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

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 $$V_{F_{\mu}}(-\infty,x]=|\mu|((-\infty,x])$$.

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 $$t_0=-n. We see $$V_F(-\infty,x]>|\mu|((-\infty,x])-\epsilon$$, and we see it for any $$\epsilon$$. Hence $$V_F\ge|\mu|$$ on this interval.
• @Beerus If $\mu$ is any measure on any measure space and $A_1\subseteq A_2\subseteq A_3\subseteq\cdots$ is an increasing tower of measurable sets, then $\mu\left(\bigcup_{j=1}^\infty A_j\right)=\lim_{j\to\infty}\mu(A_j)$. There is a result for intersections too. If $A_1\supseteq A_2\supseteq A_3\supseteq\cdots$ and (this is essential) $\mu(A_1)<\infty$, then $\mu\left(\bigcap_{j=1}^\infty A_j\right)=\lim_{j\to\infty}\mu(A_j)$. These are both known as “continuity of measure”. Commented Aug 3 at 14:48