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A uniformly bounded local martingale is a true uniformly integrable martingale: a localizing sequence gives uniformly dominated stopped values, so their conditional identities pass to the limit. A deterministic bound on each compact time interval gives a true martingale on each such interval, even when the bound grows with time.

If $h(t)={\int }_0^t{g}_{s}ds$ with deterministic $g\in L^2(0,\infty )$, the finite-time Cameron-Martin theorem densities have second moments $e^{{\int }_0^t{g}_s^{2}ds}$ bounded uniformly in time. Their uniformly integrable martingale limit is the strictly positive displayed density, giving equivalence on the full infinite continuous-path space. Finite-time densities alone do not guarantee such a global density: a nonzero constant drift has infinite energy and its infinite-horizon path law is singular, distinguished by the limiting ratio $B_t/t$.

This is an exponential-moment criterion for the stochastic exponential to be a true martingale. In the infinite-horizon stopped-moment form, a zero-starting convergent continuous local martingale satisfying ${\sup }_T\mathbb{E}{e}^{M_T/2}<\infty$ has a uniformly integrable martingale $\mathcal{E}(M)$. The half-threshold for the exponential-martingale Hölder bound first controls strict scalings. The terminal scaling inequality for stochastic exponentials and terminal expectation criterion for a nonnegative local martingale then include the endpoint scaling.

The optional stopping theorem for a right-continuous martingale $M$ on a filtered probability space satisfying the usual conditions for a filtration states that, for bounded stopping times $\sigma \le \tau$,For a submartingale the conditional equality becomes $\mathbb{E}[M_{\tau }\mid {\mathcal{F}}_{\sigma }]\ge M_{\sigma }$, and for a supermartingale it becomes the reverse inequality.

For a uniformly integrable martingale, boundedness of the stopping times can be removed, with the terminal limit used if a stopping time is infinite. A useful application is to a bounded stopped martingale $M_{t\wedge \tau }$ with $\tau <\infty$ almost surely: apply the bounded-time theorem first and then the bounded convergence theorem to obtain $\mathbb{E}{M}_{\tau }=\mathbb{E}{M}_0$. Finiteness of $\tau$ alone is not a sufficient hypothesis for an arbitrary martingale.

For $g(x)=\log |x|$ on ${\mathbb{R}}^2\setminus \{0\}$,The Itô formula, stopped in a compact annulus, therefore givesThis proves the local martingale property up to a possible hitting of the origin. For the annulus exit time $T$, $M_{t\wedge T}$ is bounded between $\log r$ and $\log R$, so it is a true uniformly integrable martingale.

The exit is finite almost surely. Indeed $|B_t{|}^2-2t$ is a martingale, so stopping at $T\wedge t$ gives $2\mathbb{E}(T\wedge t)=\mathbb{E}|B_{T\wedge t}{|}^2-1\le R^2-1$. Letting $t\to \infty$ proves $\mathbb{E}T<\infty$. Bounded convergence in the logarithmic martingale now yieldsIf $p$ is the probability of reaching radius $r$ before radius $R$, continuity givesHitting zero before reaching radius $R$ would require reaching every positive radius $r$ first. Its probability is bounded by $p$, which tends to zero as $r\downarrow 0$. Any finite-time path hitting zero is bounded before that time and hence hits zero before exiting some integer-radius disk. A countable union over those disks proves $B_t\ne 0$ for every $t\ge 0$, almost surely. This is planar Brownian motion avoids a fixed point; it also makes $M$ a globally defined continuous local martingale.

To decide whether it is a true martingale, compute its expectation directly. Condition on $B_0$ if necessary and use rotation to take $B_0=(1,0)$ and write $B_t=B_0+W_t$. Conditional on $\rho =|W_t|$, the angle of $W_t$ is uniform. The angular average of a logarithmic potential isFor $\rho <1$ this follows by taking the real part of the convergent power series for $\log (1+\rho e^{i\theta })$: each nonconstant Fourier term integrates to zero. For $\rho >1$ factor out $\rho$ and apply the same argument to $1/\rho$. The equality at one follows by the integrable limiting logarithmic singularity. All interchanges are valid: the shifted Gaussian density is bounded near the origin, where ${\int }_0^{1}r|\log r|dr<\infty$, and its Gaussian tail controls logarithmic growth.

The radius $\rho$ has density $(\rho /t)e^{-{\rho }^2/(2t)}$. Consequently, for $t>0$,Since $M_0=0$, $M$ is not a martingale, despite integrability at each deterministic time. This is the logarithmic radius of planar Brownian motion example.

The optional hint is consistent with this conclusion. The upper-radius exit $T(2,0)$ is finite and never occurs at zero. After that exit the stopped square is $(\log 2{)}^2$; before it, a radius in $[1/2,2]$ also gives a logarithmic square no larger than $(\log 2{)}^2$, while a radius below $1/2$ contributes precisely the extra indicated term. This proves the hinted inequality without assuming that the unstopped process is a true martingale.

A simple predictable process can be writtenwhere each ${\xi }_j$ is bounded and ${\mathcal{F}}_{t_j}$-measurable. Define its integral to start at zero:Refining the partition leaves this sum unchanged. Every summand is a martingale, by conditioning its future increment on the information at $t_j$. Bounded coefficients and the $L^2$ bound on $M$ make the integral an L2-bounded continuous martingale. Its terminal value is the same sum with $M_{\infty }$ substituted at any infinite endpoint; $L^2$ martingale convergence supplies this limit.

We need the conditional square identity, rather than assuming independent increments. The Doob L2 maximal inequality gives $\mathbb{E}{\sup }_t|M_t{|}^2<\infty$. Localize the square identity for $M^2-[M]$ at level and bracket stopping times. The stopped identity and this maximal bound give $\mathbb{E}[M{]}_{\infty }<\infty$ by monotone convergence. Nowis integrable, so localization passes to conditional expectations and $M^2-[M]$ is a uniformly integrable martingale. Expanding a squared increment and using $\mathbb{E}(M_t\mid {\mathcal{F}}_s)=M_s$ provesThe infinite endpoint follows from $L^2$ convergence and integrability of the bracket. The initial value $M_0$ need not be zero: the integral itself is zero-starting.

In the square of the terminal integral, all cross terms vanish. For $j<k$, the $j$th increment and both coefficients in its cross term are measurable at $t_k$, whereas the conditional mean of the $k$th increment there is zero. For the diagonal terms, the preceding conditional identity can be multiplied by the bounded ${\xi }_j^2$. ThereforeThis proves the Itô isometry for these simple integrands. Ordered stopping-time intervals also work, replacing the conditional identity by optional stopping theorem, but deterministic intervals suffice for the approximation requested here.

For that approximation, put $\mu (A)=\mathbb{E}{\int }_0^1{1}_A(\omega ,s)ds$ on the predictable sigma-algebra. This is a finite measure. That sigma-algebra is generated, apart from the zero-time slice of measure zero, by rectangles $A\times (s,t]$ with $A\in {\mathcal{F}}_s$. The indicator of every such rectangle is a bounded simple predictable process.

Here is a density proof. Let $V$ be the closed linear span of these indicators in $L^2(\mu )$, and let $\mathcal{D}$ contain the predictable sets whose indicators belong to $V$. The whole space belongs to $\mathcal{D}$; complements stay in $\mathcal{D}$; and disjoint countable unions do also, because their finite partial indicator sums converge in $L^2$ under the finite measure. Thus $\mathcal{D}$ is a Dynkin system containing the generating rectangle pi-system. The Pi-lambda theorem makes it the entire predictable sigma-algebra. Bounded predictable simple functions therefore belong to $V$. Truncating any $L^2(\mu )$ function and then approximating its truncated range by finitely many levels shows that every such function belongs to $V$. A finite sum of rectangle indicators can be rewritten on one common deterministic time partition, with the coefficient on each interval measurable at its left endpoint; it is exactly an allowed simple process.

Consequently choose $H^n$ of this form with $\Vert H^n-H{\Vert }_{L^2(\mu )}<1/n$. This provesIt is density of simple predictable processes for finite measures, with a proof of the actual approximation step.

Finally apply the preceding isometry to $M_t=B_{t\wedge 1}$, whose bracket is $t\wedge 1$. The elementary integrals $I_n={\int }_0^1{H}_s^{n}d{B}_s$ satisfyDefine the Itô integral as the $L^2$ limit of these elementary integrals. The same identity shows independence of the approximating sequence, linearity, mean zero, and $\mathbb{E}({\int }_0^1{H}_{s}d{B}_s{)}^2=\mathbb{E}{\int }_0^1{H}_s^{2}ds$. Applying the Doob inequality to the difference processes also gives uniform-in-time $L^2$ convergence on $[0,1]$, with a continuous martingale version of the integral.

Let $D=d\mathbb{Q}/d\mathbb{P}$ on $\mathcal{F}$. The Radon-Nikodym derivative is nonnegative and ${\mathbb{E}}_{\mathbb{P}}D=1$. For $A\in {\mathcal{F}}_t$,Uniqueness of the restricted Radon-Nikodym derivative therefore identifiesFor $s\le t$, the tower property of conditional expectation gives ${\mathbb{E}}_{\mathbb{P}}(Z_t\mid {\mathcal{F}}_s)=Z_s$, and ${\mathbb{E}}_{\mathbb{P}}{Z}_t=1$. Thus $Z$ is a nonnegative true martingale, indeed a uniformly integrable martingale, since it consists of conditional expectations of one integrable random variable. This is the Radon-Nikodym density martingale. Under the usual conditions for a filtration, it has the standard càdlàg version; continuity is the additional assumption in part (c), not a consequence of absolute continuity alone.

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.

Use the following form of the Girsanov theorem. If $W$ is Brownian motion, $\vartheta$ is predictable, andis a uniformly integrable martingale with $Z_0=1$, then the probability measure with terminal density $Z_{\infty }$ makes $W_t-{\int }_0^t{\vartheta }_{s}ds$ a Brownian motion. The analogous finite-horizon statement uses the terminal density at that horizon.

Here take $W_t=X_t-x$ and ${\vartheta }_s=1_{\{s<\tau \}}/X_s$. Before $\tau$, $X_s\in [\epsilon ,M]$, so the stochastic integral is well-defined. The assumed uniform integrability of the stopped stochastic exponential givesThus the proposed weighting really defines a probability measure, and the Girsanov theorem gives the Brownian motionunder that measure. In particularAll calculations are first made on this compact stopped interval. The integral involving $1/X$ is naturally defined on $[0,T_0)$; at $T_0$ its quadratic variation may diverge. The density will have a zero limiting extension there, as established below, rather than a finite stochastic logarithm at the hitting time.

Continuity and adaptedness make the first boundary hit a stopping time. A continuous path starting in the open domain cannot leave it before meeting its boundary. Thus $X_{t\wedge T}\in \mathcal{D}\cup \partial \mathcal{D}$, with the usual interpretation when $T=\infty$. If $|u|\le K$ on this set, thenThe stopped process is a bounded local martingale; the bounded local martingale criterion makes it a martingale and, in fact, a uniformly integrable martingale. The martingale convergence theorem gives almost sure and $L^1$ convergence as $t\to \infty$.

Its limit can also be identified pathwise. On $\{T<\infty \}$ the stopped process is eventually constant at $e^{-\lambda T}u(X_T)$. On $\{T=\infty \}$ its absolute value is at most $K{e}^{-\lambda t}$ and hence tends to zero. Thus the limit is $e^{-\lambda T}u(X_T)1_{\{T<\infty \}}$.

Let $K={\sup }_T\mathbb{E}[e^{M_T/2}]<\infty$, with $T$ ranging over finite stopping times. To locate the half-threshold for the exponential-martingale Hölder bound, put $u=\sqrt{pq}>\sqrt{q}$. Since $u(u-1)$ is increasing for $u>1$,On the other hand, taking $q=1+\epsilon$ and $p=1+{\epsilon }^2$ gives $C(p,q)\to 1/2$ as $\epsilon \downarrow 0$. Hence

Fix $0<a<1$. Choose $p,q>1$ with $aC(p,q)<1/2$, and put $\lambda =2aC(p,q)\in (0,1)$. Part (a) and the Jensen inequality give, for every finite stopping time,Thus the entire stopped family has a uniform $L^p$ bound for some $p>1$. By uniform integrability from an Lp bound, it is uniformly integrable.

To check that this local martingale is a true martingale, stop it by a localizing sequence. At each fixed time the stopped variables are uniformly integrable by the same bound; taking limits in their conditional martingale identities proves the unstopped identity. The uniform bound over all finite stopping times then makes it a uniformly integrable martingale, with terminal expectation one. Therefore

Use deterministic times tending to infinity and the Fatou lemma:Apply the terminal scaling inequality for stochastic exponentials to $X=aM$ and $r=1/a$. Part (c) gives $\mathbb{E}[\mathcal{E}(aM{)}_{\infty }]=1$, soLetting $a\uparrow 1$ shows that the terminal expectation is at least one. The nonnegative local martingale $\mathcal{E}(M)$ starts at one and is a supermartingale, so the Fatou lemma gives the opposite inequality. Consequently

The terminal expectation criterion for a nonnegative local martingale now closes the argument. Conditional Fatou lemma applied to the supermartingale at times tending to infinity givesThe left side has expectation one and the right side at most one, so equality holds almost surely. Thusand uniform integrability of conditional expectations proves that $\mathcal{E}(M)$ is a uniformly integrable martingale. This establishes the required stopped-moment form of the Kazamaki criterion.

The angle is a bounded local martingale, hence a uniformly integrable martingale. The martingale convergence theorem gives an almost sure and $L^1$ limit ${\Theta }_{\infty }\in [-\pi /2,\pi /2]$ withIt must lie at an endpoint. Indeed, the Itô isometry and boundedness giveThus the bracket integral is finite almost surely. If the angle converged to an interior point, its squared cosine would eventually be bounded below by a positive constant, forcing this integral to be infinite. ConsequentlyThis proves endpoint convergence of a bounded angle diffusion.

Let $p_x=\mathbb{P}({\Theta }_{\infty }=\pi /2)$. Since $X_t=\tan {\Theta }_t$, this is exactly $\mathbb{P}(X_t\to +\infty )$, and the other endpoint means $X_t\to -\infty$. Taking expectations gives

Let $S=T_b\wedge T_{-a}$ and $p=\mathbb{P}(T_b<T_{-a})$. Set $R=\max (a,b)$. The exit time $S$ from $(-a,b)$ is at most the exit time from $(-R,R)$, so part (ii) gives $\mathbb{E}S<\infty$. Continuity ensures that exactly one of the two different boundary levels is reached at $S$.

The stopped local martingale $e^{X_{t\wedge S}}$ lies in $[e^{-a},e^b]$, hence is a uniformly integrable martingale. The optional sampling theorem, justified at the terminal time also by bounded convergence theorem, givesThus the boundary hitting probability from a diffusion scale function isFor integer $n\to \infty$, the events $\{T_b<T_{-n}\}$ increase to $\{T_b<\infty \}$. Indeed, on any path with finite $T_b$, continuity makes the path bounded below on $[0,T_b]$, so some lower barrier $-n$ has not yet been hit. The reverse inclusion is immediate. Taking the limit in the displayed probability gives

Take $L>0$, as needed for a nontrivial interval. Since $X$ is continuous and adapted, for every level $a$ and $t\ge 0$,The infimum is attained on this compact interval and can be computed on the countable dense set $(\mathbb{Q}\cap [0,t])\cup \{t\}$. It is therefore ${\mathcal{F}}_t$-measurable. Thus $T_a$ is a stopping time, and so is $T=T_{-L}\wedge T_L$, because $\{T\le t\}=\{T_{-L}\le t\}\cup \{T_L\le t\}$.

By continuity, $X_{t\wedge T}\in [-L,L]$, even before knowing that $T$ is finite. HenceStopping preserves the local martingale property. The bounded local martingale criterion now shows that $M^T=(M_{t\wedge T}{)}_{t\ge 0}$ is a uniformly integrable martingale; its deterministic bound also gives square integrability. In particular,

A nonnegative local martingale is a supermartingale. If its integrable terminal limit has the same expectation as its initial value, conditional Fatou lemma gives $\mathbb{E}[Z_{\infty }\mid {\mathcal{F}}_t]\le Z_t$, and equality of expectations makes this an equality. It is therefore a uniformly integrable martingale.

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.