I am self-studying measure theory using Measure Theory by Donald Cohn. I got stuck on a step in his proof of Proposition 4.4.5. Here is the statement of the proposition:
Proposition 4.4.5$\quad$Let $\mu$ be a finite signed measure on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$, and let $F_{\mu}:\mathbb{R}\to\mathbb{R}$ be defined by $F_{\mu}(x)=\mu((-\infty,x])$. Then $F_{\mu}$ is absolutely continuous if and only if $\mu$ is absolutely continuous with respect to Lebesgue measure.
Here is the step of the proof on which I got stuck:
Proof$\quad$$\dots$ Now suppose that $F_{\mu}$ is absolutely continuous. Then $V_{F_{\mu}}$ is absolutely continuous (Lemma 4.4.4), and so the functions $F_1$ and $F_2$ defined by $F_1=\frac{V_{F_{\mu}}+F_{\mu}}{2}$ and $F_2=\frac{V_{F_{\mu}}-F_{\mu}}{2}$ are absolutely continuous. Let $\mu_1$ and $\mu_2$ be the finite positive measures on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$ that correspond to $F_1$ and $F_2$. $\dots$
I don't understand why there is are finite positive measures $\mu_1$ and $\mu_2$ that correspond to $F_1$ and $F_2$. So far what I can think of is the following:
Since $F_{\mu}$ vanishes at $-\infty$ and is right-continuous, $V_{F_{\mu}}$ vanishes at $-\infty$ and is right-continuous, and thus $F_1$ amd $F_2$ both vanish at $-\infty$ and are right-continuous. Then there are finite signed measures $\mu_1$ and $\mu_2$ on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$ that correspond to $F_1$ and $F_2$ (Proposition 4.4.3).
Could someone please help me explain why $\mu_1$ and $\mu_2$ must also be positive measures? Thanks a lot in advance!
Related Definitions and Results:
Definition$\quad$ Suppose that $F$ is a real-valued function whose domain includes the interval $[a,b]$. Let $\mathscr{S}$ be the collection of finite sequences $\{t_i\}_{i=0}^n$ such that\begin{align*} a \leq t_0 < t_1 < \dots < t_n \leq b.\end{align*}Then $V_F[a,b]$, the variation of $F$ over $[a,b]$} is defined by\begin{align*} V_F[a,b] = \sup\left\{\sum_{i=1}^n|F(t_i)-F(t_{i-1})|:\{t_i\}_{i=0}^n\in\mathscr{S}\right\}.\end{align*}The function $F$ is of finite variation (or of bounded variation) on $[a,b]$ if $V_F[a,b]$ is finite.
Definition$\quad$ Suppose that $F$ is a real-valued function whose domain includes the interval $(-\infty,b]$. Let $\mathscr{S}$ be the collection of finite sequences $\{t_i\}_{i=0}^n$ such that\begin{align*} -\infty < t_0 < t_1 < \dots < t_n \leq b.\end{align*}Then $V_F(-\infty,b]$, the variation of $F$ over $(-\infty,b]$, is defined by\begin{align*} V_F(-\infty,b] = \sup\left\{\sum_{i=1}^n|F(t_i)-F(t_{i-1})|:\{t_i\}_{i=0}^{n}\in\mathscr{S}\right\}.\end{align*}The function $F$ is of finite variation (or of bounded variation) on $(-\infty,b]$ if $V_F(-\infty,b]$ is finite.
Definition$\quad$ Suppose that $F$ is a real-valued function whose domain includes $\mathbb{R}$. Let $\mathscr{S}$ be the collection of finite sequences $\{t_i\}_{i=0}^n$ such that\begin{align*} -\infty < t_0 < t_1 < \dots < t_n < +\infty.\end{align*}Then $V_F(-\infty,+\infty)$, the variation of $F$ over $\mathbb{R}$, is defined by\begin{align*} V_F(-\infty,+\infty) = \sup\left\{\sum_{i=1}^n|F(t_i)-F(t_{i-1})|:\{t_i\}_{t=0}^n\in\mathscr{S}\right\}.\end{align*}The function $F$ is of finite variation (or of bounded variation) if $V_F(-\infty,+\infty)$ is finite. If $F:\mathbb{R}\to\mathbb{R}$ is of finite variation, then the variation of $F$ is the function $V_F:\mathbb{R}\to\mathbb{R}$ defined by $V_F(x)=V_F(-\infty,x]$.
Definition$\quad$ Suppose that $\mu$ is a finite signed measure on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$. Define a function $F_{\mu}:\mathbb{R}\to\mathbb{R}$ by letting\begin{align} F_{\mu}(x) = \mu((-\infty,x])\tag1\end{align}hold at each $x$ in $\mathbb{R}$. If $\{t_i\}_{i=0}^n$ is an increasing sequence of real numbers, then\begin{align*} \sum_{i=1}^n|F_{\mu}(t_i)-F_{\mu}(t_{i-1})| = \sum_{i=1}^n|\mu((t_{i-1},t_i])| \leq |\mu|(\mathbb{R});\end{align*}it follows that $V_{F_{\mu}}(-\infty,+\infty)\leq|\mu|(\mathbb{R})$ and hence that $F_{\mu}$ is of finite variation.
Proposition$\quad$The function $F_{\mu}$ defined by (1) vanishes at $-\infty$ and is right-continuous..
Proposition$\quad$Suppose that $F:\mathbb{R}\to\mathbb{R}$ is of finite variation. If $-\infty<a<b<+\infty$, then\begin{align} V_F(-\infty,b] = V_F(-\infty,a]+V_F[a,b].\end{align}
Lemma 4.4.1$\quad$Let $F$ be a function of finite variation on $\mathbb{R}$. Then
- $V_F$ is bounded and nondecreasing,
- $V_F$ vanishes at $-\infty$, and
- if $F$ is right-continuous, then $V_F$ is right-continuous.
Proposition 4.4.2$\quad$Let $F$ be a function of finite variation on $\mathbb{R}$. Then there are bounded nondecreasing functions $F_1$ and $F_2$ such that $F=F_1-F_2$. (The functions $F_1$ and $F_2$ by $F_1=\frac{V_F+F}{2}$ and $F_2=\frac{V_F-F}{2}$ have the required properties.)
Proposition 4.4.3$\quad$Equation (1) defines a bijection $\mu\mapsto F_{\mu}$ between the set of all finite signed measures on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$ and the set of all right-continuous functions of finite variation that vanish at $-\infty$.
Lemma 4.4.4$\quad$If $F:\mathbb{R}\to\mathbb{R}$ is absolutely continuous and of finite variation, then $V_F$ is absolutely continuous.