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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.

For continuous semimartingales, the Stratonovich integral adds half the quadratic covariation to the Itô integral:One can see the factor directly from symmetric endpoint sums. On a partition of an interval, replacing $X$ at the left endpoint by the average of its two endpoint values adds $\frac{1}{2}\sum \Delta X\Delta Y$; its limit is $\frac{1}{2}[X,Y]$. Thus the correction depends only on the continuous local martingale parts of the semimartingales. If either integrator or integrand has finite variation, the quadratic covariation vanishes and these two integrals coincide. Here the printed notation $\partial Y$ denotes Stratonovich integral integration, equivalently $\circ dY$.

For deterministic step functions, define directly from Brownian motion incrementswith $s_0$ the left endpoint of the first interval and $f=0$ outside the listed intervals. The printed endpoint list omits $s_0$; the intended convention is $0\le s_0<s_1<\cdots <s_n$, usually $s_0=0$.

Given finitely many observation times, refine the deterministic partition of an interval to include all observation times and all $s_i$. Every $Z_{t_j}$ is a linear combination of the same finite family of independent centered normal Brownian motion increments. Thus every linear combination of $(Z_{t_1},\dots ,Z_{t_m})$ is normal, including possibly a degenerate constant: $Z$ is a Gaussian process. By independence, only increments in the common interval $[0,s\wedge t]$ contribute to the covariance, givingThe same calculation for two step functions $f,g$ proves, rather than assumes, the Itô isometryFor $f\in L^2({\mathbb{R}}_{+})$, choose deterministic step functions $f_n\to f$ in the L2 space. This identity makes $Z_t(f_n)$ a Cauchy sequence in the L2 space; the Riesz-Fischer theorem defines $Z_t(f)$ by mean-square convergence, independently of the approximation. The covariance formula passes to the limit by the Cauchy-Schwarz inequality. For finitely many times and coefficients $a_j$, the characteristic functions converge toso the extension remains a Gaussian process.

For completeness, the construction has a continuous version. Each approximating $Z(f_n)$ is a martingale, directly by the centered independent increments of Brownian motion; the Doob L2 maximal inequality givesTo prove the inequality used here without assuming Itô integral properties, a continuous martingale $N$ satisfies $\lambda \mathbb{P}(N_T^{\ast }\ge \lambda )\le \mathbb{E}(|N_T|1_{\{N_T^{\ast }\ge \lambda \}})$, by conditioning at the first crossing on finite deterministic grids and then using continuity; integrating in $\lambda$ and using the Cauchy-Schwarz inequality yields $\Vert N_T^{\ast }{\Vert }_2\le 2\Vert N_T{\Vert }_2$. A sufficiently fast subsequence therefore converges uniformly almost surely on every compact time interval, giving the continuous version of the stochastic integral.

For a deterministic partition of an interval $0=t_0<\cdots <t_m=t$, put $Q_{\pi }={\sum }_j{\Delta }_{j}B{\Delta }_j\tilde{B}$. Independence of the two Brownian motions and their independent increments givesThe defining sums for quadratic covariation therefore converge to zero in the L2 space and in probability. Hence, with the continuous versions of the quadratic covariation,Equality holds simultaneously for all $t$ by checking rational times and using continuity. For continuous local martingales, the angle-bracket notation here agrees with quadratic covariation.

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.

For a partition of an interval $P:0=x_0<\cdots <x_m=1$, putThe upper Darboux sum and lower Darboux sum areThe upper and lower integrals areThe bounded function is Riemann integrable when these values agree, and their common value is its Riemann integral.

For the indicator of the rational numbers, every nondegenerate interval contains both a rational number and an irrational number. Hence every upper sum is one and every lower sum is zero. The function is not Riemann integrable.

A càdlàg process $A$ is a finite-variation process when, almost surely, for every $T<\infty$,where the supremum is over every finite partition of an interval $0=t_0<\cdots <t_n=T$.

Uniform convergence on compacts in probability of $X^n$ to $X$ means that, for every $T<\infty$ and $\epsilon >0$,

Apply the realized absolute covariation theorem to each dyadic partition of an interval. More explicitly, use the continuous increasing clock $C=[M]+[N]$ and the Radon-Nikodym theorem to writeFor each time, let $(U,V)$ have the centered bivariate normal distribution with covariance matrix $\left(\begin{array}{cc}a& c\\ c& b\end{array}\right)$, and defineThis process is continuous and increasing. To prove convergence, localize $M,N$, represent the pair as stochastic integrals against a two-dimensional Brownian motion, and approximate the integrands in $L^2(dC)$ by bounded step previsible processes. For step integrands, the result is the weak law of large numbers applied on each block to independent Gaussian random variables. The Burkholder-Davis-Gundy inequality and the Cauchy-Schwarz inequality make the error uniform on each compact interval in probability. Consequentlyin the sense of uniform convergence on compacts in probability.

False. For a common sequence, each term of which is a refining deterministic partition of an interval, standard Brownian paths have quadratic variation $t$ almost surely, whereas the paths $t\mapsto B_{2t}$ have quadratic variation $2t$ almost surely. These two path properties define disjoint measurable subsets of $C([0,1])$, so the two laws are mutually singular measures. In particular, the law of $B_{2\cdot }$ is not absolutely continuous with respect to Wiener measure. This is the Pathwise quadratic variation distinguishes Brownian speeds argument.

The difference between a weighted integral and a discrete approximation is a quadrature error. For a Hölder continuous function on a partition of an interval with mesh $\Delta$, a left endpoint approximation has error of order ${\Delta }^{\alpha }$. For a normalized window indicator function, only the two endpoint cells contribute and the error is bounded by a constant times $\Delta /h$.

A quadrature rule approximates an integral by a finite weighted sum of sampled function values:The quadrature nodes $x_i$ specify where to sample, and the quadrature weights $w_i$ specify how much each sample contributes. The density $\rho (x)$ specifies the target integral, whereas the discrete quadrature weights belong to the approximation. Ordinary numerical integration has $\rho =1$. Gaussian quadrature and the midpoint quadrature rule choose interior quadrature nodes; the trapezoidal rule includes endpoints.^[ref]

For ordinary numerical integration, exactness on the constant function $1$ requires ${\sum }_i{w}_i=b-a$; exactness on $x$ additionally requires ${\sum }_i{w}_i{x}_i=(b^2-a^2)/2$. These two checks make the quadrature rule exact for every affine function. For example, both the midpoint quadrature rule and trapezoidal rule give $6$ for ${\int }_0^2(2x+1)dx$. Testing progressively higher polynomial degrees defines the degree of precision of a quadrature rule. Exactness describes a class of integrands, rather than promising exact results for every function.

Composite midpoint quadrature rules and composite trapezoidal rules apply a simple quadrature rule on every panel of a partition of an interval. This buys accuracy by shrinking the panels while keeping the elementary formula fixed. For continuous functions sampled exactly, their quadrature errors tend to zero as the largest panel width tends to zero. The familiar second-order bounds require stronger assumptions about derivatives. The signed approximation error used here is $Q(f)-I_{\rho }(f)$; some sources define quadrature error with the opposite sign, so the convention must accompany any formula.

For continuous local martingales $M,N$ and a sequence in which each term is a partition of an interval whose mesh tends to zero, the sumsconverge in the sense of uniform convergence on compacts in probability to a continuous increasing process. To identify the limit, put $C=[M]+[N]$ and choose Radon-Nikodym derivativesIf $(U,V)$ is a centered bivariate normal distribution with covariance matrix $\left(\begin{array}{cc}a& c\\ c& b\end{array}\right)$, thenLocalizing, representing the pair as stochastic integrals against a two-dimensional Brownian motion, and approximating the integrands by bounded predictable step processes proves the convergence. The step-process case follows from the weak law of large numbers for independent Gaussian increments; the Burkholder-Davis-Gundy inequality controls the approximation error. Since $|\mathbb{E}[UV]|\le \mathbb{E}|UV|\le \sqrt{\mathbb{E}{U}^2\mathbb{E}{V}^2}$,

For a path of finite variation, its total-variation process iswhere the supremum is over every partition of an interval of $[0,t]$. It is an increasing process, and the signed measure $dA$ satisfies $|dA|=dV(A)$.

If Riemann integrable functions $f_n:[a,b]\to \mathbb{R}$ converge by uniform convergence to $f$, then $f$ is Riemann integrable andThis concerns one fixed compact interval, with finite length.^[ref]

Here is a proof outline that also establishes Riemann integrability of the uniform limit. Assume $a<b$, and put $\ell =b-a$ and $E_N=\Vert f_N-f{\Vert }_{\infty }$. One bounded $f_N$ and finite $E_N$ show that $f$ is a bounded function. For any partition of an interval $P$, its upper Darboux sum and lower Darboux sum satisfyGiven $\eta >0$, choose $N$ with $E_N<\eta /(4\ell )$, then a partition of an interval with $U(f_N,P)-L(f_N,P)<\eta /2$. The displayed gap is below $\eta$, proving Riemann integrability. Finally, $|f_n-f|\le E_n$ and monotonicity of the Riemann integral give the stated error estimate. When $a=b$, every integral is zero.

For the polynomial functions $f_n(x)=x^n$ on $[0,0.8]$, direct integration and the uniform bound giveAt $n=21$, the exact integral is approximately $0.0003353953$, while the general error bound is approximately $0.0073786976$. The bound certifies the result without an exact antiderivative. On $[0,1]$, these same integrals still tend to zero, although there is no uniform convergence: sufficient conditions need not be necessary. On an unbounded interval, even uniform convergence can fail to control total area, as uniformly vanishing functions can retain a nonzero integral shows.

Uniform convergence alone also fails to preserve derivatives: $f_n(x)=\sin (n^{2}x)/n$ converges uniformly to zero on $[-1,1]$, while $f_n^{\prime }(0)=n$. The small amplitude does not control the slopes of the sine oscillations.

For a bounded function $f$ and a partition of an interval $P$, the upper Darboux sum multiplies the supremum of $f$ on each subinterval by its length and adds the results.

For observations of an Itô integral on a partition of an interval, the statistic ${\sum }_{i}g(t_{i-1})(X_{t_i}-X_{t_{i-1}}{)}^2$ estimates $\int g(t)\sigma (t{)}^{2}dt$. With a deterministic noise coefficient the squared increments are independent, and the fluctuation has variance $2{\sum }_{i}g(t_{i-1}{)}^2({\int }_{t_{i-1}}^{t_i}{\sigma }^2{)}^2$.