Distribution Function of Finite Signed Borel Measure is of Bounded Variation

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Theorem

Let $\mu$ be a finite signed Borel measure on $\R$.

Let $F_\mu$ be the distribution function of $\mu$.


Then $F_\mu$ is of bounded variation.


Proof

Let $\SS$ be a non-empty finite subset of $\R$.

Write:

$\SS = \set {x_0, x_1, \ldots, x_n}$

with:

$x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n$

Then, we have:

\(\ds \sum_{i \mathop = 1}^n \size {\map {F_\mu} {x_i} - \map {F_\mu} {x_{i - 1} } }\) \(=\) \(\ds \sum_{i \mathop = 1}^n \size {\map \mu {\hointl {-\infty} {x_i} } - \map \mu {\hointl {-\infty} {x_{i - 1} } } }\) Definition of Distribution Function of Finite Signed Borel Measure
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^n \size {\map \mu {\hointl {-\infty} {x_i} \setminus \hointl {-\infty} {x_{i - 1} } } }\) Measure of Set Difference with Subset: Signed Measure
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^n \size {\map \mu {\hointl {x_{i - 1} } {x_i} } }\)
\(\ds \) \(\le\) \(\ds \size {\map \mu {\hointl {-\infty} {x_0} } } + \sum_{i \mathop = 1}^n \size {\map \mu {\hointl {x_{i - 1} } {x_i} } } + \size {\map \mu {\hointr {x_n} \infty} }\)

Note that:

$\set {\hointl {-\infty} {x_0}, \hointl {x_0} {x_1}, \hointl {x_1} {x_2}, \ldots, \hointl {x_{n - 1} } {x_n}, \hointr {x_n} \infty}$ is a partition of $\R$ into Borel sets.

So from Definition 2 of the variation of a signed measure, we have:

$\ds \size {\map \mu {\hointl {-\infty} {x_0} } } + \sum_{i \mathop = 1}^n \size {\map \mu {\hointl {x_{i - 1} } {x_i} } } + \size {\map \mu {\hointr {x_n} \infty} } \le \map {\size \mu} \R$

where $\size \mu$ is the variation of $\mu$.

From Signed Measure Finite iff Finite Total Variation, we have:

$\map {\size \mu} \R < \infty$

So, using the notation from the definition of bounded variation, for all non-empty finite subset of $\R$, $\SS$, we have:

$\map {V_f^\ast} {\SS; I} \le M = \map {\size \mu} \R$

So $f$ is of bounded variation.

$\blacksquare$


Sources

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