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Let $\alpha(t)$ be a non-decreasing function on $\mathbb{B}$ and consider the integral $$ \int_{-\infty}^{+\infty} e^{-xt}d\alpha(t) $$ absolutely convergent in $I$.

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

I know construction of such a measure using Caratheodory's theorem, but this holds for non-decreasing function on an interval $(a,b)$, what can I say in the case that $(a,b)$ is the real line?

Thank you

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The unboundedness of the interval does not make much difference: we can still associate measures to nondecreasing functions. Indeed, the intervals $[a,b)$ with $-\infty<a\le b< \infty$ 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 $\mu(A)=\lim_{n\to\infty} \mu_n(A\cap (-n,n))$ is a Borel measure on $\mathbb R$.

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  • $\begingroup$ I was studying this construction on the book of Ash-Probability and measure theory. He construct such a measure, but he use a non-decreasing and RIGHT-CONTINUOUS (or left-cont) function. Do you know if without right-continuity is still possible to construct such a measure? Thank you $\endgroup$
    – alemou
    Commented Jul 12, 2013 at 10:06

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