add proper cupdots
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+16
@@ -162,6 +162,22 @@
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% contradiction
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\newcommand{\contr}{\text{\Large\lightning}}
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% disjoint unions: provides cupdot and bigcupdot
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\makeatletter
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\def\moverlay{\mathpalette\mov@rlay}
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\def\mov@rlay#1#2{\leavevmode\vtop{%
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\baselineskip\z@skip \lineskiplimit-\maxdimen
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\ialign{\hfil$\m@th#1##$\hfil\cr#2\crcr}}}
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\newcommand{\charfusion}[3][\mathord]{
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#1{\ifx#1\mathop\vphantom{#2}\fi
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\mathpalette\mov@rlay{#2\cr#3}
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}
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\ifx#1\mathop\expandafter\displaylimits\fi}
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\makeatother
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\newcommand{\cupdot}{\charfusion[\mathbin]{\cup}{\cdot}}
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\newcommand{\bigcupdot}{\charfusion[\mathop]{\bigcup}{\cdot}}
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\ExplSyntaxOn
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% S-tackrelcompatible ALIGN environment
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@@ -177,7 +177,7 @@
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$\forall i, j \in \N$ mit $i \neq j$. Damit folgt wegen
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$\mathcal{D}$ Dynkinsystem
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\[
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\bigcup_{i \in \N} A_i = \mathop{\dot{\bigcup_{i \in \N}}} B_i \in \mathscr{D}
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\bigcup_{i \in \N} A_i = \bigcupdot_{i \in \N} B_i \in \mathscr{D}
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.\]
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\end{enumerate}
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\end{proof}
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@@ -191,16 +191,17 @@
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\item Sei $A \in \mathscr{H}(D)$. Dann ist $A \cap D \in \mathscr{D}_0$. Da
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$\mathscr{D}_0$ Dynkinsystem folgt:
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\begin{align*}
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A^{c} \cap D = (X \setminus A) \cap D = (X \cap D) \setminus (A \cap D)
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= \left( (X \cap D)^{c} \mathop{\dot{\cup}} (A \cap D) \right)^{c}
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A^{c} \cap D
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= D \setminus (A \cap D)
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= \left( D^{c} \cupdot (A \cap D) \right)^{c}
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\in \mathscr{D}_0
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.\end{align*}
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\item Sei $A_i \in \mathscr{H}(D)$ $\forall i \in \N$ mit $A_i \cap A_j = \emptyset$
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$\forall i, j \in \N, i \neq j$. Dann folgt direkt, da die $A_i$ paarweise
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disjunkt sind und $\mathscr{D}_0$ Dynkinsystem:
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\[
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\left( \bigcup_{i \in \N} A_i \right) \cap D
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= \mathop{\dot{\bigcup_{i \in \N}}} (\underbrace{A_i \cap D}_{ \in \mathscr{D}_0})
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\left( \bigcupdot_{i \in \N} A_i \right) \cap D
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= \bigcupdot_{i \in \N} (\underbrace{A_i \cap D}_{ \in \mathscr{D}_0})
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\in \mathscr{D}_0
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.\]
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\end{enumerate}
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