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A planar Brownian motion exits a conformal image of a bounded planar domain in finite time almost surely. If the image lifetime were infinite, the inverse derivative is bounded away from zero on a small interior disc. Recurrence of planar Brownian motion and the Strong Markov property give infinite Brownian occupation time there, forcing the inverse conformal Brownian clock to diverge. It cannot exceed the finite original exit time. The expected image exit time can nevertheless be infinite.

Let $B$ be a standard planar Brownian motion started at $z$. Neighbourhood recurrence means that, with probability one, every nonempty open subset of ${\mathbb{R}}^2$ is visited at arbitrarily large times. Not hitting points means that for each fixed $w\ne z$, ${\mathbb{P}}_z(\exists t\ge 0:B_t=w)=0$. If $w=z$, time zero must be excluded, and there is still no return at any positive time. This assertion concerns each prescribed point; it does not say that the path contains no points.

We prove both claims from the planar Brownian annulus hitting probability. Fix a center $w$ and radii $0<r<\rho =|z-w|<R$, and define the stopping times ${\tau }_r=\inf \{t:|B_t-w|=r\}$ and ${\tau }_R=\inf \{t:|B_t-w|=R\}$. Their minimum $\sigma$ is finite almost surely: a coordinate of Brownian motion cannot stay forever in a bounded interval, by the Brownian reflection principle, so exit from the outer disk is finite. On the annulus,The Itô formula therefore makes $\log |B_{t\wedge \sigma }-w|$ a local martingale. It is bounded between $\log r$ and $\log R$, hence is a genuine martingale. The optional stopping theorem at $t\wedge \sigma$, followed by the bounded convergence theorem, givesThusThe exit can only be at one of the two distinct boundaries by path continuity.

For a fixed $r$, let $R\to \infty$. These events increase to $\{{\tau }_r<\infty \}$: any path up to a finite hitting time is bounded and so avoids some sufficiently large outer circle. The ratio tends to one, proving that any closed disk of positive radius is hit almost surely from any starting point outside it. An initial point inside it already counts as a hit.

To prove the stronger recurrence statement, take a closed disk contained in any prescribed nonempty open set. At each integer time $n$, the Markov property and the preceding disk-hitting result imply that some time $t\ge n$ visits this disk with probability one. Intersecting these events over all $n$ proves visits at arbitrarily large times. A countable basis of disks with rational centers and rational radii makes this simultaneous for all nonempty open sets. This is recurrence of planar Brownian motion, or equivalently neighbourhood recurrence.

For point avoidance, keep $R>\rho$ fixed and let $r\downarrow 0$. Hitting $w$ before ${\tau }_R$ requires hitting every inner circle first. The displayed ratio tends to zero, so this point-hitting event has probability zero. Any finite hit of $w$ would occur before exit from some outer circle with integer radius, again because the preceding path is bounded. A countable union now gives $\boxed{{\mathbb{P}}_z(T_w<\infty )=0(z\ne w)}$, the polar point for planar Brownian motion assertion.

Finally, even if the process starts at $w$, for each deterministic $s>0$ its normal distribution has a density, so $\mathbb{P}(B_s=w)=0$. The Markov property at $s$ and the result from other starting points exclude any visit to $w$ at or after $s$. Taking $s=1/n$ excludes all positive-time returns. Thus planar Brownian motion repeatedly approaches every point, while almost surely never hitting any prescribed point at positive time.

Put $L_r={\theta }_{T_r}$. The hitting times $T_r$ are finite almost surely by recurrence of planar Brownian motion, and increase with $r$. At $T_r$, the position is $e^{-r}{e}^{i{L}_r}$. By the Strong Markov property, Brownian scaling, and rotational invariance of planar Brownian motion, the future processis a standard planar Brownian motion started at $1$, independent of ${\mathcal{F}}_{T_r}$. The random rotation does not change its conditional law, since that law is rotationally invariant.

The first time this rescaled process reaches radius $e^{-h}$ corresponds to the original time $T_{r+h}$. Its continuous argument starts at zero and equals the increment of the original winding angle. HenceFor any ordered finite collection of levels, apply this assertion successively. Earlier winding values are measurable at the current hitting time, so all increments are independent and their laws depend only on the level differences. This proves the Lévy process property of winding at logarithmic radial passage levels.

One can also verify stochastic continuity and identify this process. The logarithmic-radius and argument martingales have the common clockThe same Cauchy-Riemann equations calculation as in part (a), applied locally to the logarithm and joined along the continuous argument, gives independent Brownian motions $U,V$ such that $\log R_t=U_{H_t}$ and ${\theta }_t=V_{H_t}$. This clock diverges: if it had a finite limit, convergence of martingales with finite total quadratic variation would make $\log R_t$ converge to a finite value, forcing the clock integral itself to diverge.

Thus, with ${\eta }_r=\inf \{u:U_u=-r\}$, we have $L_r=V_{{\eta }_r}$. The Brownian first-passage Laplace transform is $\mathbb{E}{e}^{-q{\eta }_r}=e^{-r\sqrt{2q}}$. It follows either from the Exponential martingale for Brownian motion stopped at ${\eta }_r\wedge n$, or from the one-dimensional exit equation. Conditioning on the independent process $U$ givesSo this is a symmetric Cauchy process, with level parameter $r$. The formula implies stochastic continuity at zero and, by stationary increments, everywhere. Under the usual definition requiring càdlàg paths, use the right-continuous passage clock ${\eta }_r^{+}=\inf \{u:U_u<-r\}$. It agrees with ${\eta }_r$ for each fixed level almost surely, and $V_{{\eta }_r^{+}}$ is the càdlàg modification. This distinction concerns simultaneous passage levels and does not alter any increment law.

First fix a deterministic point $y$ different from the initial point. By translation, work with $y=0$ and $B_0=x\ne 0$. For $R>|x|$, hitting $0$ before $T_R$ requires hitting every circle of radius $r\in (0,|x|)$ before $T_R$. Part (a) givesAny finite-time hit of $0$ lies before exit from some sufficiently large integer-radius circle, since a continuous path on a finite interval is bounded. A countable union over such radii therefore shows that this fixed point is never hit. This is the polar point for planar Brownian motion property; it is a statement about each fixed point, not about avoiding all points simultaneously.

For the initial point $x$, fix a deterministic $\epsilon >0$. The Gaussian distribution of $B_{\epsilon }$ has a density, so $B_{\epsilon }\ne x$ almost surely. Conditional on ${\mathcal{F}}_{\epsilon }$, the future is Brownian motion starting from $B_{\epsilon }$, and the preceding fixed-point result shows that it never hits $x$. Taking $\epsilon =1/m$ and a countable union proves there is no return to the starting point at any positive time.

In contrast, neighborhoods are revisited arbitrarily late. Choose $r>0$ so that the closed disk centered at $x$ of radius $r$ lies in the given open set $U$. From any point $z$ outside this disk, let $R\to \infty$ in part (a), centered at $x$. It givesAt each deterministic integer time $n$, condition on $B_n$. If it is in the disk there is already a visit to $U$ at time $n$; if it is outside, the Markov property and the last calculation ensure a later visit to that disk, hence to $U$. With probability one this holds for all $n$ simultaneously. Thus the set of visits to $U$ is unbounded. The recurrence of planar Brownian motion concerns neighborhoods even though every fixed point is polar.

Translate the centre of the ball to zero. Choose an inner radius $\rho$ strictly smaller than the ball's radius. If the starting point already lies in the ball there is nothing to prove; otherwise write $d=|z|>\rho$ and choose $R>d$. Let $H_{\rho },H_R$ be the first hits of the two circles.

The Brownian exit time from the annulus $\rho <|z|<R$ is finite almost surely: it is no later than exit of the first coordinate from $(-R,R)$, which is finite by part (a). The function $\log |z|$ is harmonic on this annulus. The Itô formula and optional stopping theorem apply to its bounded stopped values, yieldingSolving gives the planar Brownian annulus hitting probabilityAs $R\to \infty$ this tends to one, while the event in the formula is contained in $\{H_{\rho }<\infty \}$. Thus the inner circle is hit almost surely. That circle lies strictly inside the original open ball, proving the recurrence of planar Brownian motion in the form requested:A probability-one intersection over balls with rational centres and positive rational radii makes this conclusion simultaneous for all nonempty open balls, since every such ball contains one from that countable family.

A finite original exit time alone does not imply that the integral defining the conformal Brownian clock is finite. To establish this, let $\psi ={\varphi }^{-1}$. On the event $\{\tilde{T}=\infty \}$, the Brownian path of part (a) lives forever in $D^{\prime }$, while its inverse clock satisfiesChoose a closed disc $C$ compactly contained in $D^{\prime }$. Its radius may be decreased so that $|{\psi }^{\prime }{|}^2\ge c>0$ on $C$. Recurrence of planar Brownian motion, together with the Strong Markov property, gives infinite total Brownian occupation time in $C$: return repeatedly to a smaller concentric disc, and use the fixed positive probability of remaining in $C$ for a fixed positive duration. The successive trials imply infinitely many such durations. This is the same mechanism as the divergence of a positive planar Brownian occupation integral.

Hence, on any infinite-lifetime Brownian path,This contradicts $\tau (t)<T<\infty$. Thus $\tilde{T}$ is finite almost surely, and equality in law from part (a) gives finite exit from a conformal image of a bounded planar domain:This asserts almost-sure finiteness, not finiteness of the expected exit time.

Because $f$ is a continuous probability density, there are a point $z$, a radius $r>0$, and $ϵ>0$ such thatBy the recurrence of planar Brownian motion, the smaller disc $B(z,r/2)$ is visited at arbitrarily large times. Starting anywhere in that smaller disc, Brownian continuity and compactness give a uniform probability $\delta >0$ of staying in $B(z,r)$ for a fixed time $u>0$.

Apply the Strong Markov property at successive visits separated by at least $u$. The conditional probability of each stay event is at least $\delta$, so the conditional Borel-Cantelli lemma gives infinitely many successful stays almost surely. Every success adds at least $ϵu$ to $A_t$. Since $A_t$ is nondecreasing,

Planar Brownian motion and the initial law ${\nu }_{0,R}$ are invariant under every rotation about the origin. The hitting time ${\tau }_{0,r}$ is also rotation invariant, so the law of $B_{{\tau }_{0,r}}$ is invariant under every rotation of the circle of radius $r$. Planar Brownian motion hits that circle almost surely by recurrence of planar Brownian motion. The unique rotation-invariant probability measure on the circle is its uniform measure, and therefore

A prescribed point is almost surely never visited by planar Brownian motion started elsewhere. The harmonic function $\log |z-w|$ gives an annular hitting probability that tends to zero as the inner radius decreases to zero. This does not contradict recurrence of planar Brownian motion, which concerns neighbourhoods.