Let $a:=(a_{n})_{n\in\mathbb{N}}$ be a sequence of real numbers such that $a_{n}>0$ for all $n\in\mathbb{N}$. Is there a measure $\mu$ on $\mathbb{N}$ such that $$\int_{\mathbb{N}}f \ \text{d}\mu = \sum_{n\in\mathbb{N}}a_{n}f(n)?$$
It looks like a weighted version of the counting measure, so I think it does indeed exist. However, I don't see how I can explicitly construct $\mu$ in terms of $a$.
Any suggestions are greatly appreciated!
Of course you can. Just let $\mu = \sum_{n} a_n \delta_n$ where $\delta_n$ is the point mass at $n \in \mathbb N$ where $0 \leq a_n \leq \infty$. Then for each non-negative $\overline {\mathbb R}$-valued sequence $f$, you have $\int f d \mu = \sum a_n f(n)$ and for any sequence, it is integrable if and only if $\sum|a_n f(n)| $ is finite (and the integral is indeed given by the expected formula).