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Construct the times inductively. Suppose $(B_{T_0},\dots ,B_{T_n})$ has the law of $(S_0,\dots ,S_n)$. Conditional on the past, the martingale increment $S_{n+1}-S_n$ has mean zero and finite second moment. Apply the conditional form of the Skorokhod embedding theorem to this regular conditional law, using the fresh Brownian motion $B_{T_n+t}-B_{T_n}$ supplied by the Strong Markov property. This gives a stopping time increment ${\tau }_{n+1}$ and $T_{n+1}=T_n+{\tau }_{n+1}$ such that the next Brownian increment has the required conditional law. Induction provesfor every $k$.

A stopping time $T$ has $\{T\le n\}$ determined by $X_0,\dots ,X_n$. The Strong Markov property says that, conditional on $T<\infty$ and $X_T=i$, the post-$T$ chain is a fresh chain started at $i$, independent of the pre-$T$ history.

For simple random walk on ${\mathbb{Z}}^d$, $d\ge 3$, the Green-function decay for simple random walk on the integer lattice and the Strong Markov property giveThe series over $k$ converges, so the first of the Borel-Cantelli lemmas says that almost surely only finitely many of the points $x_k$ are ever hit. Moreover, ${\mathbb{Z}}^d$ is a transient graph for $d\ge 3$, so each of those finitely many points is visited only finitely often. Therefore the simple random walk visits $A$ only finitely often almost surely:

The displayed identity is false with the printed non-strict inequality. For example, take $A=B=\{y\}$ and let $r=P_y({\tilde{H}}_y<\infty )\in (0,1)$. The event $L_y\le {\tilde{H}}_y$ says that the walk returns to $y$ at most once, so its left side is ${\mu }_y(1-r^2)$, whereas its right side is $e_{\{y\}}(y)={\mu }_y(1-r)$.

The standard and evidently intended last-exit decomposition for a transient random walk has $L_B<{\tilde{H}}_A$. Decompose that corrected event according to $n=L_B$ and $z=X_n\in B$. The Strong Markov property at time $n$ givesReversibility of the random walk on a graph gives the path-reversal identitySince $e_B(z)={\mu }_z{P}_z({\tilde{H}}_B=\infty )$ is the equilibrium measure of a finite set, summing first over $n$ and then over $z$ yields

The Strong Markov property at the first visit to $0$ gives the singleton equilibrium potential from the Green functionLet $U_n$ be an increasing exhaustion of $G$ by finite sets containing $0$, and let $g_{U_n}(x,0)$ be the Green function of a transient weighted graph stopped on leaving $U_n$. Each $g_{U_n}(\cdot ,0)$ has finite support. The Green identity and the Markov property give, for $m\ge n$,By the monotone convergence theorem, the right side tends to zero as $m,n\to \infty$, while $g_{U_n}(x,0)\uparrow g(x,0)$. Thus $g(\cdot ,0)$ is an $H_0$ limit of finitely supported functions. Dividing by the positive number $g(0,0)$ proves $h\in H_0$.

First identify the recurrence type, rather than infer an occupation limit from a one-time probability. At time $k(a+b)$, return to zero requires exactly $kb$ upward steps. The binomial distribution giveswhere the last step is Stirling's approximation. At all other positive times the return probability is zero. Hence ${\sum }_{n\ge 0}{\mathbb{P}}_0(S_n=0)=\infty$. The recurrence criterion by return probabilities proves the recurrent property; by the irreducible Markov chain property established in (i), every state is recurrent. The equivalence of positive recurrence and existence of a stationary distribution for an irreducible Markov chain on a countable state space, together with (iii), makes this chain null recurrent.

For completeness, the zero occupation proportion follows directly by treating returns as a renewal process. Let $L_1,L_2,\dots$ be successive strictly positive return-cycle lengths to zero. The Strong Markov property makes them independent and identically distributed, the recurrent property makes them finite almost surely, and the null recurrent property gives $\mathbb{E}{L}_1=\infty$. For each positive integer $K$, the strong law of large numbers applied to $\min (L_j,K)$ givesTake a common event of probability one for these countably many $K$. By the monotone convergence theorem, the right side increases to infinity. Thus the number of completed return cycles by time $n$, divided by $n$, tends to zero. Counting the initial visit changes this ratio by at most $1/n$. Consequently,The walk visits zero infinitely often, but the asymptotic fraction of time spent there is zero.

First take the intended annulus configuration $0<\tilde{R}<R<\hat{R}$ with $R\ge 1$. The first stopping time is finite by the allowed fact, and continuity gives $|Z_{H_R}|=R$. After $H_R$, the annulus exit time is bounded by the exit time of the outer disc, which is almost surely finite; use the Strong Markov property at $H_R$. Thus $H_{R,\tilde{R},\hat{R}}<\infty$ almost surely.

The function $\log |z|$ is harmonic on this annulus. Its stopped value is a bounded martingale, so optional stopping theorem and continuity give, if $p$ is the probability of hitting the inner circle first,ConsequentlyFor fixed $\tilde{R}<1$, one may take $R=1$ and let $\hat{R}\to \infty$. The probability of hitting the inner circle before the outer one tends to one. The probability of ever hitting that inner circle is at least each of these probabilities, hence is one. Taking the countable intersection for $\tilde{R}=1/n$, $n\ge 2$, proves visits to balls of arbitrarily small radius almost surely.

The source permits $R<1$ as well. Then $H_R=0$ and the actual starting radius for the next exit is one, rather than $R$. More generally put $s=\max (1,R)$. If $\hat{R}>s$, the answer for the literal definition is $\log (\hat{R}/s)/\log (\hat{R}/\tilde{R})$. If $\hat{R}\le s$, the infimum defining the next exit is already $H_R$ almost surely and the inner-circle probability is zero. Positive $\tilde{R}$ is implicit in an annulus and is necessary for the logarithmic calculation.

planar Brownian motion is recurrence of planar Brownian motion when, almost surely, it returns to every nonempty open neighborhood infinitely often, at arbitrarily large times. This does not mean that it hits each specified point; a prescribed point is polar for planar Brownian motion.

recurrence of planar Brownian motion and continuity imply that all the circle-hitting times $S_k,T_k$ in the question are finite. Starting on radius $ϵ$, it eventually visits the exterior of radius $2ϵ$ and must cross that circle; it later visits the interior of radius $ϵ$ and must cross back. Each duration ${\Delta }_k=T_k-S_k$ is strictly positive by path continuity, because the two radii are separated.

At $S_k$, the Strong Markov property restarts Brownian motion at its current point on the inner circle. rotational invariance of Brownian motion makes the law of the subsequent first exit time from the radius-$2ϵ$ disk independent of that starting angle. Its conditional distribution given the past is therefore the same fixed law at every cycle, proving that the ${\Delta }_k$ are independent and identically distributed.

To prove divergence of the conformal clock for any nonconstant entire $f$, choose a point $z_0$ with $f^{\prime }(z_0)\ne 0$. By continuity, there exist $\rho ,c>0$ such that $|f^{\prime }(z){|}^2\ge c$ throughout the closed disk $|z-z_0|\le 2\rho$. Use the same recurrence of planar Brownian motion circle cycles centered at $z_0$, with radii $\rho ,2\rho$. On each segment from the inner circle to the outer circle, the path remains in that disk, givingThe durations are positive iid variables. If their mean has not been computed, apply the strong law of large numbers to the bounded iid variables $\min ({\Delta }_k,1)$: their mean is strictly positive, so their sum, and hence $\sum {\Delta }_k$, diverges almost surely. ThereforeCentering the disks at a noncritical point avoids an unjustified positive lower bound on $|f^{\prime }|$ around the original origin.

Write $f=u+iv$. Harmonicity and Itô's formula make $u(Z_t),v(Z_t)$ continuous local martingales. The Cauchy-Riemann equations give equal brackets ${\int }_0^t|f^{\prime }(Z_s){|}^{2}ds$ and zero mutual bracket. Consequently Lévy's characterization of Brownian motion after time change of a continuous process makes $W$ a planar Brownian motion started at $f(Z_0)$, with no finite clock lifetime because the clock just proved is unbounded. The clock is strictly increasing: otherwise $f^{\prime }$ would vanish along a nonconstant path interval, contradicting its isolated zeros.

For the notation in the source, $|\nabla f{|}^2$ is interpreted as the conformal scale $|f^{\prime }{|}^2=|\nabla u{|}^2=|\nabla v{|}^2$. If it instead means the sum of the squared magnitudes of both complex coordinate derivatives, it is $2|f^{\prime }{|}^2$, and the resulting $W$ runs at half the usual Brownian speed. This convention does not change divergence of the integral. The mechanism is recurrent disk occupation makes an entire conformal clock infinite.

Take $\lambda >0$ and let ${\sigma }_{\lambda }$ be the first time $X$ hits $\lambda$. Reflect the horizontal coordinate after ${\sigma }_{\lambda }$, leaving the vertical coordinate unchanged. The Strong Markov property and symmetry preserve the law of planar Brownian motion and leave its real-axis exit time unchanged. On paths with ${\sigma }_{\lambda }<\tau$, the reflection swaps the two terminal events $X_{\tau }<\lambda$ and $X_{\tau }>\lambda$. There is no terminal atom at $\lambda$. Hence the Brownian reflection principle givesHorizontal symmetry gives the same probability for $-\inf X>\lambda$. A union bound therefore yieldsAlso $X^{\ast }\ge |X_{\tau }|$, whose Cauchy distribution is $2{\pi }^{-1}{\int }_{\lambda }^{\infty }(1+x^2{)}^{-1}dx$. Thus the maximum tail is bounded above and below by positive constants times ${\lambda }^{-1}$ for large $\lambda$. The tail integral formula for moments for positive moments givesThe large-$\lambda$ integral is finite exactly when $0<p<1$; it diverges for $p\ge 1$. Near zero its upper bound by ${\int }_0^{1}p{\lambda }^{p-1}d\lambda$ is finite.

The vertical coordinate remains nonnegative up to exit and starts at one. The strip calculation, now with upper boundary $\lambda \ge 1$, givesThus $Y^{\ast }$ has density ${\lambda }^{-2}$ on $[1,\infty )$, and in factFor completeness, if the source's $p<1$ includes negative powers, $X^{\ast }$ also has all negative moments. Let ${\sigma }_{\eta }$ be the first horizontal exit from $(-\eta ,\eta )$. ThenThe first bound follows from the Gaussian reflection bound for the vertical hit of zero; the second follows by splitting time into blocks of length ${\eta }^2$, each having a fixed positive chance of exiting the horizontal interval. This small-value estimate integrates against every negative power. At $p=0$ both moments equal one.

Continuity and the regularity of zero for a one-dimensional Brownian coordinate make $S$ the first contact with the imaginary axis. The Strong Markov property and symmetry of the subsequent real increments imply that reversing those increments after $S$ preserves the Brownian law. The imaginary coordinate is unchanged, so $\tilde{Z}$ is again planar Brownian motion started at one.

Reflection preserves distance to the origin, hence ${\tilde{T}}_{D_r}=T_{D_r}$ pathwise. On $T_{D_r}<S$, the path exits through the right semicircle and reflection has not yet acted. On $S<T_{D_r}$, reflection reverses the sign of the exit real part. ThusThe two circular points on the imaginary axis have zero exit probability, so they cause no extra term. Subtracting proves the reflection identity for Brownian exit from a half-disc requested in the question.

Both processes have the same exit law. With $p={\pi }^{-1}{\tan }^{-1}((r^2-1)/(2r))$ from part (b),The final trigonometric identity follows from ${\tan }^{-1}((r^2-1)/(2r))=\pi /2-2{\tan }^{-1}(1/r)$ for $r>1$.

Suppose first that $d=|z_0|>ϵ$. Apply part (a) with inner radius $ϵ$ and outer radius $R>d$:Each event implies finite hitting time of the $ϵ$ circle, so that hitting probability is one. If $d=ϵ$, the hitting time is zero. If $d<ϵ$, it is the exit time from a bounded disk. The martingale $|Z_t-z_0{|}^2-2t$, stopped before that exit, gives a finite mean exit-time bound $({ϵ}^2-d^2)/2$. Thus the asserted hitting time is almost surely finite in this case too.

For alternating visits, choose two closed smaller disks $K_i$ strictly inside the given open disks $D_i$, so their distance is positive. Define successive first hitting times of $K_1,K_2,K_1,K_2,\dots$. Translation invariance and the Strong Markov property apply the just-proved hitting result from every successive starting point. A countable intersection makes all these times finite almost surely, with the required disk memberships.

These increasing times cannot accumulate at a finite time. On a compact time interval a continuous Brownian path is uniformly continuous, whereas every consecutive visit must cross the fixed positive distance between $K_1$ and $K_2$. ThereforeThis is the recurrence of planar Brownian motion in neighborhoods; it does not assert that every individual point is hit.

For planar Brownian motion starting on the positive real axis inside a disk, let $T$ be circular exit and $S$ first contact with the imaginary axis. Reflecting the path after $S$ fixes the disk and swaps positive and negative circular exits. The Strong Markov property makes their probabilities equal on $S<T$. Therefore $\mathbb{P}(T<S)=\mathbb{P}(\mathrm{Re}{B}_T>0)-\mathbb{P}(\mathrm{Re}{B}_T<0)$.

For planar Brownian motion started at $1$, continuous winding angles sampled at the displayed radii have independent stationary increments. At $T_r$, apply the Strong Markov property, rotate by the current angle, and rescale space by $e^r$ and time by $e^{-2r}$. The future radial-level winding increment has the original law and is independent of the past. The logarithmic radius and winding angle are independent Brownian coordinates on the common clock $\int |B_s{|}^{-2}ds$. At radial passage they are therefore $U_{{\eta }_r}=-r$ and $V_{{\eta }_r}$, where ${\eta }_r$ is a Brownian first-passage time. Its Laplace transform gives the displayed characteristic function. The level process is a symmetric Cauchy process; replacing the passage clock by its right-continuous version supplies a càdlàg modification without changing any fixed-level law.