Cleanup

src/measure/lebesgue-integral/simple.tex

@@ -12,6 +12,7 @@
     \[
     \int f d\mu = \int f(x) \mu(dx) = \sum_{y \in f(X)}y \cdot \mu(\bracs{f = y})
     \]
+
     is the \textbf{Lebesgue integral} of $f$.
 \end{definition}
 
@@ -30,11 +31,13 @@
     \[
     \alpha f = \sum_{y \in f(X)} (\alpha y) \cdot \one_{\bracs{f = y}}
     \]
+
     is the standard form of $\alpha f$. Therefore
     \[
     \int \alpha f d\mu = \sum_{y \in f(X)}(\alpha y) \cdot \mu({\bracs{f = y}}) = \alpha \int f d\mu
     \]
 
+
     (2): Since $\mu$ is finitely additive,
     \begin{align*}
         \int f d\mu + \int g d\mu &= \sum_{y \in f(X)}y \cdot \mu(\bracs{f = y}) + \sum_{y \in g(X)}y \cdot \mu(\bracs{g = y}) \\
@@ -53,12 +56,15 @@
     \[
     \one_A \cdot f = \sum_{y \in f(X)}y \cdot \one_{A \cap \bracs{f = y}}
     \]
+
     so by (1),
     \[
     \int \one_A \cdot f d\mu = \sum_{y \in f(X)}y \cdot \mu(A \cap \bracs{f = y})
     \]
+
     Since $\mu$ is countably additive, for any $\seq{E_n} \subset \cm$ pairwise disjoint with $E = \bigcup_{n \in \natp}E_n$,
     \[
     \int \one_E \cdot f d\mu = \sum_{y \in f(X)}y \cdot \one_{E \cap \bracs{f = y}} = \sum_{y \in f(X)}\sum_{n \in \natp}y \cdot \one_{E_n \cap \bracs{f = y}} = \sum_{n \in \natp}\int \one_{E_n} \cdot f d\mu
     \]
+
 \end{proof}