[proofplan] We verify four properties of the total variation process $V$: that it is increasing, cadlag, of finite variation, and adapted. The first three are immediate from the pathwise definition -- each sample path $V(\omega, \cdot)$ is the total variation function of the BV path $A(\omega, \cdot)$, which is monotone increasing and inherits right-continuity with left limits from the definition. The non-trivial property is adaptedness: we express $V_t$ as a limit of sums involving $|A_{t_i} - A_{t_{i-1}}|$ over increasingly fine partitions of $[0,t]$, and use the fact that each summand is $\mathcal{F}_t$-measurable because $A$ is adapted. [/proofplan] [step:Verify that $V$ is increasing] For $0 \leq s \leq t$, every partition of $[0,s]$ can be extended to a partition of $[0,t]$ by appending the point $t$. Therefore the supremum over partitions of $[0,t]$ is at least as large as the supremum over partitions of $[0,s]$: \begin{align*} V_s(\omega) = \sup_{\pi \text{ of } [0,s]} \sum_{i} |A_{t_i}(\omega) - A_{t_{i-1}}(\omega)| \leq \sup_{\pi \text{ of } [0,t]} \sum_{i} |A_{t_i}(\omega) - A_{t_{i-1}}(\omega)| = V_t(\omega). \end{align*} Hence $V_s \leq V_t$ for every $\omega$, so $V$ is increasing. [/step] [step:Verify that $V$ has cadlag paths] Fix $\omega \in \Omega$. The path $t \mapsto A_t(\omega)$ is cadlag and in $BV[0,T]$ by hypothesis. We use the additive representation \begin{align*} V_t(\omega) - V_s(\omega) = \sup \Bigl\{ \sum_{i=1}^{n} |A_{r_i}(\omega) - A_{r_{i-1}}(\omega)| : s = r_0 < r_1 < \cdots < r_n = t \Bigr\} \end{align*} for $s < t$, which expresses $V_t - V_s$ as the total variation of $A(\omega, \cdot)$ on the interval $(s, t]$. **Right-continuity.** Fix $t_0 \in [0, T)$. We show $V_t(\omega) - V_{t_0}(\omega) \to 0$ as $t \downarrow t_0$. For any partition $t_0 = r_0 < r_1 < \cdots < r_n = t$ of $(t_0, t]$, the triangle inequality gives \begin{align*} \sum_{i=1}^{n} |A_{r_i}(\omega) - A_{r_{i-1}}(\omega)| \leq \sum_{i=1}^{n} \bigl(|A_{r_i}(\omega) - A_{t_0}(\omega)| + |A_{r_{i-1}}(\omega) - A_{t_0}(\omega)|\bigr). \end{align*} Since $A(\omega, \cdot)$ is right-continuous at $t_0$, for any $\varepsilon > 0$ there exists $\delta > 0$ such that $|A_u(\omega) - A_{t_0}(\omega)| < \varepsilon$ for all $u \in (t_0, t_0 + \delta)$. When $t \in (t_0, t_0 + \delta)$, every partition point $r_i$ lies in $(t_0, t_0 + \delta)$, so each $|A_{r_i}(\omega) - A_{r_{i-1}}(\omega)| \leq 2\varepsilon$. But we can do better: the total variation of $A(\omega, \cdot)$ on $(t_0, t]$ is bounded by the oscillation of $A$ on that interval. Specifically, $V_t(\omega) - V_{t_0}(\omega) \leq \text{Var}(A(\omega, \cdot); (t_0, t])$, and since $A(\omega, \cdot)$ is right-continuous at $t_0$, the variation on $(t_0, t]$ tends to zero as $t \downarrow t_0$. To see this rigorously: for any $\varepsilon > 0$, choose $\delta > 0$ so that $|A_u(\omega) - A_{t_0}(\omega)| < \varepsilon / 2$ for $u \in [t_0, t_0 + \delta]$. Then for any partition $t_0 = r_0 < \cdots < r_n = t$ with $t < t_0 + \delta$, the variation sum satisfies \begin{align*} \sum_{i=1}^{n} |A_{r_i} - A_{r_{i-1}}| \leq V_T(\omega), \end{align*} and since $A(\omega, \cdot)$ is BV, the set function $\text{Var}(A(\omega, \cdot); (s, u])$ defines a finite positive measure on $[0, T]$. A finite measure assigns vanishing mass to intervals shrinking to a point, so $\text{Var}(A(\omega, \cdot); (t_0, t]) \to 0$ as $t \downarrow t_0$. Hence $V_{t_0+}(\omega) = V_{t_0}(\omega)$. **Existence of left limits.** Since $V(\omega, \cdot)$ is increasing and bounded above by $V_T(\omega) < \infty$, for every $t_0 \in (0, T]$ the left limit $V_{t_0-}(\omega) = \sup_{s < t_0} V_s(\omega)$ exists and is finite. This is a standard property of bounded increasing functions: they have left limits at every point. [/step] [step:Verify that $V$ has finite variation] Since $V$ is increasing (as shown above), its total variation on $[0,T]$ equals $V_T(\omega) - V_0(\omega) = V_T(\omega)$. By hypothesis, $A(\omega, \cdot) \in BV[0,T]$, so $V_T(\omega) < \infty$ for every $\omega$. Therefore $V(\omega, \cdot) \in BV[0,T]$ for every $\omega$. [/step] [step:Prove adaptedness by expressing $V_t$ as a limit of $\mathcal{F}_t$-measurable sums] Fix $t \geq 0$. Let $\bigl(\pi_m\bigr)_{m \geq 1}$ be a nested sequence of partitions of $[0,t]$, where $\pi_m = \{0 = t_0^{(m)} < t_1^{(m)} < \cdots < t_{n_m}^{(m)} = t\}$, with $\max_i (t_i^{(m)} - t_{i-1}^{(m)}) \to 0$ as $m \to \infty$. Define \begin{align*} V_t^{(m)}(\omega) := \sum_{i=1}^{n_m} \bigl|A_{t_i^{(m)}}(\omega) - A_{t_{i-1}^{(m)}}(\omega)\bigr|. \end{align*} Each $A_{t_i^{(m)}}$ is $\mathcal{F}_{t_i^{(m)}}$-measurable because $A$ is adapted, and $t_i^{(m)} \leq t$ for all $i$. Since the filtration is increasing, $A_{t_i^{(m)}} \in \mathcal{F}_{t_i^{(m)}} \subseteq \mathcal{F}_t$. The map $x \mapsto |x|$ is Borel measurable, so each summand $|A_{t_i^{(m)}} - A_{t_{i-1}^{(m)}}|$ is $\mathcal{F}_t$-measurable. A finite sum of $\mathcal{F}_t$-measurable functions is $\mathcal{F}_t$-measurable, so $V_t^{(m)} \in \mathcal{F}_t$ for every $m$. Since the partitions are nested and their mesh tends to zero, the partial variation sums $V_t^{(m)}$ form a non-decreasing sequence (by the inclusion of partitions) converging to the total variation: \begin{align*} V_t(\omega) = \lim_{m \to \infty} V_t^{(m)}(\omega) = \sup_{m \geq 1} V_t^{(m)}(\omega). \end{align*} The pointwise limit (equivalently, supremum) of a sequence of $\mathcal{F}_t$-measurable functions is $\mathcal{F}_t$-measurable. Therefore $V_t \in \mathcal{F}_t$. Since $t \geq 0$ was arbitrary, $V$ is adapted to the filtration $(\mathcal{F}_t)_{t \geq 0}$. [guided] This is the substantive step. Why is adaptedness non-trivial? The total variation $V_t$ involves a supremum over all partitions of $[0,t]$, and while each partition sum is $\mathcal{F}_t$-measurable, a supremum over an uncountable family of measurable functions need not be measurable. The resolution is to restrict to a countable family that achieves the same supremum. Fix $t \geq 0$. We construct a nested sequence of partitions $(\pi_m)_{m \geq 1}$ of $[0,t]$ with $|\pi_m| \to 0$. For example, take $\pi_m$ to be the dyadic partition $\{k \cdot t / 2^m : k = 0, 1, \ldots, 2^m\}$, and define \begin{align*} V_t^{(m)}(\omega) := \sum_{i=1}^{n_m} \bigl|A_{t_i^{(m)}}(\omega) - A_{t_{i-1}^{(m)}}(\omega)\bigr|. \end{align*} **Why is each $V_t^{(m)}$ measurable with respect to $\mathcal{F}_t$?** Each partition point satisfies $t_i^{(m)} \leq t$. Since $A$ is adapted, $A_{t_i^{(m)}} \in \mathcal{F}_{t_i^{(m)}} \subseteq \mathcal{F}_t$ (using the increasing property of the filtration). The difference $A_{t_i^{(m)}} - A_{t_{i-1}^{(m)}}$ is $\mathcal{F}_t$-measurable as the difference of two $\mathcal{F}_t$-measurable functions. The absolute value and finite sum preserve measurability. **Why does $V_t^{(m)} \to V_t$?** Since the partitions are nested ($\pi_m \subseteq \pi_{m+1}$), the sums $V_t^{(m)}$ are non-decreasing in $m$ (adding points to a partition can only increase the variation sum). The limit equals the supremum over all partitions because the dyadic partitions are dense in $[0,t]$, and the cadlag property of $A$ ensures that no variation is "hidden" between partition points of a sufficiently fine partition. **Why does the limit inherit measurability?** The $V_t^{(m)}$ are non-decreasing, so $V_t = \sup_m V_t^{(m)} = \lim_m V_t^{(m)}$. The supremum of a countable family of $\mathcal{F}_t$-measurable functions is $\mathcal{F}_t$-measurable (this is a standard result in measure theory, following from the identity $\{\sup_m V_t^{(m)} \leq c\} = \bigcap_m \{V_t^{(m)} \leq c\} \in \mathcal{F}_t$). [/guided] [/step]