omvamen71

## Answered question

2022-09-07

Let $\alpha \left(t\right)$ be a non-decreasing function on B and consider the integral
${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-xt}d\alpha \left(t\right)$
absolutely convergent in I.
Does exist a measure ${\mu }_{\alpha }$ related to the non-decreasing function $\alpha \left(t\right)$ and how it can be constructed?

### Answer & Explanation

pereishen9g

Beginner2022-09-08Added 7 answers

The unboundedness of the interval does not make much difference: we can still associate measures to nondecreasing functions. Indeed, the intervals [a,b) with $-\mathrm{\infty } form a semiring which generates the Borel $\sigma$-algebra on $\mathbb{R}$. Given $\alpha$, define the pre-measure ${\mu }_{\alpha }\left(\left[a,b\right)\right)=\alpha \left(b-\right)-\alpha \left(a-\right)$. Then extend it by the Carathéodory's extension theorem to a Borel measure.
Alternatively, you can try a gluing argument. On each interval (−n,n) we have a measure ${\mu }_{n}$ constructed from $\alpha$. These measures are consistent in the sense that ${\mu }_{n}\left(A\right)={\mu }_{m}\left(A\right)$ if $A\subseteq \left(-n,n\right)$ and $n. It takes some work to verify that $\mu \left(A\right)=\underset{n\to \mathrm{\infty }}{lim}{\mu }_{n}\left(A\cap \left(-n,n\right)\right)$ is a Borel measure on $\mathbb{R}$

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