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A continuous martingale is a continuous-time martingale whose paths are continuous outside one null event. Continuity does not remove the integrability or conditional-expectation requirements in the martingale definition.

For Brownian motion started at zero, let $f_n$ equal $|x|$ outside $(-1/n,1/n)$ and $n{x}^2/2+1/(2n)$ inside. Its derivative is the clipped sign function $g_n(x)=\max \{-1,\min \{1,nx\}\}$. The Itô integral $M_t^{(n)}={\int }_0^t{g}_n(B_s)d{B}_s$ converges uniformly on bounded time intervals in $L^2$ to a continuous martingale $M$, by the Itô isometry and Doob L2 maximal inequality: the squared maximal error is at most $8\sqrt{2T/\pi }/n$. The Itô formula gives $\frac{n}{2}{\int }_0^t{1}_{\{|B_s|<1/n\}}ds=f_n(B_t)-f_n(0)-M_t^{(n)}$. Along $n=k^2$, the maximal errors are summable, so First Borel-Cantelli lemma gives uniform almost-sure convergence. The ratio-one interpolation for shrinking-window occupation integrals extends it to every integer $n$. The limit is $|B_t|-M_t=L_t^0$, the local time of a semimartingale in the Tanaka formula convention. The unnormalized factor $n$ gives twice this limit.

For a bounded simple predictable process $H$ and a continuous martingale $M\in {\mathcal{H}}^{2,c}$, the Itô isometry statesHere $⟨M⟩$ is the predictable quadratic variation, equal to $[M]$ for a continuous martingale. On the whole half-line, the terminal form isThe Itô integral starts at zero even when $M_0\ne 0$. For Brownian motion, $d⟨B{⟩}_s=ds$.

For completeness, cross terms of the finite increment sum have expectation zero by conditioning on the earlier endpoint of the later increment. Also $M^2-⟨M⟩$ is a martingale, so the remaining terms satisfyThis proves the displayed identity without a change-of-variables formula. Completing bounded simple predictable processes in the norm $(\mathbb{E}\int H^{2}d⟨M⟩{)}^{1/2}$ defines the Itô integral of every square-integrable stochastic integrand. This completion will justify the unbounded Brownian integrands below.

Let $0=t_0<t_1<\cdots <t_m<\infty$ and let ${\xi }_j$ be bounded ${\mathcal{F}}_{t_j}$-measurable random variables. For the bounded simple predictable processthe Itô integral against a continuous martingale $M\in {\mathcal{H}}^{2,c}$ is defined by the increment sumThe same definition applies to elementary predictable processes with stopping-time intervals, replacing $t_j$ by ordered stopping times $T_j$ and taking bounded ${\mathcal{F}}_{T_j}$-measurable coefficients. Refinement of the intervals leaves the increment sum unchanged. The initial value of the integral is zero, and its martingale property follows from the conditional expectation identity for subsequent increments of $M$.

Let $f(y)=|y|$ away from zero. Its first and second partial derivatives areThus its Laplace operator is $\Delta f(y)=(n-1)/|y|$. The quadratic covariations of the independent Brownian coordinates satisfy $d[B_j,B_k{]}_s={\delta }_{jk}ds$. The stopped Itô formula therefore givesFor $t<\tau$, this is the required integral representation.

To identify the driving Brownian motion, the stopped stochastic term alone is insufficient: its quadratic variation is $t\wedge \tau$. Extend its unit direction after exit. Define a vector predictable process $u(s)$ byPredictability follows from the continuous adapted direction before stopping and the left-continuous adapted indicator $1_{\{s\le \tau \}}$. This $W$ is a zero-starting continuous martingale, andThe Lévy characterization of Brownian motion identifies $W$ as a one-dimensional Brownian motion. Before $\tau$, the preceding representation becomes the stochastic differential equation of the dimension-$n$ Bessel process:The fully stopped equation has both integrals stopped at $\tau$.

For $n=2$, apply the one-dimensional Itô formula to the logarithm. Before $\tau$, $d[X{]}_t=dt$, so the two drift terms cancel:Consequently,Indeed $X_s\ge r$ before stopping, so the Itô isometry bounds the stochastic term's second moment by $t/r^2$, and $\log X_{t\wedge \tau }\in [\log r,\log R]$. This is the martingale assertion under the paper's standing stopping assumption. Without stopping, the same calculation establishes only a local martingale; the stopped qualification is essential for the stated argument.

Take a standard Brownian motion $B$ and defineThe Itô formula gives $N_t=2{\int }_0^{t\wedge 1}{B}_{s}d{B}_s$, so both are continuous martingales starting at zero. Their second moments are $\mathbb{E}{M}_t^2=t\wedge 1$ and $\mathbb{E}{N}_t^2=2(t\wedge 1{)}^2$, making them L2-bounded continuous martingales. At equal times, the odd moments of a centered normal random variable vanish, soThey are consequently weakly orthogonal continuous martingales by solution.

Their product, for $t<1$, is $B_t^3-t{B}_t$. The Itô formula givesThe continuous finite-variation term $2{\int }_0^t{B}_{s}ds$ is not identically zero: it has variance $4t^3/3$ at every $t>0$. Uniqueness of the continuous semimartingale decomposition therefore shows that $MN$ is not even a local martingale. Equivalently $[M,N{]}_t=2{\int }_0^{t\wedge 1}{B}_{s}ds$ is not zero, so the pair is not orthogonal continuous local martingales.

The previsible sigma-algebra $\mathcal{P}$ on $\Omega \times [0,\infty )$ is generated by $E\times \{0\}$ with $E\in {\mathcal{F}}_0$ and by $E\times (s,t]$ with $E\in {\mathcal{F}}_s$, $0\le s<t$. Equivalently it is the smallest sigma-algebra making every left-continuous adapted process measurable. A previsible process is a process measurable for $\mathcal{P}$.

A bounded simple predictable process has finitely many time intervals and bounded coefficients known at their left endpoints. One may use deterministic endpoints, or the customary more general elementary predictable process with stopping-time intervals:A bounded ${\mathcal{F}}_0$-measurable value at time zero may be added; it does not affect integration against a continuous martingale. A process with finite support is included by adding a zero coefficient on its terminal interval. Both versions of simple processes satisfy the arguments below. The elementary stochastic integral isWhen a stopping time is infinite, use the terminal value $M_{\infty }$, which exists in $L^2$ and almost surely for an L2-bounded continuous martingale. Refining intervals leaves this definition unchanged.

For a continuous local martingale $M$, its quadratic variation $[M]$ is the unique continuous adapted nondecreasing process starting at zero such that $(M_t-M_0{)}^2-[M{]}_t$ is a local martingale. It is predictable because it is continuous and adapted. To recover it from the path, use refining deterministic grids, for example $t_k^{(n)}=k{2}^{-n}$, and formOnly finitely many increments are nonzero on a compact time interval. ThenIn particular each fixed-time sum converges in probability, and a suitable subsequence converges uniformly on compact intervals almost surely. The partial last grid interval is included in this path approximation.

The intended setting is $X=M+A$ with $A$ continuous and adapted. These conditions are needed for the requested continuous-semimartingale conclusion; finite variation alone does not supply them. For example, $M=0$, $A_t=1_{\{t\ge 1\}}$ satisfies the printed deterministic bound and starts at zero, but $f_n(X)$ has a jump for sufficiently large $n$. We use the continuous adapted interpretation throughout this question.

Define $f_n(x)={\int }_0^x\varphi (ny)dy$. Then $f_n\in C^2$, $|f_n^{\prime }|\le 1$ and $f_n^{\prime \prime }(x)=n{\varphi }^{\prime }(nx)\ge 0$. The Itô formula gives the canonical continuous semimartingale decompositionThe initial value is zero. The first term is a continuous martingale, since $M$ is bounded and the integrand is bounded. The second is continuous adapted finite variation, hence predictable. This decomposition is unique because a continuous finite-variation local martingale is constant. The phrase Doob-Meyer here must mean this semimartingale decomposition: $V^{(n)}$ need not be nondecreasing, and the hypotheses do not make $f_n(X)$ a submartingale as required by the Doob-Meyer decomposition theorem. Indeed $f_n^{\prime }(0)=-1$, so the bounded deterministic drift $A_t=t\wedge \delta$ makes $f_n(A_t)$ initially decrease, excluding the submartingale property.

The sign process is predictable because it is a Borel function of the continuous adapted, and therefore predictable, process $X$. Its paths need not themselves be left-continuous. Pointwise, including at $x=0$,For a fixed horizon $T$, $\mathbb{E}[M{]}_T=\mathbb{E}{M}_T^2\le K^2$. The dominated convergence theorem for the finite measure $\mathbb{E}d[M]$ on $[0,T]$ givesThe Itô isometry, extended from simple predictable integrands by their density, and the Doob L2 maximal inequality now implyHence

For $f(x)=x$, the continuous martingale is $M_t^f=B_t$. Since $B_t$ has the centered normal distribution of variance $t$,For $f(x)=x^2$, the continuous martingale is $B_t^2-t$. The Gaussian fourth moment givesThis verifies both cases. The time integral in the printed identity is understood with its omitted $ds$.

Write $M^{(n)}=M^{f_n}$. The derivatives of the approximations areThus $|g_n|\le 1$. The Itô formula gives $M_t^{(n)}={\int }_0^t{g}_n(B_s)d{B}_s$. This use of the formula is valid despite the two second-derivative discontinuities: mollify the $C^1$ function $f_n$, whose derivative is Lipschitz. Its smooth derivatives converge boundedly, while the second derivatives are bounded by $n$ and converge except at $\pm 1/n$. For every $s>0$, the normal distribution of $B_s$ assigns those points zero probability. Dominated convergence for the drift integral and the Itô isometry for the stochastic integral pass the smooth formula to $f_n$. Equivalently, this identifies the martingale in the assumed decomposition and permits its cross-covariances to be calculated.

If $m\ge n$, then $|g_m-g_n|\le 1_{\{|x|<1/n\}}$. The Itô isometry and Doob L2 maximal inequality therefore give, for every fixed $T$,The Gaussian density is bounded by $(2\pi s{)}^{-1/2}$, soConsequently the processes are Cauchy in the complete space of continuous random paths with norm $(\mathbb{E}{\sup }_{s\le T}|Z_s{|}^2{)}^{1/2}$. They have a continuous-path limit $M$ there. Limits obtained for different integer horizons agree on overlaps; choose these versions consistently. Each $M_t$ is adapted, and passing to the $L^2$ limit in $\mathbb{E}[M_t^{(n)}\mid {\mathcal{F}}_s]=M_s^{(n)}$ shows that $M$ is a continuous martingale. Letting $m\to \infty$ in the preceding bound proves the requested convergence, with an explicit rate:The same Itô isometry identifies $M_t={\int }_0^t\mathrm{sgn}(B_s)d{B}_s$, with $\mathrm{sgn}(0)=0$.

There is a genuine factor-of-two error in the last printed conclusion. The stated $f_n$ has second derivative $n{1}_{\{|x|<1/n\}}$ almost everywhere, and the generator in the decomposition is one half of the second derivative. Hence the exact identity isBecause ${\sup }_x|f_n(x)-|x||\le 1/(2n)$ and $f_n(0)=1/(2n)$, this already proves uniform $L^2$ convergence on bounded intervals to $L_t=|B_t|-M_t$. It does not by itself prove almost-sure convergence of the full sequence, so we supply that additional step.

Take $n_k=k^2$. For each fixed integer $T$ and each $\epsilon >0$, the maximal-error bound and Markov's inequality implyThe first Borel-Cantelli lemma, applied to rational positive $\epsilon$ and integer $T$, gives uniform almost-sure convergence of this subsequence on every bounded interval. Therefore $L_{k^2}\to L$ uniformly there almost surely.

For $k^2\le n\le (k+1{)}^2$, the indicator windows are nested. The ratio-one interpolation for shrinking-window occupation integrals givesBoth ratios tend uniformly to one as $k\to \infty$. The two bounds converge uniformly to the continuous nonnegative process $L$, so the entire sequence does too. Thus the correct normalized conclusion isand with the factor actually printed in the question the limit isThese are the occupation approximations for Brownian local time. In the Tanaka formula convention, $L_t$ is the local time of a semimartingale at zero. It is not identically zero: $\mathbb{E}{L}_t=\mathbb{E}|B_t|=\sqrt{2t/\pi }>0$ for $t>0$, since $M$ starts at zero and is a martingale. Thus the discrepancy cannot be removed by treating the two limits as equal. The standard normalization is also given in Lalley's notes on the Itô calculus and local time.

For every $t$, subtraction of the definitions givesThe difference of the two supplied continuous martingales is a square-integrable martingale on $[0,1]$; for each fixed mesh its finite sums are bounded. For this dyadic quadratic variation of a bounded continuous martingale, the Doob L2 maximal inequality therefore givesThus the dyadic approximations to quadratic variation are Cauchy for the expected squared uniform norm, as required. No monotonicity in time of the partially completed squared-increment sums is needed.

Write ${\delta }_n=2^{-n}$ and let $Q_n(U)$ denote the printed squared-increment sum for a process $U$, on grid times $k{\delta }_n$. We construct its limit first for bounded martingales, then use localizing sequences and the finite variation part of a semimartingale.

Continuity and monotonicity of the bounded-martingale limit. Let $U$ be a uniformly bounded continuous martingale. The given result supplies a limit $R^U$ in uniform convergence on compacts in probability. Each $Q_n(U)$ is continuous and adapted. A subsequence converges almost surely uniformly on each compact time interval, by choosing summable error probabilities and diagonalizing. Its limit therefore has a continuous version; in the usual completed filtration this version is adapted.

The sums $Q_n(U)$ themselves can decrease between grid points, because the last partial increment is being squared. To establish monotonicity, use instead the quadratic variation from completed grid incrementsThese step processes are nondecreasing, and pathwise uniform continuity giveswhere ${\omega }_U(\delta ;T)=\sup \{|U_t-U_s|:s,t\le T,|t-s|\le \delta \}$. Thus the same almost surely uniform subsequence of $S_n$ converges to $R^U$, proving that $R^U$ is nondecreasing. Also $R_0^U=0$.

Localization of a continuous local martingale. Subtract the initial value, which does not affect increments, so that $M_0=0$. SetContinuity gives ${\tau }_m\uparrow \infty$ almost surely and makes $M^{{\tau }_m}$ bounded. The bounded local martingale criterion makes it a true martingale. Let $R^m$ be its continuous, adapted, nondecreasing limit.

The discrete sums commute exactly with stopping:For $\ell \le m$, uniqueness of the probability limit and stability of uniform convergence on compacts in probability under stopping giveup to indistinguishability. Taking a common null set for the countably many pairs, patch these processes into $R^M$ by setting $R_t^M=R_t^m$ when $t\le {\tau }_m$. Compatibility makes this definition independent of $m$. It is continuous, adapted, and nondecreasing. For each $T,\epsilon >0$,Let $n\to \infty$ and then $m\to \infty$. This proves the localization and patching of quadratic variation.

Adding finite variation. Decompose the continuous semimartingale as $X=X_0+M+A$, where $M$ is a continuous local martingale starting at zero and $A$ is continuous, adapted, and locally of finite variation. For every compact interval, the sum of the absolute increments of $A$ is at most its total variation of a function. Hence, pathwise,The Cauchy-Schwarz inequality bounds the mixed increment sum byThe first factor is bounded in probability, since $Q_n(M)\to R^M$ uniformly on compacts in probability; the second tends to zero almost surely. Thus the mixed term tends to zero in probability. Expanding the square provesThe constructed $R=R^M$ is continuous, nondecreasing, and adapted. This is the quadratic variation of $X$, and expresses the fact that finite-variation terms do not change quadratic variation.

A continuous martingale is an adapted process with almost surely continuous paths, $\mathbb{E}|M_t|<\infty$ for every $t$, andA continuous local martingale is continuous and adapted and has a localizing sequence of stopping times $T_n\uparrow \infty$ almost surely such that every $M_t^{T_n}=M_{t\wedge T_n}$ is a true martingale. The initial value is normally integrable; the frequently used class ${\mathcal{M}}_{c,\mathrm{loc}}$ is normalized to start at zero.

The usual conditions for a filtration are completeness and right continuity. Completeness means that ${\mathcal{F}}_0$ contains every subset of each $\mathbb{P}$-null set in the ambient $\mathcal{F}$; right continuity means ${\mathcal{F}}_t={\bigcap }_{u>t}{\mathcal{F}}_u$. These conditions concern the filtered probability space, whereas continuity of a martingale concerns its sample paths.

A martingale whose values over its whole time interval form a uniformly integrable family has an integrable terminal limit and is closed by that limit. Conversely, conditional expectations of an integrable terminal variable form a uniformly integrable martingale, by uniform integrability of conditional expectations. For continuous martingales, one may equivalently use uniform integrability of the family stopped at finite stopping times.