Defined the Lebesgue integral.
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Bokuan Li
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\section{Integration of Complex-Valued Functions}
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\label{section:lebesgue-complex}
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\begin{definition}[Integrable]
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\label{definition:lebesgue-integrable}
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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
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\[
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\int \abs{f} d\mu < \infty
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\]
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The set
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\[
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\mathcal{L}^1(X, \cm, \mu; \complex) = \mathcal{L}^1(X, \cm, \mu) = \mathcal{L}^1(X; \complex) = \mathcal{L}^1(X) = \mathcal{L}^1(\mu; \complex) = \mathcal{L}^1(\mu)
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\]
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is the vector space of \textbf{$\mu$-integrable functions} on $X$.
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\end{definition}
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\begin{proof}
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Let $f, g \in \mathcal{L}^1(X)$ and $\lambda \in \complex$, then
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\[
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\int \abs{\lambda f + g}d\mu \le \int \abs{\lambda} \abs{f} + \abs{g}d\mu = \lambda \int \abs{f}d\mu + \int \abs{g}d\mu
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\]
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by \ref{proposition:lebesgue-non-negative-properties}.
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\end{proof}
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\begin{definition}[Positive and Negative Parts]
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\label{definition:positive-negative-parts}
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Let $X$ be a set and $f: X \to \real$ be a function, then
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\[
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f^+ = \max(f, 0) \quad f^- = -\min(f, 0)
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\]
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are the \textbf{positive} and \textbf{negative} parts of $f$, and $f = f^+ - f^-$.
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\end{definition}
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\begin{definition}[Integral]
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\label{definition:lebesgue-complex}
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Let $(X, \cm, \mu)$ be a measure space and $f \in \mathcal{L}^1(X)$. If $f$ is $\real$-valued, then
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\[
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\int f d\mu = \int f^+ d\mu - \int f^- d\mu
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\]
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is the \textbf{integral} of $f$. If $f$ is $\complex$-valued, then
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\[
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\int f d\mu = \int \text{Re}(f)d\mu + i\int \text{Im}(f)d\mu
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\]
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is the \textbf{integral} of $f$.
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\end{definition}
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\begin{proposition}[{{\cite[Proposition 2.21, 2.22]{Folland}}}]
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\label{proposition:lebesgue-integral-properties}
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Let $(X, \cm, \mu)$ be a measure space, then the integral is a linear functional on $\mathcal{L}^1(X)$ such that for any $f \in \mathcal{L}^1(X)$, $\abs{\int f d\mu} \le \int \abs{f}d\mu$.
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\end{proposition}
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\begin{proof}
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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,
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\[
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\int \lambda f d\mu = \int (\lambda f)^+ d\mu - \int (\lambda f)^- d\mu
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= \begin{cases}
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\lambda\int f^+ d\mu - \lambda\int f^- d\mu &\lambda \ge 0 \\
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-\lambda\int f^- d\mu + \lambda\int f^+ d\mu &\lambda < 0
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\end{cases}
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\]
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by \ref{proposition:lebesgue-non-negative-properties}, so $\int \lambda f d\mu = \lambda \int f d\mu$.
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Let $h = f + g$, then $h = h^+ - h^- = f^+ + g^+ - f^- - g^-$, so
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\begin{align*}
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h^+ + f^- + g^- &= h^- + f^+ + g^+ \\
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\int h^+d\mu + \int f^- d\mu + \int g^-d\mu &= \int h^- d\mu + \int f^+ d\mu + \int g^+ d\mu \\
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\int h^+ d\mu - \int h^- d\mu &= \int f^+ d\mu - \int f^- d\mu + \int g^+ d\mu - \int g^- d\mu \\
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&= \int f d\mu + \int g d\mu
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\end{align*}
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by \ref{proposition:lebesgue-non-negative-properties}, so $\int f + g d\mu = \int f d\mu + \int g d\mu$.
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Now suppose that $f, g$ are $\complex$-valued and $\lambda = \alpha + \beta i \in \complex$ with $\alpha, \beta \in \real$, then
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\begin{align*}
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\int (\alpha f)d\mu &= \int \text{Re}(\lambda f)d\mu + i\int \text{Im}(\lambda f)d\mu \\
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&= \int \alpha \text{Re}(f) - \beta \text{Im}(f)d\mu + i\int \beta\text{Re}(f) + \alpha \text{Im}(f)d\mu \\
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&= \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 \\
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&= \lambda \int f d\mu
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\end{align*}
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and
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\begin{align*}
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\int f + g d\mu &= \int \text{Re}(f + g)d\mu + i\int \text{Im}(f + g)d\mu \\
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&= \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 \\
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&= \int f d\mu + \int g d\mu
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\end{align*}
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so the integral is a linear functional on $\mathcal{L}^1(X)$.
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Finally, if $f$ is $\real$-valued, then
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\[
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\int f = \int f^+ d\mu - \int f^-d\mu \le \int f^+ + \int f^-d\mu = \int \abs{f}d\mu
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\]
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and if $f$ is $\complex$-valued and $\alpha = \ol{\sgn(\int f d\mu)}$, then
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\begin{align*}
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\abs{\int f d\mu} &= \alpha \int f d\mu = \int \alpha f \\
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&= \text{Re}\paren{\int \alpha f d\mu} = \int \text{Re}(\alpha f)d\mu \\
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&\le \int \abs{\text{Re}(\alpha f)}d\mu \le \int \abs{\alpha f}d\mu = \int \abs{f}d\mu
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\end{align*}
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so $\abs{\int f d\mu} \le \int \abs{f}d\mu$.
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\end{proof}
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\begin{remark}
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\label{remark:integral-lack-simple}
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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.
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\end{remark}
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\begin{theorem}[Dominated Convergence Theorem {{\cite[Theorem 2.24]{Folland}}}]
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\label{theorem:dct}
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Let $(X, \cm, \mu)$ be a measure space, $\seq{f_n} \subset \mathcal{L}^1(X)$, and $f:X \to \complex$ such that
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\begin{enumerate}
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\item $f_n \to f$ pointwise.
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\item There exists $g \in \mathcal{L}^1(X)$ such that $\abs{f_n} \le \abs{g}$ for all $n \in \natp$.
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\end{enumerate}
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then $\int fd\mu = \limv{n}\int f_n d\mu$.
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\end{theorem}
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\begin{proof}
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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},
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\begin{align*}
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\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 \\
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\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
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\end{align*}
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so
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\[
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\limsup_{n \to \infty}\int f_n d\mu \le \int f d\mu \le \liminf_{n \to \infty}\int f_n d\mu
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\]
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and $\int f d\mu = \limv{n}\int f_n d\mu$.
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\end{proof}
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@@ -3,3 +3,4 @@
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\input{./src/measure/lebesgue-integral/simple.tex}
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\input{./src/measure/lebesgue-integral/non-negative.tex}
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\input{./src/measure/lebesgue-integral/complex.tex}
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@@ -5,14 +5,14 @@
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\label{definition:measurable-non-negative}
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Let $(X, \cm)$ be a measure space, then
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\[
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L^+(X, \cm) = \bracs{f: X \to \real| f \ge 0, f \text{ is } (\cm, \cb_\real) \text{-measurable}}
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\mathcal{L}^+(X, \cm) = \bracs{f: X \to \real| f \ge 0, f \text{ is } (\cm, \cb_\real) \text{-measurable}}
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\]
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is the space of non-negative $\real$-valued measurable functions on $(X, \cm)$.
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\end{definition}
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\begin{definition}[Integral of Non-Negative Function]
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\label{definition:lebesgue-non-negative}
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Let $(X, \cm, \mu)$ be a measure space and $f \in L^+(X, \cm)$, then
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Let $(X, \cm, \mu)$ be a measure space and $f \in \mathcal{L}^+(X, \cm)$, then
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\[
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\int f d\mu = \int f(x)\mu(dx) = \sup\bracs{\int \phi d\mu \bigg | \phi \in \Sigma^+(X, \cm), \phi \le f}
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\]
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@@ -21,7 +21,7 @@
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\begin{lemma}
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\label{lemma:lebesgue-non-negative-strict}
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Let $(X, \cm, \mu)$ be a measure space and $f \in L^+(X, \cm)$, then
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Let $(X, \cm, \mu)$ be a measure space and $f \in \mathcal{L}^+(X, \cm)$, then
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\[
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\int f d\mu = \sup\bracs{\int \phi d\mu \bigg | \phi \in \Sigma^+(X, \cm), \phi|_{\bracs{f > 0}} < f|_{\bracs{f > 0}}, \phi \le f}
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\]
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@@ -36,7 +36,7 @@
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\begin{theorem}[Monotone Convergence Theorem, {{\cite[Theorem 2.14]{Folland}}}]
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\label{theorem:mct}
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Let $(X, \cm, \mu)$ be a measure space, $\seq{f_n} \subset L^+(X, \cm)$, and $f \in L^+(X, \cm)$ such that $f_n \upto f$ pointwise, then
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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
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\[
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\limv{n}\int f_n d\mu = \int f d\mu
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\]
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@@ -53,7 +53,7 @@
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\begin{lemma}[Fatou, {{\cite[Lemma 2.18]{Folland}}}]
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\label{lemma:fatou}
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Let $(X, \cm, \mu)$ be a measure space, $\seq{f_n} \subset L^+(X, \cm)$, then
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Let $(X, \cm, \mu)$ be a measure space, $\seq{f_n} \subset \mathcal{L}^+(X, \cm)$, then
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\[
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\int \liminf_{n \to \infty}f_n d\mu \le \liminf_{n \to \infty}\int f_nd\mu
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\]
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@@ -68,7 +68,7 @@
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\begin{lemma}
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\label{lemma:lebesgue-simple-monotone}
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Let $(X, \cm)$ be a measurable space and $f \in L^+(X, \cm)$, then there exists $\seq{f_n} \subset \Sigma^+(X, \cm)$ such that $f_n \upto f$ pointwise.
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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.
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\end{lemma}
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\begin{proof}
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By \ref{proposition: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.
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@@ -79,12 +79,12 @@
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\label{proposition:lebesgue-non-negative-properties}
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Let $(X, \cm, \mu)$ be a measure space, then
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\begin{enumerate}
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\item For any $f, g \in L^+(X, \cm)$ and $\alpha \ge 0$, $\int \alpha f + g d\mu = \alpha \int f d\mu + \int g d\mu$.
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\item For any $\seq{f_n} \subset L^+(X, \cm)$, $\sum_{n \in \natp}\int f_nd\mu = \int \sum_{n \in \natp}f_n d\mu$.
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\item For any $f, g \in L^+(X, \cm)$ with $f \le g$, $\int f d\mu \le \int g d\mu$.
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\item For any $f \in L^+(X, \cm)$ with $\int f d\mu < \infty$, $\mu(\bracs{f = \infty}) = 0$.
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\item For any $f \in L^+(X, \cm)$ with $\int f d\mu < \infty$, $\bracs{f > 0}$ is $\sigma$-finite.
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\item For any $f \in L^+(X, \cm)$ with $\int f d\mu = 0$, $f = 0$ $\mu$-almost everywhere.
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\item For any $f, g \in \mathcal{L}^+(X, \cm)$ and $\alpha \ge 0$, $\int \alpha f + g d\mu = \alpha \int f d\mu + \int g d\mu$.
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\item For any $\seq{f_n} \subset \mathcal{L}^+(X, \cm)$, $\sum_{n \in \natp}\int f_nd\mu = \int \sum_{n \in \natp}f_n d\mu$.
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\item For any $f, g \in \mathcal{L}^+(X, \cm)$ with $f \le g$, $\int f d\mu \le \int g d\mu$.
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\item For any $f \in \mathcal{L}^+(X, \cm)$ with $\int f d\mu < \infty$, $\mu(\bracs{f = \infty}) = 0$.
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\item For any $f \in \mathcal{L}^+(X, \cm)$ with $\int f d\mu < \infty$, $\bracs{f > 0}$ is $\sigma$-finite.
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\item For any $f \in \mathcal{L}^+(X, \cm)$ with $\int f d\mu = 0$, $f = 0$ $\mu$-almost everywhere.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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