If $\mu: B \rightarrow [0,1]$ is a measure on a boolean algebra $\mathcal{B}$, What is the best way to prove that $\mu(b \vee c) \leq \mu(b) + \mu(c)$? If $b \wedge c = 0$, this is an equality. What about when $b \wedge c > 0$?
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I'm using this definition of measure on a Boolean Algebra.
Notice that, in a Boolean Algebra, \begin{equation} x < y \;\Longrightarrow \; x \wedge (x' \wedge y) = 0 \text{ and } x \vee (x' \wedge y) = y.\tag{$*$} \end{equation}
Let us see that $\mu$ is order-preserving.
If $x \leq y$, then $$\mu(y) = \mu(x \vee y) = \mu(x \vee (x' \wedge y)) =_{\because (*)} \mu(x) + \mu(x' \wedge y) \geq \mu(x),$$
where the inequality follows from $\mu(x' \wedge y) \geq 0$.
Now, $$\mu(b \vee c) = \mu(b \vee (b' \wedge c)) =_{\because (*)} \mu(b) + \mu(b' \wedge c) \leq \mu(b) + \mu(c),$$ where the inequality follows from the $b'\wedge c \leq c$.
amrsa
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The step I'm missing is how do you get $x \leq y \Leftrightarrow x = x \wedge y \textrm{ (by def) } \Leftrightarrow y = x \vee y$? (I wonder about the second equivalence). – Ordinalz Apr 16 '17 at 23:43
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In your question, you make use of the binary operations, $\wedge$ and $\vee$ and also of the order relation. So I suppose you know about the connection between these. The operations may be defined in terms of the order relation by $x \wedge y = \inf{x,y}$ and $x \vee y = \sup{x,y}$. So, if $x \leq y$, then what are $x \wedge y$ and $x \vee y$. – amrsa Apr 17 '17 at 08:44
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Conversely, $x \wedge y \leq x$ and $x \leq x \vee y$, so if $x = x \wedge y$, then $x \leq y$, yielding $x \vee y = y$. – amrsa Apr 17 '17 at 08:45
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If that isn't enough, and since you seem to understand the first equivalence, I suppose this post can help you. – amrsa Apr 17 '17 at 09:36