Does the signed measure based on a Jordan decomposition of a function with bounded variation depend on the decomposition? Let $g_1, g_2, h_1, h_2 : \mathbb{R} \rightarrow \mathbb{R}$ be non-decreasing and right-continuous. Define
$$
\begin{align}
f_1 & := g_1 - h_1 \\
f_2 & := g_2 - h_2
\end{align}
$$
and suppose $f_1 = f_2$. In other words, for every $a, b \in \mathbb{R}$ with $a < b$, the restriction of $g_1 - h_1$ and $g_2 - h_2$ to $[a, b]$ are two Jordan decompositions of the same bounded-variation function.
Denote with $\mu_1, \mu_2$ the (positive) Lebesgue-Stieltjes measures engendered by $g_1, g_2$, respectively, and with $\nu_1, \nu_2$ the (positive) Lebesgue-Stieltjes measures engendered by $h_1, h_2$, respectively. Suppose that either $\mu_1$ or $\nu_1$ is finite (so we may define the signed measure $\mu_1 - \nu_1$, as we do below).


*

*Is it necessarily the case that either $\mu_2$ or $\nu_2$ is finite? (so we may define the signed measure $\mu_2 - \nu_2$, as we do below.)

*Suppose that either $\mu_2$ or $\nu_2$ is finite. Define the signed measures
$$
\begin{align}
\varphi_1 & := \mu_1 - \nu_1 \\
\varphi_2 & := \mu_2 - \nu_2
\end{align}
$$
Is it the case that $\varphi_1 = \varphi_2$?
 A: Actually no, a function has at most one Jordan decomposition.
You have two decompositions, but at most one is the Jordan decomposition.
Anyway, finiteness for one decomposition does not imply finiteness for the other. Consider $$0=0-0=x^+-x^+.$$
If $f$ has bounded variation and all your measures happen to be finite then yes, $\phi_1=\phi_2$. Because for example $\phi_j([x,y))=f(y)-f(x)$.
A: I'd like to expand upon David C. Ullrich's answer to question #2. The answer, as he wrote, is: yes, $\phi_1 = \phi_2$. Indeed, let $B$ be a Borel set on the real line. Then for $i \in \{1, 2\}$,
$$
\phi_i(B) = \phi_i\left(\cup_{n \in \mathbb{Z}}(B \cap (-n, n])\right) = \sum_{n \in \mathbb{Z}} \phi_i(B \cap (-n, n])
$$
Therefore, it suffices to show, for any $n \in \mathbb{Z}$, that
$$
\phi_1(B \cap (-n, n]) = \phi_2(B \cap (-n, n])
$$
Let then $n \in \mathbb{Z}$ be some integer. Since, restricted to the interval $(-n, n]$, $\phi_1$ and $\phi_2$ are finite, Dynkin's $\pi$-$\lambda$ theorem can be used to show that $\phi_1\big|_{(-n, n]} = \phi_2\big|_{(-n, n]}$ (use the easily verified fact that for every $a, b \in \mathbb{R}$ with $a < b$ we have $\phi_1((a, b]) = f_1(b) - f_2(a) = \phi_2((a, b])$). Hence, in particular, $\phi_1(B \cap (-n, n]) = \phi_2(B \cap (-n, n])$, Q.E.D.
