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The stopping-interval indicator is an elementary predictable process with stopping-time intervals. Its stochastic integral isIt is zero before $S$, equals $M_t-M_S$ between $S$ and $T$, and equals $M_T-M_S$ after $T$. For a martingale with jumps, the jump at $S$ is excluded and the jump at $T$ is included, exactly as required by the half-open interval.

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.

Use just the parameters $1$ and $-1$. Positivity and adaptation of their local martingales giveThus both processes are adapted. A positive continuous local martingale has a semimartingale logarithm, by Itô formula after localization away from zero. Since $X=\log Z^{(1)}+A/2$ and $A$ has finite variation, $X$ is a continuous semimartingale. Write $X=N+V$, where $N$ is a zero-starting continuous local martingale and $V$ is continuous finite variation, also starting from zero.

Apply Itô formula to each exponential:Because $Z^{(\theta )}$ is itself a local martingale, the finite-variation term is zero by part (a). Divide its finite signed measure by the strictly positive $Z^{(\theta )}$. For $\theta =1,-1$, respectively,Adding and subtracting show $V=0$ and $[N]=A$. Therefore the exponential criterion for a continuous local martingale and its bracket givesThe semimartingale property was established before applying Itô's formula to $X$; it was not assumed from the outset.

For $h(x,y)=\frac{1}{2}\log (x^2+y^2)$ away from the origin,The Itô formula for planar Brownian motion therefore has no drift term before the radius reaches $r$. Since the initial logarithm is zero,The integrand's squared norm is $|B_s{|}^{-2}\le r^{-2}$ on this stopped interval, so this is a square-integrable martingale on every finite horizon, in particular a local martingale. One may first stop in an outer annulus to apply the formula on a compact smooth domain and then remove that localization.

Let $\tau =S_r\wedge T_R$. The stopped logarithm lies between $\log r$ and $\log R$, so the bounded local martingale criterion makes it a true bounded martingale. Also $\tau$ is finite almost surely. Indeed $|B_t{|}^2-2t$ is a martingale, and applying its stopped identity through $t\wedge T_R$ givesmonotone convergence theorem implies $\mathbb{E}{T}_R\le (R^2-1)/2<\infty$, hence $\tau \le T_R<\infty$ almost surely. Boundedness permits dominated convergence theorem in the stopped logarithm as $t\to \infty$. ConsequentlyThe standard results used are Itô's formula, the bounded-local-martingale criterion, the stopped martingale identity at bounded stopping times, and dominated/monotone convergence theorem. At exit the radius is either $r$ or $R$ by continuity. Solving $p\log r+(1-p)\log R=0$ also gives the planar Brownian annulus hitting probability

On a compact time interval a continuous path avoiding the origin has a strictly positive minimum radius. Thus the times $S_{1/n}$ tend to infinity almost surely. Part (a) localizes $M_t=\log |B_t|$, so $M$ is a continuous local martingale.

To show that it is not a martingale, compute its expectation at a positive deterministic time. In distribution $B_t=(1,0)+W_t$, where $W_t$ is a centered isotropic Gaussian vector. Conditional on $\rho =|W_t|$, its angle is uniform. The angular average of a logarithmic potential isFor $\rho <1$ this follows by averaging the real part of the power series for $\log (1+\rho e^{i\theta })$; for $\rho >1$ factor out $\rho$ and apply the same argument to ${\rho }^{-1}$. The value at $1$ follows by the integrable logarithmic singularity, or is irrelevant for the continuously distributed radius.

The radius has density $(\rho /t)e^{-{\rho }^2/(2t)}$. The logarithm of $|B_t|$ is absolutely integrable: near the origin its Gaussian density is bounded and ${\int }_0^{1}r|\log r|dr<\infty$; away from the origin its positive part is bounded by a constant plus $|B_t|$. Therefore conditioning is legitimate, and integration by parts yieldsSince $M_0=0$, this contradicts preservation of expectation by a martingale. Thus the logarithmic radius of planar Brownian motion is a strict local martingale. It is not a nonnegative local martingale; its negative values allow its expectation to increase.

The stochastic exponential of $M$ isItô's formula gives $d\mathcal{E}(M{)}_t=\mathcal{E}(M{)}_{t}d{M}_t$, so it is a positive continuous local martingale starting from one. For $p>1$,Every nonnegative local martingale is a supermartingale. Hence, under $[M{]}_{\infty }\le C$,The bound applies also to stopped versions. Localize the exponential to true martingales. Their uniform $L^p$ bound permits $L^1$ passage to the limit, proving that the unstopped exponential is a true martingale of mean one. The same bound makes its entire time-indexed family uniformly integrable. This proves the bounded-bracket criterion for a stochastic exponential, without needing an additional exponential-moment assumption.

One continuous Girsanov theorem formulation is as follows. Let $L$ be a zero-starting continuous local martingale such that $Z=\mathcal{E}(L)$ is a strictly positive uniformly integrable martingale, and define $d\mathbb{Q}=Z_{\infty }d\mathbb{P}$. For every continuous $\mathbb{P}$-local martingale $N$,On a finite horizon the same assertion holds with terminal density $Z_T$, provided $Z$ is a true martingale through $T$. In particular, if $L_t={\int }_0^t{\theta }_{s}d{B}_s$, thenis a Brownian motion under $\mathbb{Q}$, on the specified horizon. Its quadratic variation remains $t$ under this absolutely continuous measure (equivalent on every finite-horizon sigma-algebra), and the Lévy characterization of Brownian motion identifies it as Brownian. The density sign and drift subtraction are paired: choosing the exponential of $-\int \theta dB$ instead produces $B+\int \theta ds$.