Borel measure supported on $\mathbb{Q}$ Let $\mu$ be a Borel measure supported on $\mathbb{Q} \subset \mathbb{R}$. Must $\mu$ be a sum of Dirac measures?
 A: Yes, because by definition of the support (depending on your definition), you have $\mu ( \Bbb{R} \setminus \Bbb{Q}) \leq \mu(\Bbb{R} \setminus {\rm supp}(mu)) = 0$ and thus for every measurable set $M$:
$$
\mu(M) = \mu(M \cap \Bbb{Q}) = \sum_{q \in \Bbb{Q} \cap M} \mu({q}) = (\sum_{q \in \Bbb{Q}} \mu({q}) \delta_q ) (M),
$$
so that $\mu$ is a (countable, nonnegative) linear combination of Dirac measures. 
If you meant a finite sum, this is not correct, take e.g. $\sum_n 1/n^2 \delta_{1/n}$. 
A: I guess you mean infinite sums. The following statement holds true, which you may think of more general (or not). If $A$ is a countable set, then any positive, say $\sigma$-finite measure defined on $2^A$ is a sum of Dirac measure. This follows directly from $\sigma$-additivity of the measure. Consider any such measure $\mu$ on $A$, and define $\mu_a:=\mu(\{a\})$ for each $a\in A$. Then $\mu(B) = \mu(\bigcup_{a\in B}\{a\}) = \sum_{a\in B} \mu_a$ for every subset $B$ of $A$. Clearly, the very same expression you'll get for a measure $\sum_{a\in A}\mu_a\delta_a$, so the two measures are equal, so $\mu$ is a sum of Dirac measures.
A somewhat converse statement is also true. If $\mu$ is a positive $\sigma$-finite measure on $\Bbb R$ and $\mu$ is a sum of Dirac measures $\mu = \sum_{r\in \Bbb R}\mu_r\delta_r$, then $\{r\in \Bbb R:\mu_r>0\}$ is countable. To prove this, you represent $\Bbb R$ as a union of $R_n$ on each of which $\mu$ is finite, and show that $\{r\in R_n:\mu_r>0\}$ is countable for each $n$, hence the original set is countable as well as a countable union of countable sets.
