We have the following indirect implication of form equivalence classes:

310 \(\Rightarrow\) 310
given by the following sequence of implications, with a reference to its direct proof:

Implication Reference
310 \(\Rightarrow\) 310

Here are the links and statements of the form equivalence classes referenced above:

Howard-Rubin Number Statement
310:

The Measure Extension Theorem: Suppose that \(\cal A_0\) is a subring (that is, \(a,b \in \cal A_0  \to a\vee b \in \cal A_0\) and \(a-b \in \cal A_0\)) of a Boolean algebra \(\cal A\) and \(\mu\) is a measure on \(\cal A_0\) (that is, \(\mu:\cal A \to [0,\infty]\), \(\mu(a\vee b) =\mu(a)+\mu(b)\) for \(a\land b = 0\), and \(\mu(0) = 0\).) then there is a measure on \(\cal A\) that extends \(\mu\).

310:

The Measure Extension Theorem: Suppose that \(\cal A_0\) is a subring (that is, \(a,b \in \cal A_0  \to a\vee b \in \cal A_0\) and \(a-b \in \cal A_0\)) of a Boolean algebra \(\cal A\) and \(\mu\) is a measure on \(\cal A_0\) (that is, \(\mu:\cal A \to [0,\infty]\), \(\mu(a\vee b) =\mu(a)+\mu(b)\) for \(a\land b = 0\), and \(\mu(0) = 0\).) then there is a measure on \(\cal A\) that extends \(\mu\).

Comment:

Back