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Write a positive continuous local martingale as $M=\exp (X-⟨X⟩/2)$ using its stochastic logarithm. If $M_t\to 0$, its logarithmic bracket must diverge. Otherwise the finite-bracket convergence lemma makes $X$ converge finitely and the exponential has a positive limit.

Let $V_t=⟨M{⟩}_t$, and first suppose $V_{\infty }=\infty$ almost surely in addition to strict increase. The Dambis-Dubins-Schwarz theorem says thatdefine a Brownian motion $B$ in the time-changed filtration and giveEach $T_u$ is a stopping time and is finite. Continuity and strict increase of $V$ make $u\mapsto T_u$ continuous, $V_{T_u}=u$ and $T_{V_t}=t$.

Here are the martingale details behind this inverse-clock proof of the Dambis-Dubins-Schwarz theorem. Stopping a continuous local martingale when its bracket reaches $n$ makes it an L2-bounded continuous martingale. This follows from the stopped Itô isometry or from the $p=2$ estimate proved in Question 1(a), applied after localization. In particular it is uniformly integrable, and optional sampling is valid even at an unbounded stopping time by taking limits. Applying this to $M$ stopped at $T_n$ shows that $(M_{T_{u\wedge n}}{)}_{u\ge 0}$ is a martingale. Thus $B$ is a continuous local martingale. Time-changing $M^2-V$ in the same way shows $B_u^2-u$ is a local martingale, so $⟨B{⟩}_u=u$.

For completeness, the Lévy characterization of Brownian motion follows directly from the Itô formula. If a continuous local martingale $N$, starting at zero, has bracket $u$, thenis a complex local martingale. Its modulus is bounded on each deterministic finite horizon, so it is a true martingale there. ConsequentlyConditional characteristic functions give Gaussian increments independent of the past. Iterating this identity gives independent increments, and continuity completes the Brownian characterization. The identical vector argument proves the Lévy characterization of multidimensional Brownian motion when the bracket matrix is $uI$.

The printed strict-increase hypothesis does not imply $V_{\infty }=\infty$. For example, $M_t={\int }_0^t{e}^{-s}d{W}_s$ has strictly increasing bracket $(1-e^{-2t})/2$. To state the theorem under exactly the printed hypothesis, allow an independent enlargement of the probability space if the terminal clock $L=V_{\infty }$ can be finite.

On $\{L<\infty \}$ the martingale $M$ has a finite terminal limit. Indeed, stopping at each bracket level $n$ gives an L2-bounded continuous martingale which converges; on $\{V_{\infty }<n\}$ the stopped process is the original one. This proves the finite-bracket convergence lemma. Set $T_u=\infty$ when $u\ge L$ and continue $N_u=M_{T_u}$ by that terminal limit. The optional-sampling argument just given makes $N$ a continuous local martingale with bracket $u\wedge L$. Moreover $L$ is a stopping time in ${\mathcal{G}}_u={\mathcal{F}}_{T_u}$.

On a product extension add an independent Brownian motion $\beta$ in clock time, and putThe two summands have zero quadratic covariation, and their brackets are $u\wedge L$ and $(u-L{)}^{+}$. Thus $⟨\tilde{B}{⟩}_u=u$; the proved characterization makes $\tilde{B}$ Brownian. Since $V_t<L$ at every finite $t$ when $L$ is finite, $M_t={\tilde{B}}_{V_t}$ still holds. This is the finite-lifetime extension of the Dambis-Dubins-Schwarz theorem. An infinite clock gives Brownian motion on the original space; a finite clock may require the independent extension.

Define the radial martingale and its clock byThe Itô formula gives $R_t^2=2H_t+2t$. Independence of the coordinate Brownian motions makes their cross variation zero, soThe last identity is orthogonality of the radial martingale and planar Brownian area.

The clock is adapted and continuous, and it is strictly increasing almost surely. Otherwise the two coordinate paths would both vanish throughout a nontrivial interval. Such an interval contains a rational subinterval, while a Brownian increment over each fixed rational subinterval is a nondegenerate Gaussian and cannot be zero with positive probability.

Also $A(\infty )=\infty$ almost surely. If it were finite, the finite-bracket convergence lemma would make $H_t$ converge to a finite limit. Then $R_t^2=2t+O(1)$ on that event, forcing ${\int }_0^{\infty }{R}_t^{2}dt=\infty$, a contradiction. Thus no finite-lifetime extension is needed here.

Use the same inverse clock $T_u=\inf \{t:A(t)>u\}$ for both martingales, and set $W_u=H_{T_u}$, $B_u=Z_{T_u}$. Their bracket matrix isThe vector characterization proved in part (a) makes $(W,B)$ a two-dimensional Brownian motion; in particular its two coordinate processes are independent. Reversing the common clock givesThis is a common-clock Brownian representation of radius and area. Independence follows from the joint time change and identity bracket matrix; no independence of either Brownian motion from $A$ is asserted.

Strict positivity lets us define the continuous local martingaleThe integrand is locally bounded because a positive continuous path has positive minimum on every compact time interval. The Itô formula for $\log M$ givesThis is the stochastic exponential representation of a positive continuous local martingale.

If $⟨X{⟩}_{\infty }$ were finite on an event of positive probability, the finite-bracket convergence lemma proved in Question 2(a) would make $X_t$ converge to a finite limit there. The exponential would then have a strictly positive limit, contradicting the assumed $M_t\to 0$. ThereforeThis is the divergent logarithmic clock for a positive local martingale tending to zero.