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For fixed $\epsilon >0$, the derivative of ${\sigma }_{\epsilon }$ is zero on $(-\infty ,\epsilon /2]$, bounded on the compact transition interval $[\epsilon /2,\epsilon ]$, and equals $1/(2\sqrt{x})\le 1/(2\sqrt{\epsilon })$ for $x\ge \epsilon$. Thus ${\sigma }_{\epsilon }$ is globally Lipschitz, with at most linear growth. The global existence theorem for stochastic differential equations with Lipschitz coefficients supplies a global strong stochastic solution and pathwise uniqueness for the modified equation.

The solution stays nonnegative. More precisely, if $z>\epsilon /2$, it cannot cross $\epsilon /2$: on reaching this level the constant continuation solves the equation, and pathwise uniqueness forces that continuation. If $z\le \epsilon /2$, it is constant from the outset. In either case $Z^{\epsilon }$ is a nonnegative local martingale, so $\mathbb{E}{Z}_s^{\epsilon }\le z$.

For ${\epsilon }^{\prime }<\epsilon <z$, both coefficients agree with $\sqrt{x}$ on $[\epsilon ,\infty )$. Stop the two solutions when either first reaches $\epsilon$, and also at a common upper bound. On the resulting compact state interval, the coefficient is Lipschitz. The Itô isometry applied to their difference, followed by the Gronwall inequality, gives zero expected squared difference. Remove the upper bound using the nonexplosion of each modified solution. The two solutions agree until the first of these lower exits, and by continuity both have value $\epsilon$ there, so their exits at that level coincide. In particular,up to indistinguishability. If $\epsilon \ge z$, the first exit time is zero and the same assertion is immediate from their initial values.

Now take a decreasing sequence $0<{\epsilon }_n<z$ with ${\epsilon }_n\downarrow 0$. Write $Z^n=Z^{{\epsilon }_n}$ and ${\tau }_n=T_{{\epsilon }_n}$. Compatibility gives ${\tau }_n\le {\tau }_{n+1}$; indeed, a continuous path cannot reach the smaller level before reaching the larger one. Let $\tau ={\lim }_n{\tau }_n$, a stopping time. On $[0,\tau )$ the compatible solutions define an adapted continuous process $Z$ by setting $Z_t=Z_t^n$ whenever $t\le {\tau }_n$. We still need to justify the finite limiting endpoint and the integral equation there, rather than assume they exist.

For this purpose use the compatible predictable integrandsTheir squares increase pointwise with $n$: on the smaller interval compatibility makes the values equal, and outside it the old integrand is zero. Their limit $H$ is predictable and equals $\sqrt{Z_s}$ before $\tau$, and zero after $\tau$, apart from irrelevant single-time endpoints. For every finite $t$,The monotone convergence theorem gives $\mathbb{E}{\int }_0^t{H}_s^{2}ds\le zt$. Moreover, compatibility makes $(H-H^n{)}^2=H^2-(H^n{)}^2$ away from endpoints, so its expected time-integral tends to zero. The Itô isometry, and the Doob L2 maximal inequality if uniform convergence on a finite time interval is desired, therefore give a continuous stochastic-integral limitBefore ${\tau }_n$, the integrand agrees with $\sqrt{Z^n}$, which is the original square-root coefficient there. Locality of the stochastic integral yields $L_{t\wedge {\tau }_n}=Z_{t\wedge {\tau }_n}^n$. A countable intersection in $n$ and path continuity give this equality simultaneously for all times.

On $\{\tau <\infty \}$, every ${\tau }_n$ is finite and $L_{{\tau }_n}={\epsilon }_n$. Continuity of $L$ thus gives $L_{\tau }=0$. The limiting integrand is zero after $\tau$, so $L$ remains zero afterward. On $\{\tau =\infty \}$, every finite time is covered by one of the compatible positive solutions. Consequently $L$ is nonnegative everywhere and $H_s=\sqrt{L_s}$ for almost every $s$, including after absorption. The cutoff construction of an absorbed square-root diffusion has producedAll modified solutions, stopping times and integrands were constructed from the same prescribed Brownian motion; hence this is a strong stochastic solution. The integral is square-integrable on every finite horizon by the bound $zt$, which also rules out a hidden finite-time explosion in the patching argument.

Put $R_{\epsilon }={\int }_0^{\epsilon }(H_u-H_0)d{B}_u$ and $d_{\epsilon }={\epsilon }^{-1}{\int }_0^{\epsilon }\mathbb{E}|H_u-H_0{|}^{2}du$. Boundedness and path continuity of $H$ make $\mathbb{E}|H_u-H_0{|}^2\to 0$ by the dominated convergence theorem, so $d_{\epsilon }\to 0$. The Itô isometry gives $\mathbb{E}{R}_{\epsilon }^2=\epsilon d_{\epsilon }$.

We must not assume independence between the integral and its Brownian denominator. Instead, the Cauchy-Schwarz inequality givesConcavity, or Jensen inequality, bounds the first factor by $(\mathbb{E}{R}_{\epsilon }^2{)}^{1/8}$. Since $B_{\epsilon }\stackrel{d}{=}\sqrt{\epsilon }N$, the second factor is ${\epsilon }^{-1/8}(\mathbb{E}|N{|}^{-1/2}{)}^{1/2}$. This negative normal moment is finite: the density is bounded near zero, where ${\int }_0^1{u}^{-1/2}du<\infty$, and has Gaussian decay at infinity. In fact $\mathbb{E}|N{|}^{-1/2}=2^{-1/4}\Gamma (1/4)/\sqrt{\pi }$. The powers of $\epsilon$ cancel, yielding the fractional-moment control of Brownian increment ratiosThe ratio may be assigned any value on the null event $B_{\epsilon }=0$.

Use the representation from part (a), and put ${\Delta }_{j}M=M_{T_{j+1}}-M_{T_j}$, $U_j=h_j{\Delta }_{j}M$ and $N=H\cdot M$. The optional stopping theorem in the form needed here says that for a uniformly integrable martingale $L$ and stopping times $\sigma \le \tau$, including terminal infinite times, $\mathbb{E}[L_{\tau }\mid {\mathcal{F}}_{\sigma }]=L_{\sigma }$. An L2-bounded continuous martingale is uniformly integrable and has the required terminal value.

Consequently $\mathbb{E}[{\Delta }_{j}M\mid {\mathcal{F}}_{T_j}]=0$. Conditioning on ${\mathcal{F}}_t$ and distinguishing whether $T_j\le t$ givesThus $N_t=\mathbb{E}[{\sum }_j{U}_j\mid {\mathcal{F}}_t]$ is a martingale. Its paths are continuous, since each stopped increment is continuous and zero until its left endpoint.

For $i<j$, $U_i$ is ${\mathcal{F}}_{T_j}$-measurable, so the same conditional identity yields $\mathbb{E}(U_i{U}_j)=0$. These martingale-difference orthogonality relations also hold for the unweighted increments. ThereforeHere $N_{\infty }={\sum }_j{U}_j\in L^2$. By conditional Jensen, $\mathbb{E}{N}_t^2\le \mathbb{E}{N}_{\infty }^2$ for every $t$, soThis proves the norm bound without presupposing the Itô isometry requested next.

First justify optional sampling for the square-minus-bracket process. The Doob L2 maximal inequality gives $\mathbb{E}{\sup }_{t\ge 0}|M_t{|}^2\le 4\mathbb{E}|M_{\infty }{|}^2<\infty$. Stop the local martingale $M^2-[M]$ using a localizing sequence and bounded bracket levels. The stopped expectation identity, followed by Fatou, bounds $\mathbb{E}[M{]}_{\infty }$ by the expected squared path supremum. Hence $M^2-[M]$ has an integrable absolute supremum, and localization with conditional dominated convergence makes it a uniformly integrable martingale. The optional stopping theorem stated in part (b) is therefore applicable to both $M$ and $M^2-[M]$.

For ordered stopping times $\sigma \le \tau$, expand the squared increment and condition on ${\mathcal{F}}_{\sigma }$. The mean increment is zero, soThis is the conditional bracket isometry for stopped martingale increments. Apply it with $\sigma =T_j$, $\tau =T_{j+1}$ and multiply by the bounded ${\mathcal{F}}_{T_j}$-measurable variable $h_j^2$. The orthogonality proved in part (b) gives the Itô isometry

To identify the whole bracket process, let $N=H\cdot M$ and $C_t={\int }_0^t{H}_s^{2}d[M{]}_s$. For deterministic $s<t$ and $E\in {\mathcal{F}}_s$, apply the proved isometry to the simple integrand $1_E{1}_{(s,t]}H$. It yieldsThus $\mathbb{E}[(N_t-N_s{)}^2\mid {\mathcal{F}}_s]=\mathbb{E}[C_t-C_s\mid {\mathcal{F}}_s]$. Since $N$ is a martingale, the cross term $2N_s(N_t-N_s)$ has conditional mean zero. Consequently $N^2-C$ is a martingale. The process $C$ is continuous, adapted, nondecreasing and starts at zero, so uniqueness in the definition of quadratic variation provesThis deduction uses only the proved elementary isometry, not a previously assumed formula for the quadratic variation of stochastic integrals.

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.

Fix the specified time $t$. With the path modulus ${\eta }_n$ on $[0,t+1]$ as above, comparison of increment powers now givesThe hypothesis makes the second factor eventually bounded almost surely, while the first tends to zero almost surely because $2-p>0$. Thus the squared-increment sums tend to zero almost surely. They also converge in probability to the quadratic variation $[X{]}_t$, so uniqueness of the limit in probability implies $[X{]}_t=0$ almost surely.

The quadratic variation is nondecreasing, hence it vanishes throughout $[0,t]$. To see explicitly that the continuous local martingale must then be constant, localize it to square-integrable stopped martingales. The Itô isometry givesRemove the localization, apply the conclusion at every rational $s\le t$, and use continuity. Since $X_0=0$, this provesThus $X$ is indistinguishable from the zero process on that interval. This is the subquadratic case of dyadic power variation of a continuous local martingale.

Let $u(t,x)=P_{1-t}f(x)$ for $t<1$. Differentiating the Brownian transition semigroup in the spatial variables is justified by the bounded gradient and givesThe heat equation for this semigroup is the backward equation ${\partial }_{t}u+\frac{1}{2}\Delta u=0$. Applying the multidimensional Itô formula yieldsThe finite-variation term vanishes. ThusInitially this calculation holds on $[0,1-\epsilon ]$. The vector of stochastic integral coefficients has norm at most $K$, so the Itô isometry lets the integrals extend to time one. Also $P_{1-t}f(X_t)\to f(X_1)$ in $L^2$ by the Lipschitz function bound and continuity of Brownian motion. The same representation therefore holds on the entire interval.

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.

An elementary predictable process with stopping-time intervals has the form $H={\sum }_{j=0}^{r-1}{h}_j{1}_{({\tau }_j,{\tau }_{j+1}]}$, where the stopping times are ordered and $h_j$ is bounded and ${\mathcal{F}}_{{\tau }_j}$-measurable. Deterministic endpoints give the usual simple predictable process. Define the zero-starting stochastic integral byCommon refinement and telescoping show that this definition is independent of the representation. Use $M_{\infty }$ for an infinite endpoint. The L2-bounded martingale convergence theorem gives convergence of $M_t$ in $L^2$ to $M_{\infty }$ and $M_t=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_t]$.

Put $D_j=M_{{\tau }_{j+1}}-M_{{\tau }_j}$. The optional sampling theorem gives $\mathbb{E}[D_j\mid {\mathcal{F}}_{{\tau }_j}]=0$. For $j<k$, the variable $h_j{D}_j{h}_k$ is ${\mathcal{F}}_{{\tau }_k}$-measurable; hence $\mathbb{E}[h_j{D}_j{h}_k{D}_k]=0$. Applying the same theorem to the assumed martingale $M^2-[M]$ givesConsequently the Itô isometry follows:The unbounded stopping times cause no optional-sampling gap. The $L^2$ terminal representation makes $M$ uniformly integrable and makes the family of stopped $M^2$ uniformly integrable, by conditional Jensen applied to $M_{\infty }^2$. Also $\mathbb{E}[M{]}_{\infty }=\mathbb{E}{M}_{\infty }^2-\mathbb{E}{M}_0^2<\infty$, from monotone convergence theorem and the assumed square-compensator identity. Thus $M^2-[M]$ has the needed uniform integrability, and truncating stopping times and passing in $L^1$ is legitimate. Bounded coefficients are part of the simple-integrand convention; more general coefficients require the displayed square-integrability condition.

Use the shifted filtration ${\mathcal{G}}_s={\mathcal{F}}_{T+s}$. Each $T+s$ is a stopping time, and the optional sampling theorem implies that $M_s^{(T)}=M_{T+s}-M_T$ is a ${\mathcal{G}}_s$-martingale. It is $L^2$-bounded, for example by $\mathbb{E}|M_{T+s}-M_T{|}^2\le 4\mathbb{E}{M}_{\infty }^2$. Its quadratic variation is $[M^{(T)}{]}_s=[M{]}_{T+s}-[M{]}_T$. The shifted process $H_{T+s}$ is continuous and ${\mathcal{G}}_s$-adapted, hence predictable. Its boundedness and the integrability of the bracket make the right-hand stochastic integral well-defined in $L^2$.

For a deterministic partition $0=s_0<\cdots <s_r=t$, use left-endpoint coefficients $H_{T+s_j}$. Both integrals have the same approximating sumOn the original clock this is the integral of the simple process on the stopping-time intervals $(T+s_j,T+s_{j+1}]$; on the shifted clock it is a deterministic-partition integral. As the mesh tends to zero, continuity of $H$ gives pathwise uniform approximation on the random compact interval $[T,T+t]$. The Itô isometry and domination by a constant times the integrable $[M{]}_{\infty }$ give $L^2$ convergence of both sums. Thus stopping-time shift of a stochastic integral yieldsThe left integral includes increments strictly after $T$, so no jump at the starting time is added.

For a continuous path $m$ of finite variation on $[0,t]$, every partition satisfiesUniform continuity makes the right side tend to zero as the mesh tends to zero. The quadratic variation of the given continuous local martingale $M$ is therefore identically zero.

Stop $M$ when its magnitude first reaches an integer $n$, so the stopped process is a bounded, square-integrable martingale. The square identity, or the Itô isometry, gives $\mathbb{E}{M}_{t\wedge {\tau }_n}^2=\mathbb{E}[M{]}_{t\wedge {\tau }_n}=0$. Hence it is zero at each rational time almost surely, and continuity extends this simultaneously to all times. Let $n$ tend to infinity. This provesThis is the continuous finite-variation local martingale is constant principle with initial value zero.

Yes: different adapted volatilities can give perfectly correlated terminal prices. The counterexample in part ii already establishes this with one nonconstant volatility. Both can be nonconstant as well. Put $h(x)=1+\frac{1}{2}\tanh x$ and takeThe Doléans-Dade exponential has volatility $h(B_t)$, bounded between $1/2$ and $3/2$. The second stock has volatility $h(B_t)S_t/(1+S_t)$, also bounded and nonconstant. Their squared volatility difference integrates toThe Novikov condition and boundedness give true square-integrable martingales, and the Itô isometry gives positive variance for each $t>0$. Yet $S_t^{\prime }=(1+S_t)/2$ gives correlation one. The mechanism is positive affine dependence, not equality of the multiplicative volatility coefficients; it also explains the failure of the literal part-ii premise.

The principle terminal correlation does not identify adapted stock volatility exposes a false printed assertion for the correlation coefficient defined in the introduction. Take $r=0$, $S_0=S_0^{\prime }=1$, andThen $d{S}_t=S_{t}d{B}_t$ andHence ${\sigma }_t=1$ and ${\sigma }_t^{\prime }=S_t/(1+S_t)$ are bounded. Both terminal prices have positive finite variance and are perfectly correlated, because one is a positive affine function of the other. Nevertheless $\boxed{{\int }_0^1({\sigma }_t-{\sigma }_t^{\prime }{)}^{2}dt={\int }_0^1\frac{dt}{(1+S_t{)}^2}>0}$ on every continuous finite stock path. Equal initial prices fix the means, not the affine slope.

Here is the exact general conclusion and the intended qualified proof. Bounded volatilities make these zero-rate stochastic exponentials true square-integrable martingales. Writing $s_0=S_0=S_0^{\prime }$, perfect terminal correlation gives $S_1-s_0=c(S_1^{\prime }-s_0)$ for some $c>0$. Conditional expectation therefore gives $S_u-s_0=c(S_u^{\prime }-s_0)$ for every $0\le u\le 1$. Comparing stochastic integrals, the Itô isometry yieldsThis relation need not imply equality of the volatilities.

The terminal proportionality identifies bounded stock volatility criterion repairs the statement. If one additionally assumes equal terminal variances, then $c=1$ and the whole stock paths coincide, so positivity gives $\sigma ={\sigma }^{\prime }$ almost everywhere and the requested integral is zero. The same repair follows if terminal proportionality $S_1=c{S}_1^{\prime }$ is assumed: equal martingale means force $c=1$, after which the conditional-expectation and Itô isometry argument applies. This proves the intended conclusion under a sufficient additional hypothesis without treating centered correlation as uncentered proportionality.

First take a bounded elementary predictable process $\alpha ={\sum }_j{K}_j{1}_{(s_j,t_j]}$, with each $K_j$ bounded and ${\mathcal{F}}_{s_j}$-measurable and with finite time support. The Itô integral is the corresponding finite sum ${\sum }_j{K}_j(W_{t_j}-W_{s_j})$. Applying part (a) term by term givesSuch elementary predictable processes are dense among predictable processes in $L^2(d\mathbb{P}du)$. The Itô isometry makes the left functional continuous, with boundThe Cauchy-Schwarz inequality makes the right functional continuous, with bound $\Vert {\varphi }^{\prime }{\Vert }_{\infty }\sqrt{T}{\left(\mathbb{E}\int {\alpha }_u^{2}du\right)}^{1/2}$. Approximation therefore proves the same identity for every allowed predictable $\alpha$. The integral over the infinite time interval is the $L^2$ limit of its finite-horizon Itô integrals.

The terminal variable $F=\varphi (W_T)$ is bounded and hence square-integrable. In the completed natural Brownian filtration, the Brownian martingale representation theorem says that any square-integrable ${\mathcal{F}}_T$-measurable variable admits a representationwith $\beta$ predictable and $\mathbb{E}{\int }_0^T{\beta }_u^{2}du<\infty$. Extend $\beta$ by zero after $T$. Thus the requested constant and integrability areExpectation determines $c$ uniquely. If two integrands give the same representation, the Itô isometry gives $\mathbb{E}{\int }_0^{\infty }({\beta }_u-{\tilde{\beta }}_u{)}^{2}du=0$. Consequently $\beta$ is unique up to $d\mathbb{P}du$-almost everywhere equality, rather than pointwise equality at every time. The corresponding integral martingales are indistinguishable.