 # Let w : <mrow class="MJX-TeXAtom-ORD"> <mi mathvariant="double-struck">N </mrow> Armeninilu 2022-07-01 Answered
Let $w:\mathbb{N}\to \left[0,\mathrm{\infty }\right)$ continuous.
For each $f:\mathbb{N}\to \mathbb{C}$ such that $\sum _{n=1}^{\mathrm{\infty }}w\left(n\right)f\left(n\right)$ is absolutely convergent we define $\mathrm{\Lambda }f=\sum _{n=1}^{\mathrm{\infty }}w\left(n\right)f\left(n\right)$
It is easy to prove that $\mathrm{\Lambda }$ is linear and satisfies that $\mathrm{\forall }f:f\left(\mathbb{N}\right)\subset \left[0,\mathrm{\infty }\right)⇒\mathrm{\Lambda }f\in \left[0,\mathrm{\infty }\right)$
(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 $\mathrm{\Lambda }f={\int }_{\mathbb{N}}fd\nu$. How can I find the measure $\nu$ that fulfills this property?
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If $\mu$ is the counting measure, consider $\nu$ defined by $d\nu =wd\mu$. This is a positive measure since $w\ge 0$. Thus $\int fd\nu =\int fwd\mu =\sum _{n}f\left(n\right)w\left(n\right)$

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Compute for $E\subseteq \mathbb{N}$
$\nu \left(E\right)=\int {\mathbf{1}}_{E}d\nu =\mathrm{\Lambda }\left({\mathbf{1}}_{E}\right)=\sum _{n}w\left(n\right){\mathbf{1}}_{E}\left(n\right)=\sum _{n\in E}w\left(n\right).$

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