\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}
