I found an interesting exercise in the preprint of Measure, integration and real analysis of Sheldon Axler. Exercise 12 of section 2C says:
Suppose $X$ is a set and $\mathcal{S}$ is the $\sigma $-algebra of all subsets $E$ of $X$ such that $E$ or $X\setminus E$ is countable. Give a complete description of the set of all measures in $(X,\mathcal{S})$.
My try: in first place every singleton is measurable so for every measure $\mu$ on $(X,\mathcal{S})$ there is a function $f_\mu :X\to [0,\infty ]$ such that
$$
\mu (\{x\})=f_\mu (x),\quad \text{ for each }x\in X\tag1
$$
and note that if $A$ is a countable set then
$$
\mu (A)=\sum_{x\in A}f_\mu (x)\tag2
$$
so it seems, at first glance, that the sets of all measures in $(X,\mathcal{S})$ can be represented by the set of functions $[0,\infty ]^X$, and indeed this would be the case if $X$ is countable.
Now I will try to go further describing more precisely any measure on $(X,\mathcal{S})$, so let $P:=\{x\in X: f_\mu (x)>0\}$ and assume that $X$ is uncountable.
Case 1: $P$ is countable and $\mu (P)<\infty $, hence $\mu (P^\complement )=\mu (X)-\mu (P)$ so the measure of $P^\complement $ is determined by choosing a value for $\mu (X)\in[\mu (P),\infty ]$.
Case 2: $P$ is countable and $\mu (P)=\infty $, hence $\mu (X )=\infty $ and the measure of $P^\complement $ can be chosen arbitrarily in $[0,\infty]$.
Case 3: $P$ is uncountable what implies that if $P$ is measurable then $\mu (P)=\mu (X)=\infty $, because it can be shown that exists some $\epsilon >0$ such that $f_\mu(P)\cap (\epsilon ,\infty )$ is uncountable. In any case, being $P$ measurable or not, this means that every uncountable measurable set have infinite measure because it have an uncountable subset such that every singleton have positive measure.
My questions:
-
Is this characterization correct?
-
For the case 2 I dont have a proof about the consistency of giving to $\mu(P^\complement)$ an arbitrary value on $[0,\infty]$. If this would be correct, how I can prove it rigorously?
Best Answer
Hard to say whether your characterization is correct, because I don't quite see a "characterization" here: In the first two cases you don't say what $\mu(E)$ is for an arbitrary co-countable set $E$.
Assume $X$ is uncountable and define $P$ as above.
Case 3 is exactly right; if $P$ is uncountable then $P\cap E$ is uncountable for every co-countable $E$, so $\mu(E)=\infty$. Note that in fact in this case we can say simply $\mu(E)=\sum_{x\in E}f_\mu(x)$ for every measurable $E$.
(Because if $E_1,\dots$ are disjoint measurable sets then all but at most one $E_j$ is countable.)
Because if $E$ is co-countable then $X=E\cup N$, $\mu(N)=0$.
Now assume $P$ is countable. Then $P$ is measurable, so we can define measures $\mu^1$ and $\mu^2$ by $\mu^1(E)=\mu(E\cap P)$, $\mu^2(E)=\mu(E\setminus P)$. Note that $E\subset X$ is a co-countable subset of $X$ is and only if $E\setminus P$ is a co-countable subset of $X\setminus P$; so the lemma, with $X\setminus P$ in place of $X$, shows there exists $\alpha\in[0,\infty]$ with $\mu^2=\mu_\alpha$. Of course we know what $\mu^1$ is, and $\mu=\mu^1+\mu^2$.
Define $I(E)=0$ if $E$ is countable, $1$ if $E$ is co-countable. It seems to me that if you consider both cases treated above we've shown that in any case
(And of course conversely, for any $\alpha\in[0,\infty]$ and $f:X\to[0,\infty]$ the expression $\alpha I(E)+\sum_{x\in E}f(x)$ definies a measure.)