[proofplan]
We prove that the Lebesgue-Stieltjes integral $H \cdot A$ of a previsible process against a finite variation process is itself a finite variation process, with the main content being the proof of adaptedness. The finite variation and cadlag properties follow from deterministic pathwise theory. For adaptedness, we proceed in two stages. First, we verify adaptedness on generators of the previsible $\sigma$-algebra: for $H(\omega, s) = \mathbb{1}_E(\omega) \cdot \mathbb{1}_{(u,v]}(s)$ with $E \in \mathcal{F}_u$, the integral $(H \cdot A)_t = \mathbb{1}_E (A_{t \wedge v} - A_{t \wedge u})$ is $\mathcal{F}_t$-measurable. Second, we extend to all bounded previsible $H$ via the monotone class theorem applied to the $\pi$-system of previsible rectangles, then to locally integrable $H$ by truncation.
[/proofplan]
[step:Establish that $H \cdot A$ has cadlag paths and finite variation pathwise]
Fix $\omega \in \Omega$. The path $t \mapsto A_t(\omega)$ is cadlag and in $BV$, so it induces a signed measure $\mu_\omega$ on $[0,\infty)$ via $\mu_\omega((s,t]) = A_t(\omega) - A_s(\omega)$. Since $H(\omega, \cdot)$ is Borel measurable (as a previsible process restricted to a single $\omega$) and satisfies $\int_0^t |H_s(\omega)|\, |dA_s(\omega)| < \infty$ by hypothesis, the Lebesgue-Stieltjes integral
\begin{align*}
(H \cdot A)_t(\omega) = \int_0^t H_s(\omega)\, dA_s(\omega) = \int_{(0,t]} H_s(\omega)\, \mu_\omega(ds)
\end{align*}
is well-defined for each $t$. Right-continuity of $t \mapsto (H \cdot A)_t(\omega)$ follows from the right-continuity of $t \mapsto \mu_\omega((0,t])$ (which is the right-continuity of $A$) together with the dominated convergence theorem: as $t_n \downarrow t$, $\mathbb{1}_{(0,t_n]} \to \mathbb{1}_{(0,t]}$ pointwise, dominated by $\mathbb{1}_{(0,t_1]}$. Left limits exist by the same argument with $t_n \uparrow t$.
The total variation of $t \mapsto (H \cdot A)_t(\omega)$ on $[0,T]$ satisfies
\begin{align*}
V_{H \cdot A}(T, \omega) \leq \int_0^T |H_s(\omega)|\, |dA_s(\omega)| < \infty,
\end{align*}
so $(H \cdot A)(\omega, \cdot) \in BV[0,T]$ for every $T$ and every $\omega$.
[/step]
[step:Verify adaptedness for generators $H = \mathbb{1}_E \cdot \mathbb{1}_{(u,v]}$ with $E \in \mathcal{F}_u$]
Consider the predictable rectangle $F = E \times (u,v]$ where $E \in \mathcal{F}_u$ and $0 \leq u < v$. The corresponding integrand is $H(\omega, s) = \mathbb{1}_E(\omega) \cdot \mathbb{1}_{(u,v]}(s)$. For any $t \geq 0$,
\begin{align*}
(H \cdot A)_t(\omega) &= \int_0^t \mathbb{1}_E(\omega) \cdot \mathbb{1}_{(u,v]}(s)\, dA_s(\omega) \\
&= \mathbb{1}_E(\omega) \cdot \bigl(A_{t \wedge v}(\omega) - A_{t \wedge u}(\omega)\bigr).
\end{align*}
We verify that this is $\mathcal{F}_t$-measurable. The factor $\mathbb{1}_E$ satisfies $E \in \mathcal{F}_u \subseteq \mathcal{F}_t$ (since $u \leq t$ when the integral is nonzero, and when $t < u$ the expression is zero hence automatically measurable). The factor $A_{t \wedge v}$ is $\mathcal{F}_{t \wedge v}$-measurable since $A$ is adapted, and $\mathcal{F}_{t \wedge v} \subseteq \mathcal{F}_t$. Similarly $A_{t \wedge u}$ is $\mathcal{F}_{t \wedge u}$-measurable with $\mathcal{F}_{t \wedge u} \subseteq \mathcal{F}_t$. The product of $\mathcal{F}_t$-measurable functions is $\mathcal{F}_t$-measurable.
The same argument applies to generators of the form $H(\omega, s) = \mathbb{1}_E(\omega) \cdot \mathbb{1}_{\{0\}}(s)$ with $E \in \mathcal{F}_0$: then $(H \cdot A)_t = 0$ for all $t$ (since the Lebesgue-Stieltjes integral does not charge the point $\{0\}$ in the half-open formulation), which is $\mathcal{F}_t$-measurable (as a constant).
[guided]
The key observation is that the integral of an indicator of a predictable rectangle reduces to the difference of stopped values of $A$. For $H = \mathbb{1}_E \cdot \mathbb{1}_{(u,v]}$, the integral $\int_0^t H_s\, dA_s$ only "sees" the increments of $A$ on the interval $(u, v] \cap (0, t] = (u, t \wedge v]$, giving $\mathbb{1}_E \cdot (A_{t \wedge v} - A_{t \wedge u})$.
Why is this $\mathcal{F}_t$-measurable? We need three things:
1. $\mathbb{1}_E \in \mathcal{F}_t$: Since $E \in \mathcal{F}_u$ and $u \leq v$, if $t \geq u$ then $\mathcal{F}_u \subseteq \mathcal{F}_t$ by the filtration property. If $t < u$, then $t \wedge v = t \wedge u = t$, so $A_{t \wedge v} - A_{t \wedge u} = 0$ and the product is zero regardless of $\mathbb{1}_E$.
2. $A_{t \wedge v} \in \mathcal{F}_t$: The stopped process $A_{t \wedge v}$ evaluates $A$ at time $t \wedge v \leq t$, and adaptedness of $A$ gives $A_{t \wedge v} \in \mathcal{F}_{t \wedge v} \subseteq \mathcal{F}_t$.
3. $A_{t \wedge u} \in \mathcal{F}_t$: Same reasoning with $u$ in place of $v$.
This step establishes the base case for the monotone class argument that follows.
[/guided]
[/step]
[step:Identify the $\pi$-system of predictable rectangles generating $\mathcal{P}$]
Let $\Pi$ denote the collection of predictable rectangles:
\begin{align*}
\Pi = \bigl\{ E \times (u,v] : E \in \mathcal{F}_u,\ 0 \leq u < v \bigr\} \cup \bigl\{ E \times \{0\} : E \in \mathcal{F}_0 \bigr\}.
\end{align*}
We verify that $\Pi$ is a $\pi$-system (i.e., closed under finite intersections). For two sets $F_1 = E_1 \times (u_1, v_1]$ and $F_2 = E_2 \times (u_2, v_2]$:
\begin{align*}
F_1 \cap F_2 = (E_1 \cap E_2) \times \bigl((u_1, v_1] \cap (u_2, v_2]\bigr).
\end{align*}
Setting $u = \max(u_1, u_2)$ and $v = \min(v_1, v_2)$, this is either empty (if $u \geq v$) or equals $(E_1 \cap E_2) \times (u, v]$. The set $E_1 \cap E_2 \in \mathcal{F}_{u_1} \cap \mathcal{F}_{u_2} = \mathcal{F}_{\min(u_1, u_2)} \subseteq \mathcal{F}_u$, so $F_1 \cap F_2 \in \Pi \cup \{\varnothing\}$. Similar verification holds for intersections involving $\{0\}$-type sets. By definition, the previsible $\sigma$-algebra is $\mathcal{P} = \sigma(\Pi)$.
[/step]
[step:Apply the monotone class theorem to extend adaptedness to all bounded previsible $H$]
Define
\begin{align*}
\mathcal{V} = \bigl\{ H: \Omega \times [0,\infty) \to \mathbb{R} \text{ bounded and } \mathcal{P}\text{-measurable} : (H \cdot A)_t \in \mathcal{F}_t \text{ for all } t \geq 0 \bigr\}.
\end{align*}
We verify the hypotheses of the monotone class theorem for functions:
1. **$\mathcal{V}$ contains indicators of $\Pi$-sets:** This was shown in the second step: $\mathbb{1}_F \in \mathcal{V}$ for every $F \in \Pi$.
2. **$\mathcal{V}$ contains the constant function $1$:** The constant process $H \equiv 1$ gives $(H \cdot A)_t = A_t - A_0$. Since $A$ is adapted, $A_t \in \mathcal{F}_t$ and $A_0 \in \mathcal{F}_0 \subseteq \mathcal{F}_t$, so $(H \cdot A)_t \in \mathcal{F}_t$.
3. **$\mathcal{V}$ is closed under bounded monotone limits:** Let $(H^n)_{n \geq 1}$ be a sequence in $\mathcal{V}$ with $0 \leq H^n \uparrow H$ pointwise and $\sup_n \|H^n\|_\infty < \infty$. For each $t$ and $\omega$,
\begin{align*}
(H^n \cdot A)_t(\omega) = \int_0^t H^n_s(\omega)\, dA_s(\omega) \to \int_0^t H_s(\omega)\, dA_s(\omega) = (H \cdot A)_t(\omega)
\end{align*}
by the dominated convergence theorem applied pathwise with dominator $\|H\|_\infty \in L^1([0,t], |\mu_\omega|)$. Each $(H^n \cdot A)_t$ is $\mathcal{F}_t$-measurable (since $H^n \in \mathcal{V}$), and the pointwise limit of $\mathcal{F}_t$-measurable functions is $\mathcal{F}_t$-measurable. Therefore $H \in \mathcal{V}$.
By the monotone class theorem, $\mathcal{V}$ contains all bounded $\sigma(\Pi)$-measurable functions, i.e., all bounded $\mathcal{P}$-measurable functions. In other words, $H \cdot A$ is adapted whenever $H$ is bounded and previsible.
[guided]
The monotone class theorem (also called the functional monotone class theorem) states: if $\Pi$ is a $\pi$-system and $\mathcal{V}$ is a vector space of bounded functions that contains $1$, contains $\mathbb{1}_F$ for every $F \in \Pi$, and is closed under bounded monotone limits, then $\mathcal{V}$ contains all bounded $\sigma(\Pi)$-measurable functions.
We apply this with $\Pi$ being the predictable rectangles and $\mathcal{V}$ being the class of bounded previsible processes for which $H \cdot A$ is adapted. The key conditions to verify are:
- **Generators:** We showed in the previous step that $\mathbb{1}_F \in \mathcal{V}$ for all $F \in \Pi$.
- **Closure under monotone limits:** This is where the dominated convergence theorem enters. If $H^n \uparrow H$ with $\|H^n\|_\infty \leq M$, then for each fixed $\omega$, the integrands $s \mapsto H^n_s(\omega)$ are dominated by $M$, which is integrable against $|dA_s(\omega)|$ because $A(\omega, \cdot) \in BV$. The DCT gives pointwise convergence of the integrals, and measurability passes to the limit.
The monotone class theorem then "propagates" the adaptedness from the generators to all bounded previsible processes. This is a standard technique in stochastic calculus for proving properties that are easy to verify on simple processes and extend to general ones.
[/guided]
[/step]
[step:Extend to locally integrable previsible $H$ by truncation]
For a general previsible $H$ satisfying $\int_0^t |H_s(\omega)|\, |dA_s(\omega)| < \infty$ for all $(\omega, t)$, define the truncations
\begin{align*}
H^n(\omega, s) = H(\omega, s) \cdot \mathbb{1}_{\{|H(\omega,s)| \leq n\}}.
\end{align*}
Each $H^n$ is bounded (by $n$) and $\mathcal{P}$-measurable (since $H$ is $\mathcal{P}$-measurable and the truncation preserves measurability). By the previous step, $(H^n \cdot A)_t \in \mathcal{F}_t$ for all $t$.
For each fixed $(\omega, t)$, $H^n_s(\omega) \to H_s(\omega)$ as $n \to \infty$ for every $s$, and $|H^n_s(\omega)| \leq |H_s(\omega)|$ with $\int_0^t |H_s(\omega)|\, |dA_s(\omega)| < \infty$. By the dominated convergence theorem,
\begin{align*}
(H^n \cdot A)_t(\omega) = \int_0^t H^n_s(\omega)\, dA_s(\omega) \to \int_0^t H_s(\omega)\, dA_s(\omega) = (H \cdot A)_t(\omega).
\end{align*}
Since each $(H^n \cdot A)_t$ is $\mathcal{F}_t$-measurable and $(H \cdot A)_t$ is their pointwise limit, $(H \cdot A)_t$ is $\mathcal{F}_t$-measurable. This holds for all $t \geq 0$, so $H \cdot A$ is adapted.
Combining with the cadlag and finite variation properties from the first step, $H \cdot A$ is a finite variation process.
[/step]
