Support of sum of Dirac measures

Question

Let $\mu$ be the measure on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$. Find $\operatorname{supp}(\mu)$, where $\mu=\sum_{k\geq 1}a_k\delta_{x_k}$ and $(x_k)_{k\geq 1}$ is an arbitrary sequence of real numbers.

Since $\operatorname{supp}(\mu)$ is the collection of closed sets whose complement has measure zero, wouldn't we have that $\operatorname{supp}(\mu)=\{x_k\}$ for $k\geq 1$?

This question seems too trivial to me, but I am afraid I might be missing something.

I'd appreciate if anyone can point out where I am wrong or if I am correct.

Thanks.

Answer 1

If $a_k \geq 0$ for all $k$ then the support is exactly the closure $E$ of $\{x_k: a_k >0\}$. Proof: By ignoring the zero terms this proof reduces to the case where $a_k >0$ for all $k$. Since $\mu \{x_k\} >0$ for each $k$ it follows that each $x_k$ is in the support. Since the support is closed (by definition ) it follows that $E$ is contained in the support. On the other hand $\mu(E^{c})=0$ (since each term in the sum is $0$) so the support is also contained in $E$.

Answer 2

Suppose all $a_k >0$ and the $x_k$ enumerate the rationals. Then supp$(\mu)=\mathbb R.$ That's because there is no nonempty subset $U$ of $\mathbb R$ such that $\mu(U)=0.$