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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).

  1. 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.)

  2. 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 $\varphi_1 = \varphi_2$?

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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 $\phi_j([x,y))=f(y)-f(x)$.

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  • $\begingroup$ Thanks. (1) According to my textbook (Yeh's "Real Analysis"), a Jordan decomposition is not unique. (2) You only addressed the case in which all the measures are finite, but my question allowed for the possibility that one of the measures is not finite. $\endgroup$ – Evan Aad Jun 30 '15 at 20:21
  • $\begingroup$ (1) You need a different book! Seriously. "Jordan decomposition" is a very standard term with a standard definition, and it's not the definition in your book. A book giving non-standard definitions for standard terms is going to cause problems, seems to me. (2) $\infty-\infty$ is not defined. In any case where enough of the measures are finite that everything you write is defined, it follows that $\phi_1=\phi_2$ for the reason I gave. $\endgroup$ – David C. Ullrich Jun 30 '15 at 20:33
  • $\begingroup$ Regarding your second point: I don't see why it holds if $\phi_1$ and $\phi_2$ are not finite. I tried to check the hypothesis that they are equal using Dynkin's $\pi$-$\lambda$ theorem, but the proof breaks when I try to prove that if $B$ is a Borel set for which $\phi_1(B) = \phi_2(B)$, then so is $B^c$. $\endgroup$ – Evan Aad Jun 30 '15 at 20:43
  • $\begingroup$ You said at the start that $f$ had bounded variation. If one of $\mu_2$, $\nu_2$ is finite it follows that they are both finite. If both are infinite we can't do the subtraction. $\endgroup$ – David C. Ullrich Jun 30 '15 at 20:52
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    $\begingroup$ If $f_2=0$ and $g_2=x$ then $f=x$. So $f$ does not have bounded variation. I quit. $\endgroup$ – David C. Ullrich Jun 30 '15 at 21:40
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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.

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