Let $\lambda$ be a positive measure on $(0;+\infty)$ satisfies $\int_{(0,+\infty)} (1+s)^{-1} d \lambda (s) < +\infty$. Does this integral converge?
$$\displaystyle\int_{(0;+\infty)} \dfrac{1}{s} d \lambda(s).$$
$\forall n \ge 0$, consider series open set $A_n = \left(\dfrac{1}{n};n\right)$. Clearly, $A_n$ is descending and $$\bigcap_n A_n = \emptyset.$$ Then, $$0 = \lambda \left( \bigcap_{n=0}^{\infty} A_n \right) = \lim_{n \rightarrow \infty} A_n.$$ Hence, $\exists n \ge 1$ such that $\lambda(A_n) \le \dfrac{1}{2}$. Let $B_n = (0;+\infty) \setminus A_n$, $B_n$ is a positive measure set and $f$ is bounded on $B_n$, therefore $$\displaystyle\int_{B_n} \dfrac{1}{s} d \lambda(s) < +\infty.$$
Is this correct? Thank all!