Defined the Lebesgue integral.

src/measure/lebesgue-integral/complex.tex

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+\section{Integration of Complex-Valued Functions}
+\label{section:lebesgue-complex}
+
+\begin{definition}[Integrable]
+\label{definition:lebesgue-integrable}
+    Let $(X, \cm, \mu)$ be a measure space and $f: X \to \complex$ be a $(\cm, \cb_\complex)$-measurable function, then $f$ is \textbf{integrable} if
+    \[
+    \int \abs{f} d\mu < \infty
+    \]
+    The set
+    \[
+    \mathcal{L}^1(X, \cm, \mu; \complex) = \mathcal{L}^1(X, \cm, \mu) = \mathcal{L}^1(X; \complex) = \mathcal{L}^1(X) = \mathcal{L}^1(\mu; \complex) = \mathcal{L}^1(\mu)
+    \]
+    is the vector space of \textbf{$\mu$-integrable functions} on $X$.
+\end{definition}
+\begin{proof}
+    Let $f, g \in \mathcal{L}^1(X)$ and $\lambda \in \complex$, then
+    \[
+    \int \abs{\lambda f + g}d\mu \le \int \abs{\lambda} \abs{f} + \abs{g}d\mu = \lambda \int \abs{f}d\mu + \int \abs{g}d\mu
+    \]
+    by \ref{proposition:lebesgue-non-negative-properties}.
+\end{proof}
+
+\begin{definition}[Positive and Negative Parts]
+\label{definition:positive-negative-parts}
+    Let $X$ be a set and $f: X \to \real$ be a function, then
+    \[
+    f^+ = \max(f, 0) \quad f^- = -\min(f, 0)
+    \]
+    are the \textbf{positive} and \textbf{negative} parts of $f$, and $f = f^+ - f^-$.
+\end{definition}
+
+
+\begin{definition}[Integral]
+\label{definition:lebesgue-complex}
+    Let $(X, \cm, \mu)$ be a measure space and $f \in \mathcal{L}^1(X)$. If $f$ is $\real$-valued, then
+    \[
+    \int f d\mu = \int f^+ d\mu - \int f^- d\mu
+    \]
+    is the \textbf{integral} of $f$. If $f$ is $\complex$-valued, then
+    \[
+    \int f d\mu = \int \text{Re}(f)d\mu + i\int \text{Im}(f)d\mu
+    \]
+    is the \textbf{integral} of $f$.
+\end{definition}
+
+
+\begin{proposition}[{{\cite[Proposition 2.21, 2.22]{Folland}}}]
+\label{proposition:lebesgue-integral-properties}
+    Let $(X, \cm, \mu)$ be a measure space, then the integral is a linear functional on $\mathcal{L}^1(X)$ such that for any $f \in \mathcal{L}^1(X)$, $\abs{\int f d\mu} \le \int \abs{f}d\mu$.
+\end{proposition}
+\begin{proof}
+    Let $f, g \in \mathcal{L}^1(X)$ and $\lambda \in \complex$. First suppose that $f, g$ are $\real$-valued and $\lambda \in \real$. In which case,
+    \[
+    \int \lambda f d\mu = \int (\lambda f)^+ d\mu - \int (\lambda f)^- d\mu
+    = \begin{cases}
+        \lambda\int f^+ d\mu - \lambda\int f^- d\mu &\lambda \ge 0 \\
+        -\lambda\int f^- d\mu + \lambda\int f^+ d\mu &\lambda < 0
+    \end{cases}
+    \]
+    by \ref{proposition:lebesgue-non-negative-properties}, so $\int \lambda f d\mu = \lambda \int f d\mu$.
+
+    Let $h = f + g$, then $h = h^+ - h^- = f^+ + g^+ - f^- - g^-$, so
+    \begin{align*}
+        h^+ + f^- + g^- &= h^- + f^+ + g^+ \\
+        \int h^+d\mu + \int f^- d\mu + \int g^-d\mu &= \int h^- d\mu + \int f^+ d\mu + \int g^+ d\mu \\
+        \int h^+ d\mu - \int h^- d\mu &= \int f^+ d\mu - \int f^- d\mu + \int g^+ d\mu - \int g^- d\mu \\
+        &= \int f d\mu + \int g d\mu
+    \end{align*}
+    by \ref{proposition:lebesgue-non-negative-properties}, so $\int f + g d\mu = \int f d\mu + \int g d\mu$.
+
+    Now suppose that $f, g$ are $\complex$-valued and $\lambda = \alpha + \beta i \in \complex$ with $\alpha, \beta \in \real$, then
+    \begin{align*}
+        \int (\alpha f)d\mu &= \int \text{Re}(\lambda f)d\mu + i\int \text{Im}(\lambda f)d\mu \\
+        &= \int \alpha \text{Re}(f) - \beta \text{Im}(f)d\mu + i\int \beta\text{Re}(f) + \alpha \text{Im}(f)d\mu \\
+        &= \alpha \braks{\int \text{Re}(f)d\mu + i\int \text{Im}(f)}d\mu + i\beta \braks{\int \text{Re}(f)d\mu + i\int \text{Im}(f)}d\mu \\
+        &= \lambda \int f d\mu
+    \end{align*}
+    and
+    \begin{align*}
+        \int f + g d\mu &= \int \text{Re}(f + g)d\mu + i\int \text{Im}(f + g)d\mu \\
+        &= \int \text{Re}(f)d\mu + i\int\text{Im}(f)d\mu + \int \text{Re}(g)d\mu + i\int \text{Im}(g)d\mu \\
+        &= \int f d\mu + \int g d\mu
+    \end{align*}
+    so the integral is a linear functional on $\mathcal{L}^1(X)$.
+
+    Finally, if $f$ is $\real$-valued, then
+    \[
+    \int f = \int f^+ d\mu - \int f^-d\mu \le \int f^+ + \int f^-d\mu = \int \abs{f}d\mu
+    \]
+    and if $f$ is $\complex$-valued and $\alpha = \ol{\sgn(\int f d\mu)}$, then
+    \begin{align*}
+        \abs{\int f d\mu} &= \alpha \int f d\mu = \int \alpha f \\
+        &= \text{Re}\paren{\int \alpha f d\mu} = \int \text{Re}(\alpha f)d\mu \\
+        &\le \int \abs{\text{Re}(\alpha f)}d\mu \le \int \abs{\alpha f}d\mu = \int \abs{f}d\mu
+    \end{align*}
+    so $\abs{\int f d\mu} \le \int \abs{f}d\mu$.
+\end{proof}
+
+\begin{remark}
+	\label{remark:integral-lack-simple}
+
+	The construction of the Lebesgue integral using positive/negative and real/imaginary parts, and the deduction of properties from this definition both appear cumbersome. This comes from the lack of usage of its linearity on integrable simple functions. A more functional analytic approach will use this linearity and the density of simple functions to construct the integral. This is the route through which the Bochner integral will be presented.
+\end{remark}
+
+\begin{theorem}[Dominated Convergence Theorem {{\cite[Theorem 2.24]{Folland}}}]
+\label{theorem:dct}
+    Let $(X, \cm, \mu)$ be a measure space, $\seq{f_n} \subset \mathcal{L}^1(X)$, and $f:X \to \complex$ such that
+    \begin{enumerate}
+        \item $f_n \to f$ pointwise.
+        \item There exists $g \in \mathcal{L}^1(X)$ such that $\abs{f_n} \le \abs{g}$ for all $n \in \natp$.
+    \end{enumerate}
+    then $\int fd\mu = \limv{n}\int f_n d\mu$.
+\end{theorem}
+\begin{proof}
+    By (1) and (2), $f$ is measurable with $\int \abs{f}d\mu \le \int \abs{g}d\mu < \infty$, so $f \in \mathcal{L}^1(X)$. Now, since $g + f, g - f \ge 0$, by Fatou's lemma (\ref{lemma:fatou}) and \ref{proposition:lebesgue-integral-properties},
+    \begin{align*}
+        \int g d\mu + \int f d\mu &\le \liminf_{n \to \infty}\int g + f_n d\mu = \int g d\mu + \liminf_{n \to \infty}f_n d\mu \\
+        \int g d\mu - \int f d\mu &\le \liminf_{n \to \infty}\int g - \int f_n d\mu = \int g d\mu - \limsup_{n \to \infty}\int f_n d\mu
+    \end{align*}
+    so
+    \[
+    \limsup_{n \to \infty}\int f_n d\mu \le \int f d\mu \le \liminf_{n \to \infty}\int f_n d\mu
+    \]
+    and $\int f d\mu = \limv{n}\int f_n d\mu$.
+\end{proof}