Total variation of Lebesgue-Stieltjes integral

Question

Setting We work on a filtered probability space. Let $A$ be a process of finite variation (FV) started from $0$ and $H$ be predictable, positive and such that $H\cdot A$ is defined $\omega$-wise as a Lebesgue-Stieltjes integral. We denote by $\|A\|$ the total variation process of $A$.

Question Does $$\|H\cdot A\|=H \cdot\|A\| \enspace (*)$$ hold? If so, is there a reference with the proof of this result?

Idea Let $A=A^{\uparrow}-A^{\downarrow}$ be the Jordan-Hahn decomposition of the FV process $A$ into the difference of two nondecreasing processes started from $0$. Then $\|A\|=A^{\uparrow}+A^{\downarrow}$. Thus $H \cdot\|A\|=H\cdot A^{\uparrow} + H\cdot A^{\downarrow}$ and both integrals in the sum are nondecreasing. Maybe one can now argue with the uniqueness of the Jordan-Hahn decomposition that $H\cdot A^{\uparrow} + H\cdot A^{\downarrow} = \|H\cdot A\|$ holds...

Any help on how to proceed with this idea or any other idea on how to prove $(*)$ are appreciated.

Answer 1

Let $A=(A_t)$ be an adapted (càdlàg) process with finite variation, then $B_t=\int_{[0,t]}|\mathrm{d}A_s|$, the variation process of $A$, is an adapted increasing process(refer to S. W. He et al., Semimartingale Theory and Stochastic Calculus, Sci. Press and CRC(1992), p.101, Theorem 3.44). Hence, both \begin{equation*} A^{\uparrow}=\frac12(A+B), \qquad A^{\downarrow}=\frac12(B-A) \end{equation*} are adapted increasing processes. Meanwhile, from the Theorem 5.14-5.15, p.142-143(ibid), there exist optional R-N derivatives $a^+,a^-$ as follows \begin{gather*} a^+=\frac{\mathrm{d}A^{\uparrow}}{\mathrm{d}\|A\|}\ge 0, \quad a^-=\frac{\mathrm{d}A^{\downarrow}}{\mathrm{d}\|A\|}\ge 0,\\ a^+ + a^-= 1,\qquad a^+ a^-=0, \qquad \text{a.s. d}\|A\|. \tag{1} \end{gather*} From (1) and $0\le a^-, a^+ \le 1$, the Hahn decomposition of $X\triangleq\Omega\times\mathbb{R}_+$(on $(X,\mathscr{O})$) wrt $A$ is \begin{equation*} X=X^+\cup X^-\triangleq\{a^+=1\}\cup \{a^-=1\}.\tag{2} \end{equation*} Since $H\ge 0$, (2) is the Hahn decomposition of $X\triangleq\Omega\times\mathbb{R}_+$ wrt $H\cdot A$ too, and the Jordan decomposition of $H\cdot A$ is \begin{equation*} (H\cdot A)^+=(a^+ H)\cdot A, \qquad (H\cdot A)^-=(a^- H)\cdot A. \end{equation*} Hence, \begin{align*} \|H\cdot A\| &= (H\cdot A)^+ + (H\cdot A)^- \\ & = (a^+ H)\cdot A +(a^- H)\cdot A = H\cdot \|A\|. \end{align*}