Example of countably additive function from a Boolean algebra to [0,infinity] that is not finitely additive

Question

This example is implied to exist by Tao’s undergrad measure theory text, exercise 1.7.4, with the claim that further assuming that the empty set getting mapped to 0 prevents such a function from existing. As far as I can tell, padding out the countable union with empty sets forces any other set to be mapped to infinity if the image of the empty set is nonzero (and thus infinite). But the constant infinity map is finitely additive, right? What am I missing here?

Answer 1

That book is online:

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&ved=2ahUKEwj_yb6Xlo_hAhWT51QKHU1YAHIQFjABegQIBBAC&url=https%3A%2F%2Fterrytao.files.wordpress.com%2F2012%2F12%2Fgsm-126-tao5-measure-book.pdf&usg=AOvVaw0GnO1oyFyno__7TnnfBVCl

The Exercise 1.7.4 does not imply existence of function that is countably additive but not finitely additive. Eric's comment on the precise definition of finitely additive is the key here.

Finitely additive

Definition 1.4.19 gives: With $\mathcal{B}$ a Boolean algebra, a map $\mu:\mathcal{B}\rightarrow [0,\infty]$ is finitely additive if

(Axiom 1) $\mu(\phi)=0$.

(Axiom 2) $\mu(A \cup B) = \mu(A) + \mu(B)$ whenever $A, B$ are disjoint and in $\mathcal{B}$.

Countably additive

The defintion of countably additive (Definition 1.4.27) keeps Axiom 1 but changes Axiom 2 to treat countable unions of disjoint sets (it also assumes $\mathcal{B}$ is a sigma-algebra rather than an algebra).

Exercise 1.7.4a

The exercise 1.7.4a just wants you to show that if $\mathcal{B}$ is a Boolean algebra and if $\mu:\mathcal{B} \rightarrow [0,\infty]$ a function, then $\mu$ is finitely additive whenever the following two properties hold:

(i) $\mu(\phi)=0$.

(ii) $\mu(\cup_{i=1}^{\infty}A_i)= \sum_{i=1}^{\infty}\mu(A_i)$ whenever $\{A_i\}_{i=1}^{\infty}$ are disjoint sets in the Boolean algebra such that $\cup_{i=1}^{\infty} A_i$ is also in the Boolean algebra.

That exercise was not trying to imagine a scenario where $\mu(\phi)>0$ (that would not fit the definition of either finitely additive or countably additive).