\section{Integration of Non-Negative Functions}
\label{section:lebesgue-non-negative}

\begin{definition}
\label{definition:measurable-non-negative}
    Let $(X, \cm)$ be a measure space, then
    \[
    \mathcal{L}^+(X, \cm) = \bracs{f: X \to [0, \infty]| f \ge 0, f \text{ is } (\cm, \cb_\real) \text{-measurable}}
    \]

    is the space of non-negative $\ol \real$-valued measurable functions on $(X, \cm)$.
\end{definition}

\begin{definition}[Integral of Non-Negative Function]
\label{definition:lebesgue-non-negative}
    Let $(X, \cm, \mu)$ be a measure space and $f \in \mathcal{L}^+(X, \cm)$, then
    \[
    \int f d\mu = \int f(x)\mu(dx) = \sup\bracs{\int \phi d\mu \bigg | \phi \in \Sigma^+(X, \cm), \phi \le f}
    \]

    is the \textbf{Lebesgue integral} of $f$.
\end{definition}

\begin{lemma}
\label{lemma:lebesgue-non-negative-strict}
    Let $(X, \cm, \mu)$ be a measure space and $f \in \mathcal{L}^+(X, \cm)$. 
    \begin{enumerate}
        \item For each $\phi \in \Sigma^+(X, \cm)$, denote $\phi \le_u f$ if there exists $\delta > 0$ such that $\phi + \delta \le f$ on $\bracs{\phi > 0}$, then
        \[
        \int f d\mu = \sup\bracs{\int \phi d\mu \bigg | \phi \in \Sigma^+(X, \cm), \phi \le_u f}
        \]

        \item If $\mu$ is semifinite, then
        \[
        \int f d\mu = \sup\bracs{\int \phi d\mu \bigg | \phi \in \Sigma^+(X, \cm) \cap L^1(X; \real), \phi \le_u f}
        \]
    \end{enumerate}
\end{lemma}
\begin{proof}
    (1): Let $\phi \in \Sigma^+(X, \cm)$ with $\phi \le f$ and $\alpha \in (0, 1)$. Since $\phi(X) \setminus \bracs{0} \subset (0, \infty)$ is finite, $\delta = \min_{y \in \phi(X) \setminus \bracs{0}}(1 - \alpha)y > 0$. Thus $\alpha \phi + \delta \le \phi \le f$ on $\bracs{\phi > 0} = \bracs{\alpha \phi > 0}$, and $\alpha \phi \le_u f$. 
    
    By \hyperref[linearity on simple functions]{proposition:lebesgue-simple-properties}, 
    \[
    \int \phi d\mu = \sup_{\alpha \in (0, 1)}\alpha \int \phi d\mu = \sup_{\alpha \in (0, 1)}\int \alpha \phi d\mu
    \]

    Thus
    \[
    \int \phi d\mu \le \sup\bracs{\int \psi d\mu \bigg | \psi \in \Sigma^+(X, \cm), \psi \le_u f}
    \]

    As the above holds for all $\phi \in \Sigma^+(X, \cm)$, 
    \[
    \int f d\mu = \sup\bracs{\int \psi d\mu \bigg | \psi \in \Sigma^+(X, \cm), \psi \le_u f}
    \]
\end{proof}

\begin{theorem}[Monotone Convergence Theorem]
\label{theorem:mct}
    Let $(X, \cm, \mu)$ be a measure space, $\seq{f_n} \subset \mathcal{L}^+(X, \cm)$, and $f \in \mathcal{L}^+(X, \cm)$ such that $f_n \upto f$ pointwise, then
    \[
    \limv{n}\int f_n d\mu = \int f d\mu
    \]

\end{theorem}
\begin{proof}[Proof {{\cite[Theorem 2.14]{Folland}}}. ]
    By \autoref{definition:lebesgue-non-negative}, $\int f_n d\mu \le \int f d\mu$ for each $n \in \natp$.

    Let $\phi \in \Sigma^+(X, \cm)$ with $\phi|_{\bracs{f > 0}} < f|_{\bracs{f > 0}}$ and $\phi \le f$. Since $f_n \upto f$, $\bracs{f_n \ge \phi} \upto X$. In which case, since $A \mapsto \int \one_A \phi d\mu$ is a measure (\autoref{proposition:lebesgue-simple-properties}),
    \[
    \limv{n}\int f_n d\mu \ge \limv{n}\int \one_{\bracs{f_n \ge \phi}} \cdot \phi d\mu = \int \phi d\mu
    \]

    by continuity from below (\autoref{proposition:measure-properties}). Therefore $\limv{n}\int f_n d\mu \ge \int f$ by \autoref{lemma:lebesgue-non-negative-strict}.
\end{proof}

\begin{lemma}[Fatou]
\label{lemma:fatou}
    Let $(X, \cm, \mu)$ be a measure space, $\seq{f_n} \subset \mathcal{L}^+(X, \cm)$, then
    \[
    \int \liminf_{n \to \infty}f_n d\mu \le \liminf_{n \to \infty}\int f_nd\mu
    \]

\end{lemma}
\begin{proof}[Proof {{\cite[Lemma 2.18]{Folland}}}. ]
    For each $n \in \natp$, $\inf_{k \ge n}f_k \le f_n$. By the monotone convergence theorem,
    \[
    \int \liminf_{n \to \infty}f_n d\mu = \limv{n}\int \inf_{k \ge n}f_k d\mu
    \le \liminf_{n \to \infty}\int f_n d\mu
    \]

\end{proof}

\begin{lemma}
\label{lemma:lebesgue-simple-monotone}
    Let $(X, \cm)$ be a measurable space and $f \in \mathcal{L}^+(X, \cm)$, then there exists $\seq{f_n} \subset \Sigma^+(X, \cm)$ such that $f_n \upto f$ pointwise.
\end{lemma}
\begin{proof}
    By \autoref{corollary:measurable-simple-separable-norm}, there exists $\seq{g_n} \subset \Sigma^+(X, \cm)$ such that $0 \le g_n \le f$ for each $n \in \natp$, and $g_n \to f$ pointwise. For each $n \in \natp$, let $f_n = \max_{1 \le k \le n}g_k$, then $f_n \in \Sigma^+(X, \cm)$, $0 \le f_n \le f$, and $f_n \upto f$ pointwise.
\end{proof}


\begin{proposition}
\label{proposition:lebesgue-non-negative-properties}
    Let $(X, \cm, \mu)$ be a measure space, then
    \begin{enumerate}
        \item For any $f, g \in \mathcal{L}^+(X, \cm)$ and $\alpha \ge 0$, $\int \alpha f + g d\mu = \alpha \int f d\mu + \int g d\mu$.
        \item For any $\seq{f_n} \subset \mathcal{L}^+(X, \cm)$, $\sum_{n \in \natp}\int f_nd\mu = \int \sum_{n \in \natp}f_n d\mu$.
        \item For any $f, g \in \mathcal{L}^+(X, \cm)$ with $f \le g$, $\int f d\mu \le \int g d\mu$.
        \item For any $f \in \mathcal{L}^+(X, \cm)$ with $\int f d\mu < \infty$, $\mu(\bracs{f = \infty}) = 0$.
        \item For any $f \in \mathcal{L}^+(X, \cm)$ with $\int f d\mu < \infty$, $\bracs{f > 0}$ is $\sigma$-finite.
        \item For any $f \in \mathcal{L}^+(X, \cm)$ with $\int f d\mu = 0$, $f = 0$ $\mu$-almost everywhere.
    \end{enumerate}
\end{proposition}
\begin{proof}
    (1): By \autoref{lemma:lebesgue-simple-monotone}, there exists $\seq{f_n}, \seq{g_n} \subset \Sigma^+(X, \cm)$ with $0 \le f_n \le f$ and $0 \le g_n \le g$ for each $n \in \natp$, $f_n \upto f$, and $g_n \upto g$. By \autoref{proposition:lebesgue-simple-properties} and the \hyperref[Monotone Convergence Theorem]{theorem:mct},
    \begin{align*}
    \int \alpha f + g d\mu &= \limv{n}\int \alpha f_n + g_n d\mu = \alpha\limv{n}\int f_n d\mu + \limv{n}\int g_n d\mu \\
    &= \alpha \int f d\mu + \int g d\mu
    \end{align*}


    (2): By (1) and the \hyperref[Monotone Convergence Theorem]{theorem:mct},
    \[
    \int \sum_{n \in \natp}f_n d\mu = \int \limv{N}\sum_{n = 1}^N f_n d\mu = \limv{N}\sum_{n = 1}^N \int f_nd\mu = \sum_{n \in \natp}\int f_n d\mu
    \]


    (3): By \autoref{definition:lebesgue-non-negative}.

    (4): Since $\int f d\mu \ge \int \infty \cdot \one_{\bracs{f = \infty}}d\mu = \infty \cdot \mu(\bracs{f = \infty})$, $\int f d\mu < \infty$ implies that $\mu(\bracs{f = \infty}) = 0$.

    (5): For each $\eps > 0$,
    \[
    \int f d\mu \ge \int \eps \cdot \one_{\bracs{f \ge \eps}} = \eps \cdot \mu(\bracs{f \ge \eps})
    \]

    so $\int f d\mu < \infty$ implies that $\mu(\bracs{f \ge \eps}) < \infty$. Since $\bracs{f > 0} = \bigcup_{n \in \natp}\bracs{f_n \ge 1/n}$, $\bracs{f > 0}$ is $\sigma$-finite.

    (6): As in the proof of (5), $\mu(\bracs{f \ge \eps}) = 0$ for all $\eps > 0$. Thus $\bracs{f > 0} = \bigcup_{n \in \natp}\bracs{f_n \ge 1/n}$ is a $\mu$-null set, so $f = 0$ $\mu$-almost everywhere.
\end{proof}
