Let $B$ be a boolean algebra. Suppose we have a finitely additive measure $\mu$ defines on a subalgebra $A\subseteq B$.

Is it possible to extend $\mu$ to a finitely additive measure $\nu$ on

$B$? Is there a related theorem about extension of measures?

## Best Answer

Yes!... But maybe No! :P

This is theorem $3.2.5$ in Rao and Rao's

Theory of Charges: A Study of Finitely Additive Measures:The authors use "charge" to mean a finitely additive measure, and this theorem is slightly more general than you need. A partial charge defined on a ring of sets is already a charge.

Unfortunately, the authors allow general charges to be signed, and their proof of this theorem crucially uses this fact. Even if $\mu$ is a positive charge, it's possible that the extension $\overline{\mu}$ will be signed. Alternatively, we can always find a positive extension $\overline{\mu}$, but we must allow it to take infinite values.

However, there are circumstances where we can guarantee a positive charge admits a finite positive extension. See, for instance theorem $3.2.9$:

The proofs are not

hard, but they are somewhat long and technical, otherwise I would include them. The sketch is to see how to add a single set $A$ to $\mathscr{C}$ and choose what $\mu(A)$ should be in a way that's consistent. Then we (transfinitely) induct on all the sets in $\mathscr{F} \setminus \mathscr{C}$ in order to (eventually) get an extension on all of $\mathscr{F}$ that's consistent.The authors actually describe

lotsof related theorems on how to extend charges off of a subalgebra, and it might be worth reading the whole chapter to get a result that's more tailored to your specific use case. The book isextremelyreadable, so you shouldn't have many issues flipping through it!I hope this helps ^_^