Measures on all subsets of $\aleph_0$

Question

A theorem of Ulam says:

A finite measure $\mu$ defined on all subsets of a set of cardinality $\aleph_1$ must be $0$ for all subsets if it sends every $1$-element subset to $0$.

Will this statement hold if we substitute $\aleph_0$ for $\aleph_1$ and require the measure to be finitely additive only? In particular, could we define a nontrivial finite measure on the set of all finite graphs that assigns each finite set a measure $0$?

Answer 1

The finite additive measures are somehow easy. There exists a non-trivial finite additive measure on any infinite set, which is $0$ on any finite subset. What makes all the problems with measures is countable additivity.

The key to the solution is the word `ultrafilter'.

A filter on set $\aleph_0$ is a system of sets $\cal F$ (system of some `big' sets, in our case that will be sets of non-zero measure) which satisfies

1. if $A\in \cal F$, and $B\supseteq A$ then $B\in \cal F$,
2. if $A,B\in \cal F$ then $A\cap B\in \cal F$,
3. $\aleph_0\in \cal F$, and $\emptyset \neq \cal F$.

An example of such system is for example the system of all infinite subsets of $\aleph_0$.

A filter is an ultrafilter if it is maximal in respect to inclusion, i.e. we have to add one condition: `for every $A$, either $A \in \cal F$, or $\aleph_0 \setminus A \in \cal F$'. By axiom of choice (Zorn's lemma) we know that every filter can be extended to an ultrafilter.

So fix an ultrafilter $\cal F$ which extends the system of all infinite subsets of $\aleph_0$. Then we can finally define the measure as: $$ \mu(A) = \cases{ 1 & if $A\in \cal F$,\cr 0 & if $A\notin \cal F$.\cr } $$

You can check that this is a finitely additive measure. Well, all the dificult steps have been done in the ultrafilter notation which can be very confusing. If you want to understand ultrafilters better, you can solve the following exercise:

Given a finitely additive measure $\mu$ defined on all subsets of $X$, such that $\mu(X) = 1$, show that the systems of sets $\{ A : \mu(A) = 1 \}$, and $\{ A : \mu(A) \neq 0\}$ are filters on $X$.

There are also many other filters on $X$ that can be defined by the measure $\mu$, unless $\mu$ gives just values $0$, or $1$, in that case there is only one such filter, which is also an ultrafilter.