8
$\begingroup$

Let $a, b \in \mathbb{R}$ with $a < b$ and define the compact interval $I := [a, b]$. Let $g, h : \mathbb{R} \rightarrow \mathbb{R}$ be non-decreasing and right-continuous on $I$ and constant on $(-\infty, a]$ and on $[b, \infty)$, separately, and define $f := g - h$, so that, over every compact interval, $g - h$ is a Jordan decomposition of the bounded-variation function $f$. Denote by $\mu_g$ the Lebesgue-Stieltjes measure on $(\mathbb{R}, \mathcal{B})$ engendered by $g$ and denote by $\mu_h$ the one engendered by $h$, and define the (finite) signed measure $\mu_f := \mu_g - \mu_h$. Denote by $\mu_f'$ the total variation measure corresponding to $\mu_f$, i.e. if $\mu_f = \mu_f^+ - \mu_f^-$ is the unique Jordan decomposition of $\mu_f$, then $\mu_f' = \mu_f^+ + \mu_f^-$. Denote by $v_f : [a, \infty) \rightarrow [0, \infty]$ the total variation function of $f$ on the ray $[a, \infty)$, i.e. for every $t \in [a, \infty)$, $$ v_f(t) := \sup \left\{\sum_{k = 1}^n \left|f(t_k) - f(t_{k - 1})\right| :\mid a = t_0 \leq t_1 \leq \cdots \leq t_n = t, n \in \{1, 2, \dots\}\right\} $$ It is well known that $v_f(a) = 0$, that $v_f$ is non-decreasing and, since $f$ is right-continuous, so is $v_f$. Extend $v_f$ to the entire real line by setting $v_f(t) := 0$ for $t \in (-\infty, a)$. The extended $v_f$ remains non-decreasing and right-continuous. Denote by $\mu_v$ the Lebesgue-Stieltjes measure engendered by $v_f$.

Is it the case that $\mu_f' = \mu_v$?


Attempts at a solution

  • Attempt #1

    Since $\mu_f'(I^c) = \mu_v(I^c) = 0$, it remains to show that $\mu_f' = \mu_v$ over $(I, \mathcal{B}(I))$. All I've managed to show so far is that for $t \in [a, b]$, $\mu_v([a, t]) \leq \mu_f'([a, t])$.

    Indeed, according to proposition 10.23 in Yeh's "Real Analysis", 2nd edition, for every $E \in \mathcal{B}(I)$, $$ \mu_f'(E) = \sup \left\{\sum_{k = 1}^n |\mu_f(E_k)| :\mid E = \bigcup_{k = 1}^n E_k; E_1, \dots, E_n \in \mathcal{B}(I) \text{ pairwise disjoint}, n \in \{1, 2, \dots\}\right\} $$ Therefore, for every $t \in I$, $$ \begin{split} \mu_v([a, t]) & = \lim_{n \rightarrow \infty} \mu_v\left((a - \frac{1}{n}, t]\right) \\ & = \lim_{n \rightarrow \infty} \left(v_f(t) - v_f(a - \frac{1}{n})\right) \\ & = v_f(t) \\ & \leq \mu_f'([a, t]) \end{split} $$

    If we can show the converse inequality, namely that for every $t \in I$, $\mu_f'([a, t]) \leq \mu_v([a, t])$, we may proceed to conclude that $\mu_v = \mu_f'$ by using Dynkin's $\pi$-$\lambda$ theorem.

  • Attempt #2

    Define, for $t \in [0, \infty)$, $$ \begin{align} v_f^+(f) & := \sup \left\{\sum_{k = 1}^n\max\left(f(t_k) - f(t_{k - 1}), 0\right) :\mid a = t_0 \leq t_1 \leq \cdots \leq t_n = t, n \in \{1, 2, \dots\} \right\} \\ v_f^-(f) & := \sup \left\{\sum_{k = 1}^n-\min\left(f(t_k) - f(t_{k - 1}), 0\right) :\mid a = t_0 \leq t_1 \leq \cdots \leq t_n = t, n \in \{1, 2, \dots\} \right\} \end{align} $$ Then it can be seen (see, for instance, lemma 5.2.3 on p. 99 and the beginning of the proof of theorem 5.2.4 on p. 100 of Royden's "Real Analysis", 2nd edition (!), Macmillan Library Reference, 1968) that $v_f^+$ and $v_f^-$ are non-decreasing on $[0, \infty)$ and that for $t \in [0, \infty)$, $$ \begin{align} f(t) & = v_f^+(t) - v_f^-(t) \\ v_f(t) & = v_f^+(t) + v_f^-(t) \end{align} $$ It can also be shown that $v_f^+$ and $v_f^-$ are right continuous, by emulating the analogous proof for $v_f$ (see for instance theorem 12.22 on p. 266 of Yeh's "Real Analysis", 2nd edition). We may extend $v_f^+$ and $v_f^-$ to the entire real line by defining $v_f^+(t) := v_f^-(t) := 0$ for $t \in (-\infty, 0)$. The extended functions remain non-decreasing and right-continuous.

    If we denote the Lebesgue-Stieltjes measures engendered by $v_f^+$ and $v_f^-$ by $\mu_v^+$ and $\mu_v^-$, respectively, we have that $\mu_f = \mu_v^+ - \mu_v^-$ (indeed, as shown here, if $\phi_1, \phi_2, \varphi_1, \varphi_2 : \mathbb{R} \rightarrow \mathbb{R}$ are non-decreasing, right-continuous functions such that $\phi_1 - \phi_2 = \varphi_1 - \varphi_2$, then, denoting the Lebesgue-Stieltjes measures they engender by $\mu_{\phi_1}, \mu_{\phi_2}, \mu_{\varphi_1}, \mu_{\varphi_2}$, respectively, we have $\mu_{\phi_1} - \mu_{\phi_2} = \mu_{\varphi_1} - \mu_{\varphi_2}$, as long as both sides of the equation are well defined, i.e. as long as at least one measure on each side of the equation is finite), and, furthermore, it may be verified using Dynkin's $\pi$-$\lambda$ theorem starting from the $\pi$-system of half open-half closed finite intervals, that $\mu_v = \mu_v^+ + \mu_v^-$.

    Therefore, if we can prove that $\mu_v^+ \perp \mu_v^-$, i.e. that $\mu_v^+$ and $\mu_v^-$ are mutually singular, then, by the uniqueness of the Jordan decomposition, $\mu_v^+ - \mu_v^-$ is the Jordan decomposition of $\mu_f$, i.e. $\mu_f^+ = \mu_v^+$ and $\mu_f^- = \mu_v^-$, and therefore we have, by definition of $\mu_f'$, that $\mu_v = \mu_f^+ + \mu_f^- = \mu_f'$, as desired.

    So, all that is left to show is that $\mu_v^+ \perp \mu_v^-$.


Related posts

A similar question was asked in this forum almost two years ago, but has remained unanswered.

$\endgroup$
4
$\begingroup$

Yes, it is the case that $\mu_f' = \mu_v$.

We will use the notation introduced in section "Attempt #2" of the original post. Additionally, we make the following two definitions that will be used throughout the proof: $$ \begin{align} Z &:= (a, b] \\ \mathcal{J} &:= \left\{(c, d] :\mid c, d\in [a, b] \right\} \end{align} $$

The proof breaks down into four steps:

1) We show that it suffices to show that $\mu_f'(E) = \mu_v(E)$ for all $E \in \mathcal{B}(Z)$ in order to conclude that $\mu_f' = \mu_v$.

2) We show that $\mu_f^\pm(J) \geq \mu_v^\pm(J)$, respectively, for all $J \in \mathcal{J}$.

3) We deduce that $\mu_f^\pm(E) = \mu_v^\pm(E)$, respectively, for all $E \in \mathcal{B}(Z)$.

4) We conclude that $\mu_f'(E) = \mu_v(E)$ for all $E \in \mathcal{B}(Z)$.

The proof hinges on two characterizations of the Jordan decomposition of signed measures given in proposition 6.21(i & ii) in Dshalalow's "Foundations of Abstract Algebra", 2nd edition and in the version of the Jordan decomposition theorem given in Doob's "Measure Theory".

===

  1. We show that it suffices to show that $\mu_f'(E) = \mu_v(E)$ for all $E \in \mathcal{B}(Z)$ in order to conclude that $\mu_f' = \mu_v$, as follows. (1.1) We show that $\mu_v(Z^c) = 0$. (1.2) We show that $\mu_f'(Z^c) = 0$. (1.3) We conclude that it suffices to show that $\mu_f'(E) = \mu_v(E)$ for all $E \in \mathcal{B}(Z)$ in order to conclude that $\mu_f' = \mu_v$.

    1.1 We will show that $\mu_v(Z^c) = 0$. Since $Z^c = (-\infty, a] \cup (b, \infty)$, it suffices to show that $\mu_v\left((-\infty, a]\right), \mu_v\left((b, \infty)\right) = 0$.

    $$ \begin{align} \mu_v\left((-\infty, a]\right) & = \mu_v\left(\bigcup_{n = 0}^\infty \left(a - (n + 1), a - n\right]\right) \\ & = \sum_{n = 0}^\infty \mu_v\left(\left(a - (n + 1), b - n\right]\right) \\ & = \sum_{n = 0}^\infty \left(v_f(a - n) - v_f(b - (n + 1))\right) \\ & = \sum_{n = 0}^\infty (0 - 0) \\ & = 0 \\ \mu_v\left((b, \infty)\right) & = \mu_v\left(\bigcup_{n = 0}^\infty \left(b + n, b + (n+1)\right]\right) \\ & = \sum_{n = 0}^\infty \mu_v\left(\left(b + n, b + (n+1)\right]\right) \\ & = \sum_{n = 0}^\infty \left(v_f(b + (n + 1)) - v_f(b + n)\right) \\ & \overset{(*)}{=} \sum_{n = 0}^\infty V_{b + n}^{b + (n + 1)}(f) \\ & \overset{(**)}{=} 0 \end{align} $$ where equality $(*)$ is a fundamental result from the theory of functions of bounded variation (see e.g. lemma 12.15(b) in Yeh's "Real Analysis", 2nd edition), and equality $(**)$ is due to the fact that, for $s, t \in [b, \infty)$ with $s \leq t$, we have $$ \begin{align} V_s^t(f) &= \sup \left\{\sum_{k = 1}^n \left|f(t_k) - f(t_{k - 1}) \right| :\mid s = t_0 \leq t_1 \leq \cdots \leq t_n = t, n \in \{1, 2, \dots\}\right\} \\ & = \sup \left\{\sum_{k = 1}^n \left|f(b) - f(b) \right| :\mid s = t_0 \leq t_1 \leq \cdots \leq t_n = t, n \in \{1, 2, \dots\}\right\} \\ & = 0 \end{align} $$

    1.2 We show that $\mu_f'(Z^c) = 0$, as follows. (1.2.1) We construct finite measures $\mu^+$ and $\mu^-$ on $(\mathbb{R}, \mathcal{B})$ such that $\mu_f = \mu^+ - \mu^-$ and for which $\mu^\pm(Z^c) = 0$. (1.2.2) We apply a certain minimality condition on the Jordan decomposition of signed measures to show that $\mu_f^\pm = \mu^\pm$, respectively. (1.2.3) We conclude that $\mu_f'(Z^c) = 0$.

    1.2.1 We will construct finite measures $\mu^+$ and $\mu^-$ on $(\mathbb{R}, \mathcal{B})$ such that $\mu_f = \mu^+ - \mu^-$ and for which $\mu^\pm(Z^c) = 0$. Define the set functions $\mu^\pm : \mathcal{B} \rightarrow [0, \infty]$ as follows: $$ \mu^\pm(E) := \mu_f^\pm(E \cap Z) $$ respectively.

    The fact that $\mu^\pm$ are measures follows from the fact that $\mu_f^\pm$ are measures. The fact that $\mu^\pm$ are finite follows from the observation that $\mu^\pm \leq \mu_f^\pm$, respectively, and from the observation that $\mu_f^\pm$ are finite, since $\mu_f = \mu_g - \mu_h$ is. As for $\mu^\pm(Z^c)$, let's calculate: $\mu^\pm(Z^c) = \mu_f^\pm(Z^c \cap Z) = \mu_f^\pm(\emptyset) = 0$.

    Finally, to show that $\mu_f = \mu^+ - \mu^-$ it suffices to show that $\mu_f(Z^c) = 0$, for suppose we have shown that $\mu_f(Z^c) = 0$, then for $E \in \mathcal{B}$, $$ \begin{align} \mu_f(E) &= \mu_f(E \cap Z) + \mu_f(E \cap Z^c) \\ &= \mu_f(E \cap Z) \\ &= \mu_f^+(E \cap Z) - \mu_f^-(E \cap Z) \\ &= \mu^+(E) - \mu^-(E) \end{align} $$

    We will now show that $\mu_f(Z^c) = 0$. Since $Z^c = (-\infty, a] \cup (b, \infty)$, it suffices to show that $\mu_f\left((-\infty, a]\right), \mu_f\left((b, \infty)\right) = 0$.

$$ \begin{align} \mu_f\left((-\infty, a]\right) & = \mu_f\left(\bigcup_{n = 0}^\infty \left(a - (n + 1), a - n\right]\right) \\ & = \sum_{n = 0}^\infty \mu_f\left(\left(a - (n + 1), a - n\right]\right) \\ & = \sum_{n = 0}^\infty \left(\mu_g\left(\left(a - (n + 1), a - n\right]\right) - \mu_h\left(\left(a - (n + 1), a - n\right]\right) \right) \\ & = \sum_{n = 0}^\infty \left(\left(g(a - n) - g(a - (n + 1))\right) - \left(h(a - n) - h(a - (n + 1))\right)\right) \\ & = \sum_{n = 0}^\infty \left(\left(g(a) - g(a)\right) - \left(h(a) - h(a)\right)\right) \\ & = 0 \\ \mu_f\left((b, \infty)\right) & = \mu_f\left(\bigcup_{n = 0}^\infty \left(b + n, b + (n+1)\right]\right) \\ & = \sum_{n = 0}^\infty \mu_f\left(\left(b + n, b + (n+1)\right]\right) \\ & = \sum_{n = 0}^\infty \left(\mu_g\left(\left(b + n, b + (n+1)\right]\right) - \mu_h\left(\left(b + n, b + (n+1)\right]\right) \right) \\ & = \sum_{n = 0}^\infty \left(\left(g(b + (n + 1)) - g(b + n)\right) - \left(h(b + (n + 1)) - h(b + n)\right)\right) \\ & = \sum_{n = 0}^\infty \left(\left(g(b) - g(b)\right) - \left(h(b) - h(b)\right)\right) \\ & = 0 \end{align} $$

1.2.2 We will show that $\mu_f^\pm = \mu^\pm$, respectively. By definition $\mu^\pm \leq \mu_f^\pm$, respectively, so we are left to show that $\mu^\pm \geq \mu_f^\pm$, respectively. According to the version of the Jordan decomposition theorem given in Doob's "Measure Theory", if $(\Omega, \mathcal{A}, \lambda)$ is a signed measure space, if $\lambda = \lambda^+ - \lambda^-$ is the unique Jordan decomposition of $\lambda$, and if $\lambda = \lambda_1 - \lambda_2$ is any representation of $\lambda$ as the difference between two measures, then $\lambda^+ \leq \lambda_1$ and $\lambda^- \leq \lambda_2$. Therefore, since by (1.2.1) $\mu_f = \mu^+ - \mu^-$, we obtain $\mu^\pm \geq \mu_f^\pm$, respectively, as desired.

1.2.3 We will show that $\mu_f'(Z^c) = 0$. Indeed, $$ \mu_f'(Z^c) = \mu_f^+(Z^c) +\mu_f^-(Z^c) = \mu^+(Z^c) + \mu^-(Z^c) = 0 $$

1.3 We will now show that it suffices to show that $\mu_f'(E) = \mu_v(E)$ for all $E \in \mathcal{B}(Z)$ in order to conclude that $\mu_f' = \mu_v$. Indeed, suppose we have shown that, for every $E \in \mathcal{B}(Z)$, $\mu_f'(E) = \mu_v(E)$. Then, using (1.1) and (1.2) we see that for $E \in \mathcal{B} = \mathcal{B}(\mathbb{R})$ we have $$ \begin{align} \mu_f'(E) & = \mu_f'\left((E \cap Z) \cup (E \cap Z^c)\right) \\ & = \mu_f'(E \cap Z) + \mu_f'(E \cap Z^c) \\ & = \mu_f'(E \cap Z) \\ & = \mu_v(E \cap Z) \\ & = \mu_v(E \cap Z) + \mu_v(E \cap Z^c) \\ & = \mu_v\left((E \cap Z) \cup (E \cap Z^c)\right) \\ & = \mu_v(E) \end{align} $$

  1. We show that $\mu_f^\pm(J) \geq \mu_v^\pm(J)$, respectively, for all $J \in \mathcal{J}$, as follows. (2.1) We first show that for every $J \in \mathcal{J}$, there is some collection of subsets $S_J \subseteq \mathcal{B}(J)$, such that $\mu_v^+(J) = \sup_{E \in S_J} \mu_f(E)$. (2.2) We deduce, by citing a suitable property of the Jordan decomposition of signed measures, that for every $J \in \mathcal{J}$, $\mu_f^+(J) = \sup_{E \in \mathcal{B}(J)} \mu_f(E) \geq \mu_v^+(J)$. (2.3) Using a similar argument we show that for every $J \in \mathcal{J}$, $\mu_f^-(J) \geq \mu_v^-(J)$.

    2.1 We will show that, for every $J \in \mathcal{J}$, there is some collection of subsets $S_J \subseteq \mathcal{B}(J)$, such that $\mu_v^+(J) = \sup_{E \in S_J} \mu_f(E)$. Let $J \in \mathcal{J}$. If $J = \emptyset$, we can take $S_J$ to be $\{\emptyset\}$. Otherwise $J = (c, d]$ for some $c, d \in [a, b]$ such that $c < d$. Define $\mathcal{P}_J$ to be the set of strict partitions of $[c, d]$, that is to say $\mathcal{P}_J$ is the collection of all finite sets of points $\{c = t_0 < t_1 < \cdots < t_n = d\}$.

    For every $Q = \{c = t_0 < t_1 < \cdots < t_n = d\} \in \mathcal{P}_J$ define $$ \begin{align} \alpha(Q) & := \bigcup_{k \in \{j \in \{1, 2, \dots, n\} \mid: f(t_j) > f(t_{j - 1})\}} (t_{k - 1}, t_k]\\ \beta(Q) & := \sum_{k = 1}^n \max\left(f(t_k) - f(t_{k - 1}), 0\right) \end{align} $$ and note that $\mu_f\left(\alpha(Q)\right) = \beta(Q)$, since for any $s, t \in \mathbb{R}$ such that $s < t$, $$ \begin{align} \mu_f\left((s, t]\right) &= \mu_g\left((s, t]\right) - \mu_h\left((s, t]\right) \\ &= \left(g(t) - g(s)\right) - \left(h(t) - h(s)\right) \\ &= \left(g(t) - h(t)\right) - \left(g(s) - h(s)\right) \\ &= f(t) - f(s) \end{align} $$

    Now define $$ S_J := \{\alpha(Q) :\mid Q \in \mathcal{P}_J\} $$

    Then $S_J \subseteq \mathcal{B}(J)$ and furthermore $$ \mu_v^+(J) = v_f^+(d) - v_f^+(c) \overset{(***)}{=} {V^+}_c^d(f) = \sup_{Q \in \mathcal{P}_J} \beta(Q) = \sup_{E \in S_J} \mu_f(E) $$ Equality $(***)$ can be proved similarly to equality $(*)$ above (in section 1.1).

    2.2 We will show that for every $J \in \mathcal{J}$, $\mu_f^+(J) \geq \mu_v^+(J)$. According to proposition 6.21(i) in Dshalalow's "Foundations of Abstract Algebra", 2nd edition, for every $D \in \mathcal{B} = \mathcal{B}(\mathbb{R})$, $\mu_f^+(D) = \sup_{E \in \mathcal{B}(D)} \mu_f(E)$. Therefore, using the result of the previous step, $$ \mu_f^+(J) = \sup_{E \in \mathcal{B}(J)} \mu_f(E) \geq \sup_{E \in S_J} \mu_f(E) = \mu_v^+(J) $$

    2.3 We can show that $\mu_f^-(J) \geq \mu_v^-(J)$ for all $J \in \mathcal{J}$ using arguments analogous to those presented in (2.1) and (2.2) (this time invoking Dshalalow's proposition 6.21(ii)).

  2. We show that $\mu_f^\pm(E) = \mu_v^\pm(E)$, respectively, for all $E \in \mathcal{B}(Z)$, as follows. (3.1) Firstly we show that $\mu_f^\pm(J) = \mu_v^\pm(J)$, respectively, for all $J \in \mathcal{J}$. We do so by applying a certain minimality property of the Jordan decomposition of signed measures to the results obtained in step (2) above. (3.2) We generalize (3.1) using Dynkin's $\pi$-$\lambda$ theorem to show that $\mu_f^\pm(E) = \mu_v^\pm(E)$, respectively, for all $E \in \mathcal{B}(Z)$.

    3.1 We will show that $\mu_f^\pm(J) = \mu_v^\pm(J)$, respectively, for all $J \in \mathcal{J}$. Recall that in step (2) above we showed that $\mu_f^\pm(J) \geq \mu_v^\pm(J)$, respectively, for all $J \in \mathcal{J}$. Therefore, it is left to show that, for all $J \in \mathcal{J}$, $\mu_f^\pm(J) \leq \mu_v^\pm(J)$. Indeed, according to the version of the Jordan decomposition theorem given in Doob's "Measure Theory", if $(\Omega, \mathcal{A}, \lambda)$ is a signed measure space, if $\lambda = \lambda^+ - \lambda^-$ is the unique Jordan decomposition of $\lambda$, and if $\lambda = \lambda_1 - \lambda_2$ is any representation of $\lambda$ as the difference between two measures, then $\lambda^+ \leq \lambda_1$ and $\lambda^- \leq \lambda_2$. Therefore, recalling from section "Attempt #2" of the original post that $\mu_f = \mu_v^+ - \mu_v^-$, we obtain that $\mu_f^\pm \leq \mu_v^\pm$, respectively.

    3.2 We will show that $\mu_f^\pm(E) = \mu_v^\pm(E)$, respectively, for all $E \in \mathcal{B}(Z)$. $\mu_f^+$ and $\mu_v^+$ are positive measures that coincide on $\mathcal{J}$. Since $\mathcal{J}$ is a $\pi$-system that generates $Z$ and that includes $Z$, we conclude (for instance, by using Dynkin's $\pi$-$\lambda$ theorem) that $\mu_f^+(E) = \mu_v^+(E)$ for all $E \in \mathcal{B}(Z)$. By the same reasoning, $\mu_f^-(E) = \mu_v^-(E)$ for all $E \in \mathcal{B}(Z)$.

  3. We will show that or all $E \in \mathcal{B}(Z)$, $\mu_f'(E) = \mu_v(E)$. Recall from section "Attempt #2" of the original post that $\mu_v = \mu_v^+ + \mu_v^-$. Therefore, using (3.2), for all $E \in \mathcal{B}(Z)$, $$ \mu_f'(E) = \mu_f^+(E) + \mu_f^-(E) = \mu_v^+(E) + \mu_v^-(E) = \mu_v(E) $$ Q.E.D.


Acknowledgements

The proof idea was suggested to me by the last sentence of section X.6, "Functions of bounded variation vs. signed measures", on p. 164 of Doob's "Measure Theory", Springer 1993.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.