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If a càdlàg function has finite variation on compact intervals, its total-variation process is finite, nondecreasing, and càdlàg. Left limits follow from monotonicity. For right continuity, approximate the variation on $[t,s]$ by one finite partition. Changing its first endpoint from $t$ to $u\downarrow t$ changes the sum by at most $|X_u-X_t|$, which tends to zero. Additivity of variation then bounds $V(u)-V(t)$ by an arbitrarily small error. Its jump is the absolute jump of $X$. Without locally finite variation, right continuity can fail even as an extended-valued assertion: $X_t=t\sin (1/t)$ with $X_0=0$ has infinite variation on every interval starting at zero.

For continuous local martingales, the total-variation process of their quadratic covariation satisfiesIt is the Cauchy-Schwarz inequality for the matrix-valued measure formed by their quadratic variations and quadratic covariation.

A continuous adapted process $M$ is a local martingale if there are stopping times ${\tau }_n\uparrow \infty$ almost surely such that each stopped process $M_t^{{\tau }_n}=M_{t\wedge {\tau }_n}$ is a martingale. In this definition $M_0$ is integrable. A continuous adapted process $A$ has finite variation if, almost surely, its paths have finite total variation of a function on each compact time interval:The permitted fact that this total-variation process is continuous and adapted makes its level-hitting times stopping times.

Suppose $M=A$, and put $N=M-M_0$. DefineThese times increase to infinity almost surely. The optional stopping theorem applied to the localized martingale shows that $N^{{\sigma }_n}$ is a martingale; it is also bounded in absolute value by $n$, with total variation at most $n$. Fix $n,t$ and write $L=N^{{\sigma }_n}$. For a deterministic partition of an interval $0=t_0<\cdots <t_r=t$, orthogonality of square-integrable martingale increments givesPathwise the sum on the right is bounded by $n{\max }_j|L_{t_j}-L_{t_{j-1}}|$. For partitions with mesh tending to zero, continuity makes this converge to zero. It is bounded by $2n^2$, so dominated convergence gives $\mathbb{E}{L}_t^2=0$. First use rational $t$, then continuity, to obtain $L\equiv 0$ almost surely. Letting $n\to \infty$ provesThis proves continuous finite-variation local martingale is constant without assuming integrability of the unstopped total variation. Continuity matters: compensated jump martingales can have finite variation without being constant.

The initial-value convention is essential. A continuous finite-variation local martingale is constant, so the conclusion is $M_t=M_0$ up to indistinguishability of stochastic processes. The stated zero conclusion uses the usual zero-start convention $M_0=0$; without it, $M_t\equiv 1$ is a counterexample.

Put $N=M-M_0$. Let $V_t$ be its total-variation process, and stop when either $|N|$ or $V$ reaches $n$. The resulting process $N^{{\tau }_n}$ is a bounded continuous local martingale, hence a true martingale, and its total variation on every time interval is bounded by $n$. Fix $t$ and take deterministic partitions $0=t_0<\cdots <t_k=t$ whose mesh tends to zero. Orthogonality of martingale increments givesPathwise,by uniform continuity of a continuous path on $[0,t]$. The sums are bounded by $2n^2$, so the dominated convergence theorem makes their expectations tend to zero. Hence $N_t^{{\tau }_n}=0$ almost surely. Take the countable intersection over rational $t$ and integer $n$, use path continuity, and then let $n\to \infty$. The stopping times tend to infinity because the original path and its finite variation are bounded on every compact time interval. Thus

Fix a sample path on the probability-one event where $X$ is continuous and the total-variation process of $A$ satisfies $V_{\infty }\le K$. The signed measure $dA$ has absolute variation measure $dV$. ThusThe integrand tends to zero at every time, including times when $X_s=0$, and is bounded by $2$. Since $dV$ is a finite measure on this path, the dominated convergence theorem makes the right-hand side tend to zero. ThereforeIn fact the convergence is uniform over the entire time axis under the given total-variation bound. This pathwise estimate establishes one common full-probability event, rather than separate exceptional events for each time.

Set $h_n=2^{-n}$ and $t_n=h_n\lceil t/h_n\rceil$. Replace the last sampled value $X_{t_n}$ by $X_t$, and call the resulting absolute-increment sum $W_n(t)$. This is a sum over a genuine partition of an interval of $[0,t]$, andby right continuity. This is the dyadic approximation of total variation. Let $V(t)$ be the total variation of a function $X$ on $[0,t]$, allowing $+\infty$. Every $W_n(t)$ is at most $V(t)$. Conversely, take any finite partition of an interval of $[0,t]$. Approximate its interior points from the right by dyadic grid points, retaining the endpoint $t$. For large $n$ these approximating points remain distinct, and right continuity makes their absolute-increment sum converge to that of the original partition. The triangle inequality makes $W_n(t)$ at least this selected sum. Taking the supremum over all partitions provesThe limit is the total-variation function; for locally finite variation it is the total-variation process.

There is a missing hypothesis in the assertion about being càdlàg. A general càdlàg function can have infinite total variation of a function arbitrarily close to zero. For example,is continuous. At the successive points $s_j=(\pi /2+j\pi {)}^{-1}$ its values alternate in sign with magnitude $s_j$. The variation along these points dominates a divergent harmonic series. Thus $V(t)=\infty$ for every $t>0$, but $V(0)=0$, so even extended-valued right continuity fails at zero.

The correct càdlàg conclusion holds when $X$ has finite variation on every compact interval. Indeed $V$ is then finite and nondecreasing, so it has finite left limits. To prove right continuity at $t$, fix $s>t$ and choose a partition of $[t,s]$ whose sum is within $\epsilon$ of its variation. If $u>t$ lies before the first interior partition point, changing the first endpoint from $t$ to $u$ changes that sum by at most $|X_u-X_t|$. Additivity of total variation of a function therefore givesLet $u\downarrow t$ and then $\epsilon \downarrow 0$. This proves right continuity. This proves càdlàg regularity of finite total variation. Under that qualification, the jump formula isIf the variation is infinite, a difference of two infinite variation values is not a defined jump.

A left-continuous adapted process is a predictable process. Define the elementary predictable approximationAt every $s>0$, its sample time approaches $s$ from the left, including when $s$ is itself a grid point. Thus $H_s^{(n)}\to H_s$ pathwise by left continuity.

Use the continuous semimartingale decomposition $X=X_0+M+A$, with $M$ a continuous local martingale and $A$ a continuous finite-variation process. Localize so that on a fixed horizon $T$, the integrands are bounded by a deterministic constant, $[M{]}_T$ is bounded, and the total-variation process of $A$ at $T$ is bounded. This is possible using local boundedness and the finite pathwise clock and variation.

For the finite-variation integral, the dominated convergence theorem givesalmost surely. For the martingale integral, the Itô isometry and the Doob L2 maximal inequality giveHere the localized bounds justify dominated convergence also in expectation. Remove localization to conclude that the full elementary integrals converge in uniform convergence on compacts in probability to $\int HdX$.

The sum of completed intervals differs from the full elementary integral at time $t$ byIts maximum on $[0,T]$ tends to zero almost surely, by the local bound on $H$ and uniform continuity of $X$. ThereforeThis proves left-endpoint approximation of a continuous semimartingale integral, including the unfinished interval convention.

Use the standard convention that the finite-variation process $X$ has càdlàg paths and that the stochastic integral starts from zero. Write $V(X)$ for its total-variation process. The necessary admissibility condition is ${\int }_{(0,t]}|H_s|d{V}_s(X)<\infty$ almost surely for each finite $t$; it holds, for example, for locally bounded $H$. Define the pathwise Lebesgue-Stieltjes integralAn atom at $s>0$ contributes $H_s\Delta X_s$, so this convention also fixes the treatment of jumps.

Fix $t$. For a generator $H_s=1_A{1}_{(a,b]}(s)$ of the predictable sigma-algebra, with $A\in {\mathcal{F}}_a$, the Lebesgue-Stieltjes integral isIt vanishes when $t\le a$ and is ${\mathcal{F}}_t$-measurable otherwise. The total-variation process is adapted: for a càdlàg path its value at $t$ is the supremum over a countable collection of partitions of an interval using rational interior points and endpoint $t$. Thus the positive and negative Lebesgue-Stieltjes measures $d{X}^{\pm }=(dV(X)\pm dX)/2$ have ${\mathcal{F}}_t$-measurable masses on the generator intervals.

The Monotone class theorem, applied separately to these positive random measures on $[0,t]$, extends this measurability from the generators of the predictable sigma-algebra to all nonnegative predictable processes. Truncation and positive/negative decomposition then cover every admissible $H$. Consequently $H\cdot X$ is adapted.

The integrability condition is essential. Taken literally, an arbitrary previsible process cannot always be integrated against a finite-variation process: with $X_t=t$ and deterministic $H_s=s^{-1}$ for $s>0$, $H_0=0$, the proposed value at every $t>0$ diverges. The usual càdlàg convention is also what identifies $dX$ by $dX((a,b])=X_b-X_a$.

The usual quadratic variation convention includes $⟨M{⟩}_0=0$. If $A,C$ are two continuous finite-variation processes with $A_0=C_0=0$ such that $M^2-A$ and $M^2-C$ are continuous local martingales, then $A-C$ is a continuous local martingale of finite variation. It must be constant, and its initial value is zero, so $A=C$ up to indistinguishability of stochastic processes.

Here is a proof of the fact used. Subtract the initial value of a continuous local martingale $N$ of finite variation, and stop so that both $|N|$ and its total-variation process are bounded by deterministic constants. The stopped local martingale is a bounded true martingale, so orthogonality of its increments gives, on any deterministic partition of an interval of $[0,t]$,Continuity and the dominated convergence theorem justify the last limit. Remove stopping and use continuity on a countable dense set of times to conclude $N\equiv 0$.

As literally printed, uniqueness is false without normalization. If $A$ works then $A+c$ also works for any deterministic constant $c$, since subtracting a constant preserves the local martingale property; this is a genuine omitted convention.

With normalization, existence always holds for a continuous local martingale. Its quadratic variation $[M]$, obtained as the limit in probability of squared increments, is continuous, increasing, starts at zero and satisfies $M^2-[M]$ is a continuous local martingale, so taking $⟨M⟩=[M]$ supplies the required finite-variation process.

With the notation from part (i), the total-variation process of $[M,N]$ isFor the centered bivariate normal distribution $(U,V)$ used there, $c=\mathbb{E}[UV]$. The integral triangle inequality givesIntegrating this pointwise inequality against $dC$ proves $V_t\le {\tilde{V}}_t$ for every $t\ge 0$.