$\require{physics}\newcommand{\cbrt}[1]{\sqrt[3]{#1}}\newcommand{\sgn}{\text{sgn}}\newcommand{\ii}[1]{\textit{#1}}\newcommand{\eps}{\varepsilon}\newcommand{\EE}{\mathbb E}\newcommand{\PP}{\mathbb P}\newcommand{\Var}{\mathrm{Var}}\newcommand{\Cov}{\mathrm{Cov}}\newcommand{\pperp}{\perp\kern-6pt\perp}\newcommand{\LL}{\mathcal{L}}\newcommand{\pa}{\partial}\newcommand{\AAA}{\mathscr{A}}\newcommand{\BBB}{\mathscr{B}}\newcommand{\CCC}{\mathscr{C}}\newcommand{\DDD}{\mathscr{D}}\newcommand{\EEE}{\mathscr{E}}\newcommand{\FFF}{\mathscr{F}}\newcommand{\WFF}{\widetilde{\FFF}}\newcommand{\GGG}{\mathscr{G}}\newcommand{\HHH}{\mathscr{H}}\newcommand{\PPP}{\mathscr{P}}\newcommand{\Ff}{\mathcal{F}}\newcommand{\Gg}{\mathcal{G}}\newcommand{\Hh}{\mathbb{H}}\DeclareMathOperator{\ess}{ess}\newcommand{\CC}{\mathbb C}\newcommand{\FF}{\mathbb F}\newcommand{\NN}{\mathbb N}\newcommand{\QQ}{\mathbb Q}\newcommand{\RR}{\mathbb R}\newcommand{\ZZ}{\mathbb Z}\newcommand{\KK}{\mathbb K}\newcommand{\SSS}{\mathbb S}\newcommand{\II}{\mathbb I}\newcommand{\conj}[1]{\overline{#1}}\DeclareMathOperator{\cis}{cis}\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}\newcommand{\norm}[1]{\left\lVert #1 \right\rVert}\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}\newcommand{\ceil}[1]{\left\lceil #1 \right\rceil}\DeclareMathOperator*{\range}{range}\DeclareMathOperator*{\nul}{null}\DeclareMathOperator*{\Tr}{Tr}\DeclareMathOperator*{\tr}{Tr}\newcommand{\id}{1\!\!1}\newcommand{\Id}{1\!\!1}\newcommand{\der}{\ \mathrm {d}}\newcommand{\Zc}[1]{\ZZ / #1 \ZZ}\newcommand{\Zm}[1]{\left(\ZZ / #1 \ZZ\right)^\times}\DeclareMathOperator{\Hom}{Hom}\DeclareMathOperator{\End}{End}\newcommand{\GL}{\mathbb{GL}}\newcommand{\SL}{\mathbb{SL}}\newcommand{\SO}{\mathbb{SO}}\newcommand{\OO}{\mathbb{O}}\newcommand{\SU}{\mathbb{SU}}\newcommand{\U}{\mathbb{U}}\newcommand{\Spin}{\mathrm{Spin}}\newcommand{\Cl}{\mathrm{Cl}}\newcommand{\gr}{\mathrm{gr}}\newcommand{\gl}{\mathfrak{gl}}\newcommand{\sl}{\mathfrak{sl}}\newcommand{\so}{\mathfrak{so}}\newcommand{\su}{\mathfrak{su}}\newcommand{\sp}{\mathfrak{sp}}\newcommand{\uu}{\mathfrak{u}}\newcommand{\fg}{\mathfrak{g}}\newcommand{\hh}{\mathfrak{h}}\DeclareMathOperator{\Ad}{Ad}\DeclareMathOperator{\ad}{ad}\DeclareMathOperator{\Rad}{Rad}\DeclareMathOperator{\im}{im}\renewcommand{\BB}{\mathcal{B}}\newcommand{\HH}{\mathcal{H}}\DeclareMathOperator{\Lie}{Lie}\DeclareMathOperator{\Mat}{Mat}\DeclareMathOperator{\span}{span}\DeclareMathOperator{\proj}{proj}$ # Functions with Finite Variation We consider continuous functions $\RR_+\to \RR$ in the sequel. >[!definition] Finite signed Measure >A finite signed measure $\mu$ is just the difference between two finite positive measures, $\mu_+ - \mu_-$, ditto for $\sigma$-finite. >[!claim] Unique Disjoint Decomposition >Over $[0,T]$, there exists a unique decomposition of any finite measure $\mu = \mu_+ - \mu_-$ such that $\mu_+$ and $\mu_-$ are supported on disjoint Borel sets $D_+$ and $D_-$. > [!proof]- Radon-Nikodym kills > Let $\nu = \mu_1 + \mu_2$. By [[Analysis/Probability/Radon-Nikodym|Radon-Nikodym]], there are two non-negative Borel functions $h_1, h_2$ on $[0,T]$ such that $\mu_1(dt) = h_1(t)\nu(dt)$ and $\mu_2(dt) = h_2(t)\nu(dt)$, thus $\mu(dt) = (h_1(t) - h_2(t))\nu(dt)$ which concludes because $\{h(t) > 0\}$ and $\{h(t) < 0\}$ are disjoint. > > This is obviously unique. Then, we define $\abs{\mu} = \mu_+ + \mu_-$, and observe $\frac{d\mu}{d\abs{\mu}} = \id_{D_+} - \id_{D_-}$. >[!definition] FVP > A continuous function $a:[0,T]\to \RR$ such that $a(0) = 0$ has ==**finite variation**== if there exists a signed measure $\mu$ on $[0,T]$ such that $a(t) = \mu([0,t])$ for $t\in [0,T]$. $\abs{\mu}$ is called the ==**total variation**== of $a$. > > A continuous function $a:\RR_+\to \RR$ has finite variation if all of its restrictions to $[0,T]$ do. In particular, $\mu$ cannot have [[atom|atoms]]; conversely, any non-atomic $\mu$ yields an FVP. There will not generically be an associated signed measure for $a:\RR_+\to \RR$ because of $\infty - \infty$; however, there is a unique $\sigma$-finite (positive) measure whose restriction is the total variation measure. >[!idea] Source of confusion >I managed to convince myself that the measures associated with FVPs were [[absolutely continuous]]. This is false, even though we require that all FVPs are continuous! Notably, any nondecreasing function is an FVP, and there exist continuous nondecreasing functions which are have derivative $0$ almost everywhere, like that cantor set construction. > >Thus, *not all FVPs are of the form $a(t) = \int_0^t f(t)dt$ for some measurable $f(t)$*. This is still reasonable intuition to have. > >>[!idea]- The mistake I made >>$t\mapsto \mu([0,t])$ is continuous iff $\mu(\{t\}) = 0$ for all singletons. But the singletons do *not* generate the null sets -- they only encompass countable sets -- and uncountable null sets exist e.g. the dyadics. # Integration against FVs >[!definition] Integration against FVs! > Start with the case of $a:[0,T]\to \RR$. If $f:[0,T]\to \RR$ is integrable under $\abs{\mu}$, we define$\int_0^T f(s)da(s) = \int_{[0,T]} f(s)\mu(ds),\qquad \int_0^T f(s)\abs{da(s)} = \int_{[0,T]} f(s)\abs{\mu}(ds),$and one can define $\int_0^t f(s)da(s)$ for $t < T$ in the obvious way. This can be extended to $a:\RR_+\to \RR$ in the obvious way. In fact, $t\mapsto \int_0^t f(s)da(s)$ is a finite variation process; your signed measure assigns $\int_0^t \id_A f(s)da(s)$ to Borel sets $A$, which can be written $\mu'(ds) = f(s)\mu(ds)$. >[!claim] Triangle Inequality >$\abs{\int_0^T f(s)da(s)}\leq \int_0^T \abs{f(s)} \abs{da(s)}.$ > [!claim] FVP Integration over continuous functions looks like Reimann Integration >Pick a mesh.$\int_0^t\abs{da(s)} = \lim\sum \abs{\Delta a}.$For any continuous $f:[0,T]\to \RR$,$\int_0^t f(s)da(s) = \lim\sum f\Delta a.$ > >>[!idea]- Expand >> For any sequence $0 = t_0^n < \dots < t^n_{p_n} = t$ of subdivisions whose [[mesh]] tends to $0$, >> $\int_0^t \abs{da(s)}=\lim_{n\to \infty} \sum_{i = 1}^{p_n} \abs{a\left(t_i^n\right) - a\left(t_{i-1}^n\right)}.$ >> For any continuous $f:[0,T]\to \RR$,$\int_0^t f(s)da(s)=\lim_{n\to \infty} \sum_{i = 1}^{p_n} f\left(t^n_{i-1}\right)\left(a\left(t_i^n\right) - a\left(t_{i-1}^n\right)\right).$ # Finite Variation Processes >[!definition] Finite Variation Process, Increasing Process >A ==**finite variation process**== is an adapted process whose sample paths have FV on $\RR_+$. A FVP $A$ with nondecreasing sample paths is called an ==**increasing process**==. Note that any IP admits a limit in $[0,\infty]$, $A_\infty$. >[!idea] >One can make extensions to the cases where $A_0\neq 0$ or $A$ is merely cadlag, but we do not do so in the sequel. This book is very nice in that way! The relation between FVP and IP is not deep. If $A$ is an FVP, then $V_t = \int_0^t \abs{d A_s}$ is IP. Then, $A_t = \frac12\left(V_t + A_t\right) - \frac12\left(V_t - A_t\right)$, so you are FVP iff you are the difference of two IPs. >[!definition] Integration against FVPs! >Let $A$ be FVP and let $H$ be progressive. $H$ is ==**integrable wrt $A$**== if$\forall t \geq 0, \forall \omega \in \Omega, \int_0^t \abs{H_s(\omega)} \abs{dA_s(\omega)} < \infty.$ >In other words, $H$ is path-integrable wrt $A$ on all $[0,t]$. > >Then, the process $H\cdot A$ is defined by >$ > (H\cdot A)_t = \int_0^t H_s dA_s. >$ >[!claim] $H\cdot A$ is FVP >[!example] >If $A_t = t$ and $H$ is progressive and path-integrable, then $\int_0^t H_s ds$ is a FVP! By taking $H_s$ to be the Radon-Nikodym derivative of the $\mu$ associated with $A_t$ wrt Lebesgue, this should characterize all FVPs. >[!idea] Obvious extension to a.s. path-integrability >Suppose $H$ is a.s. path-integrable wrt $A$, say in a.s. event $B$, and **the filtration is complete.** Then, we can make the patch $H'_t(\omega) = H_t(\omega)\id_B$, which is still progressive because $\FFF$ is complete. This lets us define $H\cdot A = H'\cdot A$. >[!claim] Associativity >If $H,K$ are progressive and the quantities are defined, then $K\cdot (H\cdot A) = (KH)\cdot A$.