> [!definitionb] Definition > > Let $(X, \cm, \mu)$ be a [[Complete Measure|complete measure space]] and $E$ be a [[Normed Vector Space|normed space]] over $F \in \bracs{\real, \complex}$. For any $(\cm, \cb(E))$-[[Measurable Function|measurable functions]] $f, g: X \to E$, define $f \sim g$ if $f = g$ [[Almost Everywhere|a.e.]], and let > $ > \norm{f}_{L^1(X, \cm, \mu; E)} = \norm{f}_{L^1(X; E)} = \int_X \norm{f(x)}_{E}d\mu(x) > $ > then $\norm{\cdot}_{L^1(X; E)}$ is a well-defined norm on the [[Equivalence Class|equivalence classes]] of $\sim$. If $E$ is a [[Banach Space|Banach space]], then > $ > L^1(X, \cm, \mu; E) = L^1(X; E) = \bracsn{f \text{ strongly measurable}:\norm{f}_{L^1(X; E)} < \infty} > $ > is a [[Banach Space|Banach space]] over $F$, known as the space of **Bochner integrable functions**. For each $f \in L^1(X; E)$, there exists $\seq{f_n} \subset L^1(X; E)$ finite-valued with $\norm{f_n(x)}_E \le \norm{f(x)}_E$ for each $n \in \nat$ and [[Almost Everywhere|a.e.]] $x \in X$, such that $f_n \to f$ a.e. and in $L^1(X; E)$. Moreover, there exists a unique [[Bounded Linear Map|bounded linear map]] $I: L^1(X; E) \to E$ such that > 1. $I(x \cdot \one_{A}) = x \cdot \mu(A)$ for all $x \in E$ and $A \in \cm$ with $\mu(A) < \infty$. > 2. $\norm{I(f)}_E \le \int_X\norm{f}_Ed\mu$ for all $f \in L^1(X; E)$. > > denoted $I(f) = \int_X f d\mu = \int f$, known as the **Bochner integral**. > > *Proof of Banach space*. Let $f, g \in L^1(X; E)$ and $\lambda \in F$, then > $ > \norm{\lambda f}_{L^1(X; E)} = \int_X \norm{\lambda f}_E d\mu = \abs{\lambda}\int_X\norm{f}_Ed\mu = \abs{\lambda} \cdot \norm{f}_{L^1(X; E)} > $ > and > $ > \begin{align*} > \norm{f + g}_{L^1(X; E)} &= \int_X\norm{f + g}_Ed\mu \le \int_X\norm{f}_Ed\mu + \int_X\norm{g}_Ed\mu \\ > &= \norm{f}_{L^1(X; E)} + \norm{g}_{L^1(X; E)} > \end{align*} > $ > so $\norm{\cdot}_{L^1(X; E)}$ is a norm on $L^1(X; E)$. Let $\seq{f_n} \subset L^1(X; E)$ such that $\sum_{n \in \nat}\norm{f_n}_{L^1(X; E)} < \infty$, then $\sum_{n \in \nat}\norm{f_n(x)}_{E} \in L^1(X; \real)$ and is a.e. finite, so the series $\sum_{n = 1}^\infty f_n$ is a.e. absolutely convergent to a strongly measurable function $F: X \to E$. From here, > $ > \norm{F}_{L^1(X; E)} = \norm{\sum_{n = 1}^\infty f_n}_{L^1(X; E)} \le \sum_{n \in \nat}\norm{f_n}_{L^1(X; E)} < \infty > $ > and > $ > \begin{align*} > \norm{F - \sum_{n = 1}^Nf_n}_{L^1(X; E)} &\le \int_X \sum_{n > N}\norm{f_n}_{E}d\mu \\ > &\le \sum_{n > N}\norm{f_n}_{L^1(X; E)} \to 0 > \end{align*} > $ > > as $N \to \infty$. Thus $L^1(X; E)$ is a Banach space. > > *Proof of approximation.* Let $f \in L^1(X; E)$, then there exists a sequence $\seq{f_n}$ of finitely-valued $(\cm, \cb(E))$-measurable functions with $\norm{f_n(x)}_E \le \norm{f(x)}_E$ for all $n \in \nat$ and a.e. $x \in X$, such that $f_n \to f$ a.e. Thus $\seq{f_n} \subset L^1(X; E)$ as well. By the [[Dominated Convergence Theorem]] with $\phi(x) = 2\norm{f(x)}_E$, $\int_X \norm{f_n - f}_Ed\mu \to 0$ as $n \to \infty$, so $f_n \to f$ in $L^1(X; E)$. > > *Construction of integral.* Let $f = \sum_{j = 1}^n x_j \one_{A_j} \in L^1(X; E)$ where $\seqf{x_j} \subset E \setminus \bracs{0}$ and $\seqf{A_j} \subset \cm$. By [[Markov's Inequality]], $\mu(A_j) < \infty$ for each $1 \le j \le n$. Let > $ > I(f) = \sum_{j = 1}^n x_j\mu(A_j) > $ > then $I$ is a well-defined [[Linear Functional|linear functional]] on the finitely-valued functions in $L^1(X; E)$, with > $ > \norm{I(f)}_E \le \sum_{j = 1}^n \norm{x_j}\mu(A_j) = \int_X \norm{f}_E d\mu > $ > By density of such functions, $I$ [[Linear Extension Theorem|extends uniquely]] to a bounded linear map $L^1(X; E) \to E$ with $\norm{I}_{L(L^1(X; E), E)} \le 1$, thus satisfying $(1)$ and $(2)$. > [!definition] > > Let $(X, \cm, \mu)$ be a complete [[Vector Measure|vector measure]] space, $E$, $G$, and $H$ be Banach spaces over $F \in \bracs{\real, \complex}$, and $\lambda \in L^2(E, G; H)$ be a [[Bounded Multilinear Map|bounded bilinear map]], then there exists a unique [[Bounded Linear Map|bounded linear map]] $I_{\lambda, \mu}: L^1(X, \cm, \abs{\mu}; E) \to H$ such that > 1. $I_{\lambda, \mu}(x \cdot \one_A) = \lambda(x, A)$ for all $x \in E$ and $A \in \cm$. > 2. $\norm{I_{\lambda, \mu}(f)}_G \le \norm{\lambda}_{L^2(E, G; H)} \cdot \norm{f}_{L^1(X, \cm, \abs{\mu}; E)}$ for all $f \in L^1(X, \cm, \abs{\mu}; E)$. > > known as the **integral** with respect to $\mu$. If the bilinear pairing between $E$ and $G$ is clear from the context, denote $I_{\lambda, \mu}(f) = \int f d\mu$.