Non-measurable sets on $\mathbb{N}$ 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 $2^{\mathbb{N}}$?
 A: Let $(\mathbb N, \mathcal F, \mu)$ be a measure space. Consider for each $x \in \mathcal F$ the set 
$$
\mathcal F_x = \underset{x \in S}{\bigcap_{S \in \mathcal F}} S.
$$
We can take this intersection to be countable, since for each $y \in \mathcal F_x$, we can pick one $S_y \in \mathcal F$ with $x,y \in \mathcal F$ and and for each $z \in \mathbb N \backslash \mathcal F_x$, we pick one $S_z \in \mathcal F$ with $\mathcal F_x \subseteq S \not\ni z$. The intersection over all $S_y$ and $S_z$ is countable and equals $\mathcal F_x$, so $\mathcal F_x \in \mathcal F$. 
For each $x,y \in \mathbb N$, either $\mathcal F_x = \mathcal F_y$ or $\mathcal F_x \cap \mathcal F_y = \varnothing$. For suppose $\mathcal F_x \neq \mathcal F_y$, so that there exists $S \in \mathcal F$ with $x \in \mathcal F$ and $y \notin \mathcal F$ (we can assume this without loss of generality, otherwise reverse the roles of $x$ and $y$). It follows that $\mathcal F_y \subseteq \mathbb N \backslash S$ by definition of $\mathcal F_y$ since $y \in \mathbb N \backslash S \in \mathcal F$, so $\mathcal F_x \cap \mathcal F_y = \varnothing$. 
So the sets $\{ \mathcal F_x \, | \, x \in \mathbb N \}$ form a partition of $\mathbb N$. Write $\{\mathcal F_1,\cdots, \} = \{ \mathcal F_i \, | \, i \in I \}$ for the set of sets that partition $\mathbb N$ ($I$ could be finite or countably infinite), and let $\{x_1,\cdots,\}$ be a sequence (might be finite or infinite) of elements with the property that $x_i \in \mathcal F_i$. By assumption, for all $y \in \mathcal F_i$, $\mathcal F_y = \mathcal F_i$, so $\mu(\{y\})$ is defined if and only if $\{y\} = \mathcal F_i = \{x_i\}$. 
Since you wish to assume $\mu$ finite, we have $0 \le \mu(\mathcal F_i) \le \mu(\mathbb N) < \infty$, so there are actually plenty of ways to extend $\mu$ to a measure to all of $\mathbb N$. Your options go like this. If you give $\mathcal F_i$ the measure subspace structure, you obtain a space $(\mathcal F_i, \{ \varnothing, \mathcal F_i \}, \mu)$ where $\mu(\varnothing) = 0$ and $\mu(\mathcal F_i)$ is what it was in $(\mathbb N, \mathcal F,\mu)$. So you extend $(\mathcal F_i, \{ \varnothing, \mathcal F_i \}, \mu)$ to $(\mathcal F_i, \mathcal P(\mathcal F_i), \mu_i)$ by assigning values $\mu(\{y\})$ to each $y \in \mathcal F_i$ such that $\mu_i(\{y\}) \ge 0$ for each $y$ and $\sum_{y \in \mathcal F_i} \mu_i(\{y\}) = \mu(\mathcal F_i)$. Then you let $(\mathbb N, \mathcal P(\mathbb N), \widetilde{\mu})$ be a measure space defined by 
$$
\widetilde{\mu}(A) \overset{def}= \sum_{i \in I} \mu_i(A \cap \mathcal F_i). 
$$
Note that $\sum_{n=1}^{\infty} \widetilde{\mu}(\{n\})$ is then absolutely convergent since in positive series, we can re-arrange the terms in any order we want, so 
$$
\sum_{n=1}^{\infty} \widetilde{\mu}(\{n\}) = \sum_{i \in I} \widetilde{\mu}(\mathcal F_i) = \sum_{i \in I} \mu(\mathcal F_i) = \mu(\mathbb N) < \infty.
$$
This extension will always work, even in the non-finite case, but of course in the non-finite case the extension will also not be finite. Also, since you need to fix new values for originally non-measurable subsets, you have uniqueness if and only if $\mathcal F = \mathcal P(\mathbb N)$, so essentially in any non-trivial case the extension is not unique.
Hope that helps,
A: 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.
