Equivalence of the Variation of F$_\mu$(x) and $| \mu |$(($-\infty, x])$ Let $\mu$ be a finite signed Borel measure. I want to see why V$_{F_{\mu}}$($-\infty, x]$ = $| \mu | $ $((-\infty , x])$ for all x in R, where F$_\mu$(x) = $\mu((-
\infty, x])$ and V$_{F_{\mu}}$($-\infty, x]$ is the variation of F$_\mu$ on the interval (-$\infty$, x].
If R = P $\cup$ N is the Hahn decomposition of $\mu$ it seems like the problem comes down to showing that we can partition a half open interval (a,b] into smaller half open intervals were each of these smaller half open intervals lie either almost entirely in P with arbitrarily small measure in N or entirely in N with arbitrarily small measure in P. I am unable to see how to do this though. 
 A: We first show that $\le$ holds. Let $\infty < t_0 < \ldots< t_n \le x$ any partition of $(-\infty,x]$, then
$$\sum_{k=0}^{n-1} |F(t_{k+1})-F(t_k)| = \sum_{k=0}^{n-1} |\mu((t_k,t_{k+1}])| \le \sum_{k=0}^{n-1} |\mu|((t_k,t_{k+1}]) = |\mu|((t_0,t_n]) \le |\mu|((-\infty,x]).$$
The other direction is not so easy: Note that the Jordan decomposition $\mu = \mu^{+} - \mu^{-}$ is minimal in the sense that for any decompistion $\mu = \nu_1 - \nu_2$ with finite measures $\nu_1$, $\nu_2$, we have $\nu_1 \ge \mu^{+}$ and $\nu_2 \ge \mu^{-}$. We can define $$f_+(x) := \mu^{+}((-\infty,x]) \quad \text{and} \quad f_-(x) := \mu^{-}((-\infty,x]).$$
Note that $$\tag{1} F(x) = \mu((-\infty,x]) = f_+(x) - f_-(x)$$ is a decompisition of $f$ as a difference of two monotone functions. On the other hand, we have
$$\mathrm{Var}_F((-\infty,x]) = \mathrm{Var}_F^+((-\infty,x])+\mathrm{Var}_F^-((-\infty,x])$$
and $$\tag{2}F(x) = \mathrm{Var}_F^+((-\infty,x]) - \mathrm{Var}_F^-((-\infty,x]),$$
where $\mathrm{Var}_F^+$ denote the positive variation, resp. $\mathrm{Var}_F^-$ the negative variation. (2) is minimal in the sense that any for any decomposition $F(x) = g(x)-h(x)$ with monotone functions $g$ and $h$ such that $g(x),h(x) \rightarrow 0$ for $x \rightarrow - \infty$, we already have $$\mathrm{Var}_F^+((-\infty,x]) \le g(x) \quad \text{and} \quad \mathrm{Var}_F^-((-\infty,x]) \le h(x).$$ As a special case, we get
$$\tag{3}F_+(x):=\mathrm{Var}_F^+((-\infty,x]) \le f_+(x) \quad \text{and} \quad F_-(x) =\mathrm{Var}_F^-((-\infty,x]) \le f^-(x).$$
On the other hand $F_+$ and $F_-$ induces measures $\nu_1$ and $\nu_2$ with $$\nu_1((-\infty,x]) = F_+(x) \quad \text{and} \quad \nu_2((-\infty,x]) = F_-(x).$$
The minimality of the Jordan decomposition shows that
$\nu_1 \ge \mu^+$ and $\nu_2 \ge \mu^-$ and, especially, $$F_+(x) = \nu_1((-\infty,x]) \ge \mu^{+}((-\infty,x])=f_+(x)$$
and 
$$F_-(x) = \nu_2((-\infty,x]) \ge \mu^{-}((-\infty,x])=f_-(x).$$
Together with (3) we get equality and thus
$$\mathrm{Var}_F((-\infty,x]) = \mathrm{Var}_F^+((-\infty,x])+\mathrm{Var}_F^-((-\infty,x]) = f_+(x) +f_-(x) = |\mu|((-\infty,x]).$$
