Prove inequality of measure in boolean algebra

Question

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 $b \wedge c > 0$?

Answer 1

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 $b'\wedge c \leq c$.