Measure of Stieltjes Function of Measure

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Theorem

Let $\mu$ be a measure on $\mathcal B \left({\R}\right)$, the Borel $\sigma$-algebra on $\R$.

Suppose that for all $n \in \N$, $\mu$ satisfies:

$\mu \left({ \left[{-n \,.\,.\, n}\right) \ }\right) < +\infty$


Let $f_\mu$ be the Stieltjes function of $\mu$.

Let $\mu_{f_\mu}$ be the measure of $f_\mu$.


Then $\mu_{f_\mu} = \mu$.


Proof

From Pre-Measure of Finite Stieltjes Function Extends to Unique Measure, it suffices to verify that:

$\mu_{f_\mu} \left({\left[{a \,.\,.\, b}\right)}\right) = \mu \left({\left[{a \,.\,.\, b}\right)}\right)$

for all half-open intervals $\left[{a \,.\,.\, b}\right)$.


Now we have:

\(\displaystyle \mu_{f_\mu} \left({\left[{a \,.\,.\, b}\right)}\right)\) \(=\) \(\displaystyle f_\mu \left({b}\right) - f_\mu \left({a}\right)\) Definition of $\mu_{f_\mu}$

If either $a = 0$ or $b = 0$, the result follows immediately from the definition of $f_\mu$.


Now suppose that $a < b < 0$.

Then:

\(\displaystyle f_\mu \left({b}\right) - f_\mu \left({a}\right)\) \(=\) \(\displaystyle \mu \left({\left[{a \,.\,.\, 0}\right)}\right) - \mu \left({\left[{b \,.\,.\, 0}\right)}\right)\) Definition of $f_\mu$
\(\displaystyle \) \(=\) \(\displaystyle \mu \left({\left[{a \,.\,.\, b}\right)}\right)\) Measure of Set Difference with Subset


Finally, let $0 < a < b$.

Then:

\(\displaystyle f_\mu \left({b}\right) - f_\mu \left({a}\right)\) \(=\) \(\displaystyle \mu \left({\left[{0 \,.\,.\, b}\right)}\right) - \mu \left({\left[{0 \,.\,.\, a}\right)}\right)\) Definition of $f_\mu$
\(\displaystyle \) \(=\) \(\displaystyle \mu \left({\left[{a \,.\,.\, b}\right)}\right)\) Measure of Set Difference with Subset


The final case $a < 0 < b$ is a trivial consequence of Measure is Finitely Additive Function.

Hence it must be that $\mu_{f_\mu} = \mu$.

$\blacksquare$


Sources