Is there a probability measure $\mu$ on $\mathbb R^n$, with all Borel sets having a measure in $\{0,1\}$ and all singleton sets having measure 0? My guess would be no, as I'm failing to find an example of such a measure. I tried the following to prove that there is no such measure:
Attempt: We can partition $\mathbb R^n$ into countably many subsets. By sigma-additivity, we find that only a single set will have measure 1, and all other will have measure 0.
I want to use this to prove that by partitioning $\mathbb R^n$ in a "finer and finer" way, we can eventually find smaller and smaller sets, of which only a single one can have nonzero measure, so eventually we find a singleton set that would have nonzero measure.
I'm not sure if this is the right approach to this problem, nor how it would be possible to put my idea in rigorous terms.
 A: Suppose that such measure exists. Then taking countable cover by open sets of diameter $\leq 1$, one of those sets, say $U_1$, must have measure $1$. Cover $U_1$ by countable amount of open sets of diameter $\leq 1/2$ and take the one having measure $1$.
This process gives us a decreasing sequence of open sets $U_k$ with diameter $\leq 1/k$ and $\mu(U_k) = 1$.
Then $\mu(\bigcap_{k = 1}^\infty U_k) = \lim_{k\to \infty} \mu(U_k) = 1$ from continuity of measure, but $\bigcap_{k = 1}^\infty U_k$ has at most one element.
A: The answer is no. The following proof readily generalizes to arbitrary dimension but becomes slightly more tedious to write down:
Proposition
Let $\mu$ be a $\sigma$-finite Borel measure on $\mathbb{R}$ which takes values in $\mathbb{N}\cup \{0\}$. Then, $\mu$ is a linear combination of Dirac-masses.
Proof: Let $I_{n,k}=[k2^{-n}, (k+1)2^{-n}]$ and notice that for every sequence of intervals $J_m:=I_{m,k_m}$ such that $J_m\supseteq J_{m+1}$, we have that $\mu(J_m)$ is decreasing and hence, eventually constant. Furthermore, every such $(J_m)_{m\in \mathbb{N}}$ defines an $x\in \mathbb{R}$ as the unique element of $\cap_{m\in \mathbb{N}} J_m$. Every $x$ is defined in this way by at most two different such sequences. As such, let $J_m^x$ denote the unique such sequence for which $x<\sup J_m^x$ eventually.
If there is an interval $[-K,K]$ containing infinitely many $x_n$ such that the corresponding $J_m^{x_n}$ all have positive measure, then by compactness, these $x_n$ have a point of condensation point $x_{\infty}$ and so, for every $\varepsilon>0$, we have $\mu([x_{\infty}-\varepsilon,x_{\infty}+\varepsilon])=\infty$ (since it contains arbitrarily many distinct $J_m^{x_n}$ that all have measure at least $1$). This contradicts the assumption of $\sigma$-finiteness.
To finish, let $(x_n)_{n\in I}$ denote the discrete set of points such that $(\mu(J^{x_n}_m))_{m\in \mathbb{N}}$ is not eventually $0$. Then, applying the uniqueness theorem for $\sigma$-finite measures, we get
$$
\mu=\sum_{n\in I} a_n \delta_{x_n},
$$
where $a_n=\lim_{m\to\infty} \mu(J^{x_n}_m)$.
