Proof of Cell-Count Form of a Compensated Counting-Path Functional Along a Time Change: Window-Discrepancy Bounds for the Time-Change and Partial-Cell Errors

Preliminaries on the counting path. By clauses 1 and 2 of Counting Path and Its Jump Times, $p$ is nondecreasing with $p(0)=0$ and takes values in $\{0\}\cup\mathbb{N}$, so $0\le p(x)\le p(R)$ for $x\in[0,R]$ and every difference $|p(u')-p(u)-(u'-u)|$ with $0\le u\le u'\le R$ is at most $p(R)+R$; hence $\mathrm{Disc}_w(p)$ is a well-defined real number (the set is nonempty, containing the value for $u=u'=0$). For a natural number $k\ge1$ let $\tau_k=\tau_k(p)\in[0,\infty]$ be the $k$-th jump time of that definition, the greatest lower bound of $\{t\ge0:p(t)\ge k\}$. We claim that for $x\ge0$, $p(x)\ge k$ if and only if $x\ge\tau_k$. If $p(x)\ge k$ then $x$ belongs to the set whose greatest lower bound is $\tau_k$, so $x\ge\tau_k$. Conversely, let $x\ge\tau_k$ (so $\tau_k<\infty$). For every $t>\tau_k$ there is, by the definition of the greatest lower bound, some $t'\in[\tau_k,t)$ with $p(t')\ge k$, hence $p(t)\ge p(t')\ge k$ by monotonicity; by right-continuity (clause 3) $p(\tau_k)$ is the greatest lower bound of $\{p(t):t>\tau_k\}$, so $p(\tau_k)\ge k$, and $p(x)\ge p(\tau_k)\ge k$.

Claim 1. Let $\mathsf{C}'$ be measurable. For $k\ge1$ the set $\{u\in[0,s]:p(\mathsf{C}'_u)\ge k\}$ equals $\{u:\mathsf{C}'_u\ge\tau_k\}$ by the preliminary claim; if $\tau_k=+\infty$ this set is empty, and otherwise it is the preimage under the measurable map $\mathsf{C}'$ of the closed set $[\tau_k,\infty)$; in both cases it is measurable; so each indicator $u\mapsto\mathbf{1}\{p(\mathsf{C}'_u)\ge k\}$ is measurable. Since $p(\mathsf{C}'_u)\le p(R)$, one has $p(\mathsf{C}'_u)=\sum_{k=1}^{p(R)}\mathbf{1}\{p(\mathsf{C}'_u)\ge k\}$ (an integer $n$ with $0\le n\le p(R)$ equals the number of $k\in\{1,\dots,p(R)\}$ with $k\le n$), a finite sum of bounded measurable indicator functions, hence bounded measurable by claims 1 and 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions. Then $u\mapsto\bar{M}(\mathsf{C}'_u)=p(\mathsf{C}'_u)-\mathsf{C}'_u$ is bounded measurable (claim 2 of that lemma), and so is each component of $u\mapsto H_u\bar{M}(\mathsf{C}'_u)$, a product of bounded measurable functions (claim 3 of the same lemma). Likewise $\{u:\bar{\mathsf{C}}_u\ge b_j\}=\bar{\mathsf{C}}^{-1}([b_j,\infty))$ is measurable, so $u\mapsto\mathbf{1}\{b_j\le\bar{\mathsf{C}}_u\}H_u$ is bounded measurable.

For the decomposition fix $x\in[0,R]$ and let $j^{-}$ be the largest index $j\in\{0,1,\dots,J\}$ with $b_j\le x$ (it exists since $b_0=0\le x$). Then $\mathbf{1}\{b_j\le x\}=1$ exactly for $j\le j^{-}$, and, telescoping with $p(b_0)=p(0)=0$ and $b_0=0$,

$$\sum_{j=1}^{J}(\mathsf{K}_j-\mu_j)\mathbf{1}\{b_j\le x\}=\sum_{j=1}^{j^{-}}\bigl(p(b_j)-p(b_{j-1})-(b_j-b_{j-1})\bigr)=p(b_{j^{-}})-b_{j^{-}} .$$

Hence $\bar{M}(x)-\sum_j(\mathsf{K}_j-\mu_j)\mathbf{1}\{b_j\le x\}=p(x)-p(b_{j^{-}})-(x-b_{j^{-}})$. If $j^{-}=J$ then $x=b_J=R$ and this vanishes. Otherwise $b_{j^{-}}\le x<b_{j^{-}+1}$, so $0\le x-b_{j^{-}}<\mu_{j^{-}+1}\le\mu_{\max}$, and the window $[b_{j^{-}},x]$ has length at most $\mu_{\max}$; by the definition of the window discrepancy the absolute value is at most $\mathrm{Disc}_{\mu_{\max}}(p)$.

Claim 2. Fix $u\in[0,s]$ and let $x\le x'$ be the two numbers $\mathsf{C}_u$, $\bar{\mathsf{C}}_u$ in increasing order; then $x'-x\le w_1$ and $\bar{M}(x')-\bar{M}(x)=p(x')-p(x)-(x'-x)$, whose absolute value is at most $\mathrm{Disc}_{w_1}(p)$ by definition. Consequently

$$\Lambda(\mathsf{C})-\Lambda(\bar{\mathsf{C}})=v\bigl(\bar{M}(\mathsf{C}_s)-\bar{M}(\bar{\mathsf{C}}_s)\bigr)+\int_{[0,s]}H_u\bigl(\bar{M}(\mathsf{C}_u)-\bar{M}(\bar{\mathsf{C}}_u)\bigr)\,du$$

by linearity of the integral, and by the triangle inequality, Norm Bound for a Vector-Valued Lebesgue Integral over a Compact Interval and monotonicity, its norm is at most $|v|\,\mathrm{Disc}_{w_1}(p)+\int_{[0,s]}|H_u|\,\mathrm{Disc}_{w_1}(p)\,du=(|v|+\lVert H\rVert_1)\,\mathrm{Disc}_{w_1}(p)$.

Claim 3. For $x\in[0,R]$ write $\mathrm{r}(x)=\bar{M}(x)-\sum_{j}(\mathsf{K}_j-\mu_j)\mathbf{1}\{b_j\le x\}$, so that $|\mathrm{r}(x)|\le\mathrm{Disc}_{\mu_{\max}}(p)$ by claim 1. The map $u\mapsto\mathrm{r}(\bar{\mathsf{C}}_u)$ is bounded measurable, being $\bar{M}(\bar{\mathsf{C}}_u)$ minus a finite linear combination of the measurable indicators of claim 1. Substituting $\bar{M}(\bar{\mathsf{C}}_u)=\sum_j(\mathsf{K}_j-\mu_j)\mathbf{1}\{b_j\le\bar{\mathsf{C}}_u\}+\mathrm{r}(\bar{\mathsf{C}}_u)$ into $\Lambda(\bar{\mathsf{C}})$ and using linearity of the integral for the finite sum,

$$\Lambda(\bar{\mathsf{C}})=\sum_{j=1}^{J}(\mathsf{K}_j-\mu_j)\Bigl(v\,\mathbf{1}\{b_j\le\bar{\mathsf{C}}_s\}+\int_{[0,s]}\mathbf{1}\{b_j\le\bar{\mathsf{C}}_u\}H_u\,du\Bigr)+v\,\mathrm{r}(\bar{\mathsf{C}}_s)+\int_{[0,s]}H_u\,\mathrm{r}(\bar{\mathsf{C}}_u)\,du ,$$

and the first sum is $\sum_j(\mathsf{K}_j-\mu_j)\alpha_j$. The remaining two terms have norm at most $|v|\,\mathrm{Disc}_{\mu_{\max}}(p)+\lVert H\rVert_1\,\mathrm{Disc}_{\mu_{\max}}(p)$ by the same argument as in claim 2.

Claim 4. Combine claims 2 and 3 with the triangle inequality.