Lebesgue-Stieltjes measure and the Hahn-Jordan decomposition

Question

Let $f:[0,\infty)\to\mathbb R$ be right-continuous and of locally bounded variation$^1$ with $f(0)=0$. There are unique nondecreasing and right-continuous $f^\pm:[0,\infty)\to\mathbb R$ with $f^\pm(0)=0$ and $$f=f^+-f^-\tag2\;.$$ In particular, $$f^\pm=\frac12\left(T_f\pm f\right)\tag3\;.$$ Thus, there are unique $\sigma$-finite measures $\mu^\pm_f$ on $\mathcal B\left(\left(0,\infty\right)\right)$ with $$\mu^\pm_f\left(\left(a,b\right]\right)=f^\pm(b)-f^\pm(a)\;\;\;\text{for all }a,b\in[0,\infty)\text{ with }a<b\tag4\;.$$

Note that $\mu^\pm$ is singular in respect to $\mu^\mp$. Let $$\mu_f:=\mu^+_f-\mu^-_f\;.$$ How can we show that $\mu_f$ is a finite signed measure on $\mathcal B\left(\left(0,\infty\right)\right)$?

Any nondecreasing and right-continuous function $g:\mathbb R\to\mathbb R$ (called a distribution function) yields a unique $\sigma$-finite measure $\nu$ on $\mathcal B(\mathbb R)$ with $\nu((a,b])=g(b)-g(a)$ for all $a,b\in\mathbb R$ with $a<b$. That's what I've used to ensure the existence of $\mu_f^\pm$. However, it seems like we need finiteness of $\mu^\pm_f$ in order to ensure that $\mu_f$ is even well-defined (otherwise, we might have $\infty-\infty$).

So, does the special form of $f^\pm$ ensure that $\mu_f^\pm$ is finite (and hence $\mu_f$ is a finite signed measure)?

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$^1$ Let $$\mathcal P_{[a,\:b]}:=\left\{\left(t_0,\ldots,t_k\right):k\in\mathbb N\text{ and }a=t_0<\cdots<t_k=b\right\}$$ and $$\left|\varsigma\right|:=\max_{1\le i\le k}\left(t_i-t_{i-1}\right)\;\;\;\text{for }\varsigma=\left(t_0,\ldots,t_k\right)\in\mathcal P_{[a,\:b]}\;.$$ for $b>a\ge0$. Moreover, let $$V_\varsigma(f):=\sum_{i=1}^k\left|f(t_i)-f(t_{i-1})\right|\;\;\;\text{for }\varsigma=\left(t_0,\ldots,t_k\right)\in\mathcal P_{[a,\:b]}$$ and $$V_{[a,\:b]}(f):=\sup_{\varsigma\:\in\:\mathcal P_{[a,\:b]}}V_\varsigma(f)$$ for $f:[a,b]\to\mathbb R$ with $b>a\ge0$. $f:[0,\infty)\to\mathbb R$ is said to be of locally bounded variation $:\Leftrightarrow$ $$T_f(t):=V_{[0,\:t]}(f)<\infty\;\;\;\text{for all }t>0\;.\tag1$$ Let $T_f(0):=0$.

Answer 1

Consider $f(x)=\sin x$ as an example. Then both $\mu_f^\pm((0,\infty))$ are infinite and $\mu_f$ is not a signed measure on $(0, \infty)$:

• $\mu_f(A)=+\infty$ for $A=\bigcup_{n=1}^\infty (2\pi n, 2\pi n + \pi/2)$
• $\mu_f(B)=-\infty$ for $B=\bigcup_{n=1}^\infty (2\pi n + \pi/2, 2\pi n + \pi)$

And so we get into all kinds of $\infty-\infty$ trouble.

To avoid this, you need at least one of $f^+$ and $f^-$ to be bounded on $(0,\infty)$. This condition is strictly in between "locally BV" (too weak) and "globally BV" (gives a finite signed measure).