Every measure on a countable discrete space is its weighted sum of Dirac measures

If $X=\mathrm{\varnothing }$, then $\mu$ is the zero measure and the asserted expression is the empty weighted sum.

If $X$ is finite and nonempty, choose a bijection $e:n\to X$ for some $n\ge 1$. Every $E\subseteq X$ is the finite disjoint union of the singletons $\{e(i)\}$ with $e(i)\in E$, so $\mu (E)={\sum }_{i<n,e(i)\in E}\mu (\{e(i)\})$.

If $X$ is countably infinite, choose a bijection $e:\mathbb{N}\to X$. Every $E\subseteq X$ is the disjoint union of the sequence whose $k$-th term is $\{e(k)\}$ when $e(k)\in E$ and $\mathrm{\varnothing }$ otherwise, so $\mu (E)={\sum }_{k:e(k)\in E}\mu (\{e(k)\})$.

Evaluating any asserted representation at the singleton $\{x\}$ leaves only the Dirac term at $x$, so its coefficient must be $\mu (\{x\})$.

In the finite case, the weighted Dirac sum with coefficients $w_{e(i)}=\mu (\{e(i)\})$ has the value computed in step 1.2 on every $E$; the explicit positive-infinity branch gives $0$ off $e(i)$ and $+\mathrm{\infty }$ on sets containing it.

In the countably infinite case, the countable weighted Dirac sum has the value computed in step 1.3 on every $E$, with the same interpretation of infinite coefficients.

Steps 1.1, 2.1 and 2.2 prove the representation in the empty, finite nonempty, and countably infinite cases, and step 1.4 proves uniqueness of every coefficient.