Measure of Stieltjes Function of Measure

Theorem

Let $\mu$ be a measure on $\map \BB \R$, the Borel $\sigma$-algebra on $\R$.

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

$\map \mu {\hointr {-n} n} < +\infty$

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

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

$\map {\mu_{f_\mu} } {\hointr a b} = \map \mu {\hointr a b}$

for all half-open intervals $\hointr a b$.

Now we have:

$\ds \map {\mu_{f_\mu} } {\hointr a b}$

$=$

$\ds \map {f_\mu} b - \map {f_\mu} a$

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

Now suppose that $a < b < 0$.

Then:

$\ds \map {f_\mu} b - \map {f_\mu} a$

$=$

$\ds \map \mu {\hointr a 0} - \map \mu {\hointr b 0}$

$\ds $

$=$

$\ds \map \mu {\hointr a b}$

Finally, let $0 < a < b$.

Then:

$\ds \map {f_\mu} b - \map {f_\mu} a$

$=$

$\ds \map \mu {\hointr 0 b} - \map \mu {\hointr 0 a}$

$\ds $

$=$

$\ds \map \mu {\hointr a b}$

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

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

$\blacksquare$