omvamen71

2022-09-07

Let $\alpha (t)$ be a non-decreasing function on B and consider the integral

$${\int}_{-\mathrm{\infty}}^{+\mathrm{\infty}}{e}^{-xt}d\alpha (t)$$

absolutely convergent in I.

Does exist a measure ${\mu}_{\alpha}$ related to the non-decreasing function $\alpha (t)$ and how it can be constructed?

$${\int}_{-\mathrm{\infty}}^{+\mathrm{\infty}}{e}^{-xt}d\alpha (t)$$

absolutely convergent in I.

Does exist a measure ${\mu}_{\alpha}$ related to the non-decreasing function $\alpha (t)$ and how it can be constructed?

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 $-\mathrm{\infty}<a\le b<\mathrm{\infty}$ form a semiring which generates the Borel $\sigma $-algebra on $\mathbb{R}$. Given $\alpha $, define the pre-measure ${\mu}_{\alpha}([a,b))=\alpha (b-)-\alpha (a-)$. 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 ${\mu}_{n}(A)={\mu}_{m}(A)$ if $A\subseteq (-n,n)$ and $n<m$. It takes some work to verify that $\mu (A)=\underset{n\to \mathrm{\infty}}{lim}{\mu}_{n}(A\cap (-n,n))$ is a Borel measure on $\mathbb{R}$

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 ${\mu}_{n}(A)={\mu}_{m}(A)$ if $A\subseteq (-n,n)$ and $n<m$. It takes some work to verify that $\mu (A)=\underset{n\to \mathrm{\infty}}{lim}{\mu}_{n}(A\cap (-n,n))$ is a Borel measure on $\mathbb{R}$

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My steps:

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