Show $V_{F_{\mu}}(-\infty,x]=|\mu|((-\infty,x])$ for all $x\in\mathbb{R}$ ($\mu$ a finite signed measure on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$).

Question

My Question

I want to prove the following:

Claim$\quad$ Let $\mu$ be a finite signed measure on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$. Show that $V_{F_{\mu}}(-\infty,x]=|\mu|((-\infty,x])$ for all $x\in\mathbb{R}$.

My Attempt

Here is my attempt so far:

Proof$\quad$ Let $\mu$ be a finite signed measure on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$. Let $x\in\mathbb{R}$. Let $\mathscr{S}$ be the collection of finite sequences $\{t_i\}_{i=0}^n$ such that $$ -\infty < t_0 < t_1 < \dots < t_n \leq x. $$ We first prove that $V_{F_{\mu}}(-\infty,x]\leq|\mu|((-\infty,x])$. Let $\{t_i\}_{i=1}^n\in\mathscr{S}$. Then \begin{align*} \sum_{i=1}^n|\mu((t_{i-1},t_i])| &\leq \sum_{i=1}^n|\mu|((t_{i-1},t_i])\\ &= |\mu|\left(\bigcup_{i=1}^n(t_{i-1},t_i]\right)\\ &\leq |\mu|((-\infty,x]). \end{align*} So $|\mu|((-\infty,x])$ is an upper bound of the set $\left\{\sum_{i=1}^n|\mu((t_{i-1},t_i])|:\{t_i\}_{i=0}^n\in\mathscr{S}\right\}$. But \begin{align*} V_{F_{\mu}}(-\infty,x] &= \sup\left\{\sum_{i=1}^n|F_{\mu}(t_i)-F_{\mu}(t_{i-1})|:\{t_i\}_{i=1}^n\in\mathscr{S}\right\} \\ &= \sup\left\{\sum_{i=1}^n|\mu((-\infty,t_i])-\mu((-\infty,t_{i-1}])|:\{t_i\}_{i=0}^n\in\mathscr{S}\right\}\\ &= \sup\left\{\sum_{i=1}^n|\mu((t_{i-1},t_i])|:\{t_i\}_{i=0}^n\in\mathscr{S}\right\}. \end{align*} Thus, $V_{F_{\mu}}(-\infty,x]\leq|\mu|((-\infty,x])$.

Where I Got Stuck:

I couldn't prove the other direction. Here is my attempt, but I don't think it is correct, because $V_{F_{\mu}}$ is not necessarily a measure.

Next we show that $V_{F_{\mu}}(-\infty,x]\geq|\mu|((-\infty,x])$. Consider $\{t_0\}\in\mathscr{S}$ where $t_0=x$. Then $$ |\mu((-\infty,x])| = |F_{\mu}(t_0)| \leq V_{F_{\mu}}(-\infty,x]. $$ But $|\mu|$ is the smallest of those positive measure $\nu$ that satisfy $|\mu((-\infty,x])|\leq\nu((-\infty,x])$. Thus $|\mu|((-\infty,x])\leq V_{F_{\mu}}(-\infty,x]$. Hence we have proved that $V_{F_{\mu}}(-\infty,x]=|\mu|((-\infty,x])$.

Could someone please help me out with the proof? Thanks a lot in advance!

Answer 1

I'm happy with your proof $V_F(-\infty,x]\le|\mu|((-\infty,x])$. In the converse, note $|\mu|((-\infty,x])=\lim_{n\to\infty}|\mu|((-n,x])$ by the upper continuity of measure; we also have assumed this is a finite quantity. Fix $\epsilon>0$; I may find $n$, $|\mu|((-n,x])>|\mu|((-\infty,x])-\epsilon$. But clearly $V_F(-\infty,x]\ge|\mu|((-n,x])$ from considering the straightforward finite partition $t_0=-n<x=t_1$. We see $V_F(-\infty,x]>|\mu|((-\infty,x])-\epsilon$, and we see it for any $\epsilon$. Hence $V_F\ge|\mu|$ on this interval.

Combining that with your previous inequality, you have a total equality.