TITLE: Non-measurable sets on $\mathbb{N}$
QUESTION [7 upvotes]: I'm familiar with the "construction" of non-measurable sets on $\mathbb{R}$. But of interest to me is if there is a way to construct a countably additive probability measure $\mu$ on $\mathbb{N}$ such that we can't extend $\mu$ to $2^{\mathbb{N}}$. Apparently there's no way to do it when $\mu$ is defined on all singletons, as for any set $S \subseteq \mathbb{N}$, we have
\begin{align*}
\mu(S) & = \sum_{k \in S} \mu( \{k \}) \\
& = \sum_{k = 1}^{\infty} \chi_{S} \mu( \{k\} )
\end{align*}
Clearly we have that $\mu(\mathbb{N}) = \sum_{k = 1}^{\infty} \mu ( \{k\}) = 1$, so we know that $\lim_{N \to \infty} \sum_{k = N + 1}^{\infty} \mu( \{ k \} ) = 0$. But then we use this to show that $\mu(S)$ is well-defined, as the sum converges. Let $\epsilon > 0$, and let $N$ be large enough that $\sum_{k = N + 1}^{\infty} \mu( \{k \}) < \epsilon$.
\begin{align*}
\left| \left( \sum_{k \in S} \mu( \{k\}) \right) - \left( \sum_{k = 1}^{N} \mu( \{ k \} ) \right) \right| & = \left| \sum_{k = N + 1}^{\infty} \chi_{S} \mu(\{k\}) \right| \\
& \leq \sum_{k = 1}^{\infty} \| \chi_{S} \mu(\{k\}) \| \\
& \leq \sum_{k = N + 1}^{\infty} \mu( \{k\} ) \\
& < \epsilon .
\end{align*}
So is there a way to define a measure on $\mathbb{N}$ that is finite, but cannot be extended to measure the singletons? Is it possible if we drop the assumption that $\mu$ be finite? Can we show instead that any countably additive measure will extend to all of $2^{\mathbb{N}}$?

REPLY [2 votes]: Let $\sim$ be any equivalence relation on $\mathbb N$. Define a measure on $\mathbb N/\sim$. Then $\mathbb N$ inherits this measure, and the measurable sets are exactly unions of equivalence classes. In particular, if $\sim$ is not the equality relation, then there are non-measurable sets.
For example, we can define $x\sim y$ if $x,y$ have the same prime factors. 
But what this really means is that your probability space has indistinguishable events.
As another answer proves, this is essentially the only sort of measure that can be of this kind. 
Given such a measure, define: $x\sim y$ if every measurable set that contains $x$ contains $y.$ 
This is an equivalence relation since 

If there is a measurable set $S$ containing $y$ but not $x$ then $\mathbb N\setminus S$ is a measurable set containing $x$ not containing $y$.
If $x\sim y$ and $y\sim z$ then any set containing $x$ contains $y$ and hence contains $z$.
$x\sim x$ fairly obviously.

If $A_x$ is the equivalence class of $x$, then $A_x=\int_{y\not\sim x} A_{x,y}$ of measurable sets not containing the elements not equivalent to $x$, and this is a countable intersection.
Such measures can always be extended pretty trivially to $\mathcal P(S)$, by either picking a single element of each set as the "real" element, leaving all others with measure zero, or by divying up each equivalence class's measure amongst the individual elements in some way. If every measurable set has non-zero measure, then you can ensure that every element of $\mathbb N$ has non-zero value.