Stieltjes Function of Measure is Stieltjes Function

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Theorem

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

Suppose that for every $n \in \N$:

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


Then $F_\mu: \R \to \overline \R$, the Stieltjes function of $\mu$, is a Stieltjes function.


Proof

By definition, $F_\mu$ is a Stieltjes function if and only if it is increasing and left-continuous.


$F_\mu$ is Increasing

Let $x, y \in \R$ such that $x \le y$.

It is apparent that for all $z \in \R$:

$z \le 0 \implies \map {F_\mu} z \le 0$
$z \ge 0 \implies \map {F_\mu} z \ge 0$

Therefore, only the cases $x \le y \le 0$ and $0 \le x \le y$ remain.


In the first of these cases, note that:

$\hointr y 0 \subseteq \hointr x 0$

Hence from Measure is Monotone, it follows that:

$\map {F_\mu} x = - \map \mu {\hointr x 0} \le - \map \mu {\hointr y 0} = \map {F_\mu} y$


The remaining case is analogous, using the observation that:

$\hointr 0 x \subseteq \hointr 0 y$

Hence $F_\mu$ is increasing.

$\Box$


$F_\mu$ is Left-Continuous

Suppose that $x > 0$.

Suppose further that $x_k \uparrow x$ is an increasing sequence with limit $x$, and that $0 < x_1 < x$.

Then as an increasing sequence of sets, we have:

$\hointr 0 {x_k} \uparrow \hointr 0 x$

Now we compute:

\(\ds \lim_{k \mathop \to \infty} \map {F_\mu} {x_k}\) \(=\) \(\ds \lim_{k \mathop \to \infty} \map \mu {\hointr 0 {x_k} }\) Definition of Stieltjes Function of Measure
\(\ds \) \(=\) \(\ds \map \mu {\hointr 0 x}\) Characterization of Measures: $(3)$
\(\ds \) \(=\) \(\ds \map {F_\mu} x\)


Next, suppose $x = 0$.

Suppose further that $x_k \uparrow 0$ is an increasing sequence with limit $0$.

Now, we have the decreasing sequence of sets $\hointr {x_k} 0 \downarrow \O$.

Reasoning as above, but applying Characterization of Measures: $(3)$ instead yields the result in this case.


Finally, suppose $x < 0$ and $x_k \uparrow x$ is an increasing sequence with limit $x$.

Observe the decreasing sequence of sets $\hointr {x_k} 0 \downarrow \hointr x 0$.

This time, applying Characterization of Measures: $(3')$ yields the result.


Note the additional assumption on $\mu$ was in fact used, in combination with Measure is Monotone.

This is necessary to establish the conditions for the last two applications of Characterization of Measures.


From the Trichotomy Law for Real Numbers and Proof by Cases, the result now follows.

$\blacksquare$


Sources

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