X)}\sum_{z \in \psi(X)}(y + z) \cdot \mu\bracs{\phi = y, \psi = z}\\&= \sum_{y \in (\phi + \psi)(X)}\sum_{{z \in \phi(X) \atop {z' \in \psi(X) \atop z + z' = y}}}(z + z') \cdot \mu\bracsn{\phi = g, \psi = z'}\\&= \sum_{y \in (\phi + \psi)(X)}y \cdot \mu\bracs{\phi + \psi = y}= I\phi + I\psi\end{align*}

so $I$ is a linear operator on $\Sigma(X; E) \cap L^{1}(X; E)$ that satisfies (1).

(2): For any $\phi \in \Sigma(X; E) \cap L^{1}(X; E)$,

\[\norm{I\phi}_{E} \le \sum_{y \in \phi(X)}\norm{y}_{E} \cdot \mu\bracs{\phi = y}= \int \norm{\phi}_{E} d\mu = \norm{\phi}_{L^1(X; E)}\]

By Proposition 12.1.9, $\Sigma(X; E) \cap L^{1}(X; E)$ is dense in $L^{1}(X; E)$. Therefore by the Linear Extension Theorem, $I$ admits a unique norm-preserving extension to $L^{1}(X; E)$.$\square$

Theorem 20.2.2 (Dominated Convergence Theorem). Let $(X, \cm, \mu)$ be a measure spacs, $E$ be a Banach space over $K \in \RC$, $\seq{f_n}\subset L^{1}(X; E)$, and $f \in L^{1}(X; E)$. If

  1. $f_{n} \to f$ strongly pointwise.

  2. There exists $g \in L^{1}(X) \cap L^{+}(X)$ such that $\norm{f_n}_{E} \le g$ for all $n \in \natp$.

then $\int f d\mu = \limv{n}\int f_{n} d\mu$.

Proof. By the classical Dominated Convergence Theorem, $f_{n} \to f$ in $L^{1}(X; E)$. Since $h \mapsto \int h d\mu$ is a bounded linear operator, $\int f d\mu = \limv{n}\int f_{n} d\mu$.$\square$

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