Distribution Function of Finite Signed Borel Measure is of Bounded Variation

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 $\mathcal S$ be a non-empty finite subset of $\R$.

Write:


 * $\mathcal S = \set {x_0, x_1, \ldots, x_n}$

with:


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

Then, we have:

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 $\R$, $\mathcal S$, we have:


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

So $f$ is of bounded variation.