Generating a positive measure $\nu$ such that $\displaystyle\sum_{n=1}^{\infty}w(n)f(n)=\int_{\mathbb N}fd\nu$

Question

Let $w:\mathbb N\to [0,\infty)$ continuous.

For each $f:\mathbb N\to\mathbb C$ such that $\displaystyle\sum_{n=1}^{\infty}w(n)f(n)$ is absolutely convergent we define $\Lambda f=\displaystyle\sum_{n=1}^{\infty}w(n)f(n)$

It is easy to prove that $\Lambda$ is linear and satisfies that $\forall f:f(\mathbb N)\subset [0,\infty)\Rightarrow\Lambda f \in[0,\infty)$

(I think this last property has a name but I don't know what it is)

By the Riesz representation theorem there is only one positive measure $\nu$ such that $\Lambda f=\displaystyle\int_{\mathbb N}fd\nu$. How can I find the measure $\nu$ that fulfills this property?

Answer 1

If $\mu$ is the counting measure, consider $\nu$ defined by $d\nu = w d \mu$. This is a positive measure since $w \geq 0$. Thus $\int f d \nu = \int fw d \mu = \sum_{n} f(n)w(n)$

Answer 2

Compute for $E \subseteq \mathbb N$ $$ \nu(E) = \int \mathbf1_E d\nu = \Lambda(\mathbf1_E) =\sum_n w(n)\mathbf1_E(n) = \sum_{n \in E} w(n) . $$