Cleanup

src/measure/lebesgue-integral/complex.tex

@@ -7,10 +7,12 @@
     \[
     \int \abs{f} d\mu < \infty
     \]
+
     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}
@@ -18,7 +20,8 @@
     \[
     \int \abs{\lambda f + g}d\mu \le \int \abs{\lambda} \abs{f} + \abs{g}d\mu = \lambda \int \abs{f}d\mu + \int \abs{g}d\mu
     \]
-    by \ref{proposition:lebesgue-non-negative-properties}.
+
+    by \autoref{proposition:lebesgue-non-negative-properties}.
 \end{proof}
 
 \begin{definition}[Positive and Negative Parts]
@@ -27,6 +30,7 @@
     \[
     f^+ = \max(f, 0) \quad f^- = -\min(f, 0)
     \]
+
     are the \textbf{positive} and \textbf{negative} parts of $f$, and $f = f^+ - f^-$.
 \end{definition}
 
@@ -37,10 +41,12 @@
     \[
     \int f d\mu = \int f^+ d\mu - \int f^- d\mu
     \]
+
     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}
 
@@ -58,7 +64,8 @@
         -\lambda\int f^- d\mu + \lambda\int f^+ d\mu &\lambda < 0
     \end{cases}
     \]
-    by \ref{proposition:lebesgue-non-negative-properties}, so $\int \lambda f d\mu = \lambda \int f d\mu$.
+
+    by \autoref{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*}
@@ -67,7 +74,7 @@
         \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 $\int f + g d\mu = \int f d\mu + \int g d\mu$.
+    by \autoref{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*}
@@ -88,6 +95,7 @@
     \[
     \int f = \int f^+ d\mu - \int f^-d\mu \le \int f^+ + \int f^-d\mu = \int \abs{f}d\mu
     \]
+
     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 \\
@@ -113,7 +121,7 @@
     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},
+    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 \hyperref[Fatou's lemma]{lemma:fatou} and \autoref{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
@@ -122,12 +130,13 @@
     \[
     \limsup_{n \to \infty}\int f_n d\mu \le \int f d\mu \le \liminf_{n \to \infty}\int f_n d\mu
     \]
+
     and $\int f d\mu = \limv{n}\int f_n d\mu$.
 \end{proof}
 
 \begin{remark}[There is no dominated convergence theorem for nets]
 \label{remark:dct-no-net}
-    In analysis, one frequently encounters places where only sequential continuity is provided or required. It is my opinion that a good portion of this comes from the lack of an extension of the dominated convergence theorem (\ref{theorem:dct}) to nets. This limitation arises from the monotone convergence theorem (\ref{theorem:mct}), where continuity from below is used.
+    In analysis, one frequently encounters places where only sequential continuity is provided or required. It is my opinion that a good portion of this comes from the lack of an extension of the \hyperref[dominated convergence theorem]{theorem:dct} to nets. This limitation arises from the \hyperref[monotone convergence theorem]{theorem:mct}, where continuity from below is used.
 
     For an example, consider the Lebesgue measure on $[0, 1]$. Let $A$ be the net of all finite subsets of $[0, 1]$, directed by inclusion, then $\lim_{\alpha \in A}\one_\alpha = 1$ pointwise. However, $\int \one_\alpha = 0$ for all $\alpha \in A$.
 \end{remark}