Cleaned up MCT in measure.

src/measure/lebesgue-integral/simple.tex

@@ -18,7 +18,7 @@
 
 \begin{proposition}
 \label{proposition:lebesgue-simple-properties}
-    Let $(X, \cm, \mu)$ be a measure space and $f, g \in \Sigma^*(X, \cm)$, then:
+    Let $(X, \cm, \mu)$ be a measure space and $f, g \in \Sigma^+(X, \cm)$, then:
     \begin{enumerate}
         \item For any $\alpha \ge 0$, $\int \alpha f d\mu = \alpha \int f d\mu$.
         \item $\int f + g d\mu = \int f d\mu + \int g d\mu$.

src/measure/measurable-maps/in-measure.tex

@@ -103,19 +103,32 @@
 \begin{proof}
     By \autoref{definition:lebesgue-non-negative}, $\int f_\alpha d\mu \le \int f d\mu$ for each $\alpha \in A$. 
 
-    Assume without loss of generality that $f \ne 0$. Let $\phi \in \Sigma^+(X, \cm)$ with $\phi|_{\bracs{f > 0}} < f|_{\bracs{f > 0}}$ and $\phi \le f$. Since $\mu$ is semifinite, assume further without loss of generality that $\phi \in L^1(X, \cm; \real)$ as well. 
+    On the other hand, using \autoref{lemma:lebesgue-non-negative-strict}, it is sufficient to show that
-    
-    Let $\eps > 0$ and $\lambda \in (0, 1)$, then $\delta = \min_{y \in \phi(X) \setminus \bracs{0}}(1 - \lambda)y > 0$. By (b), there exists $\alpha \in A$ such that
     \[
-    \mu\bracs{f_\alpha + \delta < f} < \frac{\epsilon}{\norm{\phi}_u}
+    \lim_{\alpha \in A}\int f_\alpha d\mu = \sup_{\alpha \in A}\int f_\alpha d\mu \ge \int \phi d\mu
     \]
     
-    In which case, 
+    for any $\phi \in \Sigma^+(X, \cm)$ satisfying:
+    \begin{enumerate}[label=(\roman*)]
+        \item There exists $\delta > 0$ such that $\phi + \delta \le f$ on $\bracs{\phi > 0}$. 
+        \item $\phi \in L^1(X, \cm)$. 
+    \end{enumerate}
+
+    To this end, let $\eps > 0$. Since $\mu\bracs{\phi > 0} < \infty$, by (b), there exists $\alpha \in A$ such that
+    \[
+    \mu\bracs{\phi > 0, f_\alpha + \delta < \phi} \le \mu\bracs{\phi > 0, f_\alpha + \delta < f} < \frac{\eps}{\norm{\phi}_u}
+    \]
+
+    In which case, by \hyperref[linearity]{proposition:lebesgue-simple-properties},
     \begin{align*}
-        \int f_\alpha d\mu &\ge \int_{\bracs{f_\alpha + \delta \ge f}} f_\alpha d\mu \ge \int_{\bracs{f_\alpha + \delta \ge f}}\lambda\phi d\mu \\
+        \int \phi d\mu &= \int_{\bracs{\phi > 0}}\phi d\mu = \int_{\bracs{\phi > 0, f_\alpha + \delta \ge \phi}} \phi d\mu + \int_{\bracs{\phi > 0, f_\alpha + \delta < \phi}} \phi d\mu \\
-        &\ge \lambda\int \phi d\mu - \frac{\eps \norm{\phi}_u}{\norm{\phi}_u} = \lambda \int \phi d\mu - \eps
+        &\le \int_{\bracs{\phi > 0, f_\alpha + \delta \ge \phi}} f_\alpha d\mu +\norm{\phi}_u\mu \bracs{\phi > 0, f_\alpha + \delta < \phi} \\
+        &\le \int f_\alpha d\mu + \norm{\phi}_u \frac{\eps}{\norm{\phi}_u} = \int f_\alpha d\mu + \eps
     \end{align*}
 
-    As the above holds for all $\eps > 0$, $\lambda \in (0, 1)$, $\sup_{\alpha \in A}\int f_\alpha d\mu \ge \int \phi d\mu$. Therefore $\sup_{\alpha \in A}\int f_\alpha d\mu \ge \int f d\mu$ by \autoref{lemma:lebesgue-non-negative-strict}. 
+    As the above holds for all $\eps > 0$, $\int \phi d\mu \le \sup_{\alpha \in A}\int f_\alpha d\mu$. Therefore
+    \[
+    \int f d\mu = \sup_{\alpha \in A}\int f_\alpha d\mu = \lim_{\alpha \in A}\int f_\alpha d\mu
+    \]
 \end{proof}