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If $u$ is a bounded harmonic function on a domain $D$, continuous on its closure, and the Brownian exit time $T_D$ is almost surely finite from $z$, then $u(z)={\mathbb{E}}_z[u(B_{T_D})]$. The Itô formula and compact localization make $u(B_{t\wedge T_D})$ a bounded martingale. bounded convergence theorem at the exit time proves the representation even if the domain is unbounded. Removing boundedness of $u$ can invalidate this conclusion through loss of uniform integrability.

Unit-radius Brownian exit time has finite variance, since independent unit-time increments give an exponential upper tail. The Brownian exit-time skeleton clock after $2^{2n}$ radius-$2^{-n}$ exits has mean $\mathbb{E}{\sigma }_1$ and variance $2^{-2n}\mathrm{Var}({\sigma }_1)$. Chebyshev's inequality and the Borel-Cantelli first lemma yield the displayed pathwise convergence. Independence between different rows indexed by $n$ is unnecessary.

For a connected open set $D$, finiteness of the Brownian exit time expectation at one point implies finiteness everywhere. The Brownian exit-time ball averaging identity gives integrability on a ball centred at any finite point. Every point of that ball has a smaller centred ball contained in it; the same identity then gives a finite value there. The set of finite points is open. It is also relatively closed: near a limit point in $D$, choose a fixed ball radius smaller than a quarter of its distance to $D^c$ and centre it at a nearby finite point. The limit point lies in that ball and hence is finite. Connectedness completes the proof.

Suppose an unbounded domain contains the upper strip $(-1,1)\times (1,\infty )$ with its two vertical sides on the boundary. A bounded harmonic function with constant values $a$ on the right side and $b$ on the left side, and with valid bounded harmonic representation with almost sure Brownian exit, satisfies $u(x,y)\to a(1+x)/2+b(1-x)/2$ as $y\to \infty$. The horizontal Brownian exit time is finite; on that finite random interval the vertical Brownian path has a finite minimum. Consequently reaching the bottom before a side has probability tending to zero. The remaining two exit probabilities are the one-dimensional interval probabilities.

For a bounded domain whose boundary points are all regular boundary points, continuous Dirichlet boundary data $f$ have a unique solutionHere ${\tau }_D$ is the Brownian exit time. The Strong Markov property gives the mean value property for harmonic functions; a barrier for the Dirichlet problem gives continuity at the boundary. This is an existence theorem as well as the Brownian representation of the Dirichlet problem for an already known solution.

Allow $g_D$ to take the value $+\infty$. Fix a closed ball $\overline{B(z,r)}\subset D$, independently choose a radius $R$ with density $n{s}^{n-1}/r^n$ on $(0,r)$, and let $S$ be the first time the Brownian motion from $z$ reaches that radius. Conditional on $R=s$, the Strong Markov property at $S$ decomposes the remaining Brownian exit time from $D$. Since $S<T_D$ and the conditional exit time has mean $s^2/n$, Tonelli's theorem and part (a) give the Brownian exit-time ball averaging identityThis identity is valid in the extended nonnegative reals; no finiteness was assumed to derive it.

If $g_D(z)<\infty$, the integral over $B(z,r)$ is finite. For any $y\in B(z,r)$ choose $\rho >0$ with $\overline{B(y,\rho )}\subset B(z,r)$. The same identity at $y$ has a finite right side because its integral is over a subset of the integrable ball. Thus $g_D(y)<\infty$. The set $F=\{z\in D:g_D(z)<\infty \}$ is therefore open relative to $D$.

It is also relatively closed. If $z_j\in F$ converges to $y\in D$, choose $r>0$ with $\overline{B(y,3r)}\subset D$. For large $j$, $|z_j-y|<r$, and $\overline{B(z_j,2r)}\subset D$. The preceding argument shows that every point in $B(z_j,2r)$, including $y$, belongs to $F$. Hence $F$ is closed in $D$. Since $F$ is nonempty and $D$ is connected, $g_D(y)<\infty$ for every $y\in D$. This proves finiteness propagation of Brownian mean exit times without assuming beforehand that $g_D$ solves a smooth Poisson equation.

Use the intrinsic boundary of a simply connected domain, so different approaches to the two sides of a slit remain distinct. The mapping-out function $g_K$ extends as a homeomorphism from this intrinsic compactification to that of the complex upper half-plane. Set $T=T(H)$. The imaginary coordinate of planar Brownian motion hits zero in finite time almost surely, and $T$ is no larger than that time. Thus $T<\infty$ and continuity gives a finite Euclidean exit point.

By conformal invariance of planar Brownian motion, $g_K(B_t)$ is a planar Brownian motion in the complex upper half-plane, run with clockThe terminal clock cannot be infinite: that would make the transformed Brownian motion stay in the upper half-plane forever. It cannot stop while the transformed path is in the interior either, since continuity of $g_K^{-1}$ would then put the original exit point inside $H$. Hence the terminal clock is precisely the transformed Brownian exit time. The transformed path converges to a real boundary point, and applying the extended inverse proves almost sure convergence to a point of the intrinsic boundary.

Write $g_K(x+iy)=u+iv$, and let $E=g_K(S)\cap \mathbb{R}$. The point at infinity has zero harmonic measure. Conformal invariance of planar Brownian motion and the Poisson kernel for the upper half-plane giveThe hydrodynamic normalization at infinity gives $g_K(z)=z+O(1/z)$, so along the specified approach $v/y\to 1$ and $u/y\to 0$. For each fixed real $t$,If $\mathrm{Leb}(E)<\infty$, dominated convergence applies because $y/v$ is eventually bounded. If $\mathrm{Leb}(E)=\infty$, Fatou's lemma makes the limit infinite. This proves the harmonic-measure asymptotic at infinitywith the equality understood in the extended nonnegative reals.

The construction is well defined. Since $D$ is contained in a ball, its Brownian exit time is finite almost surely: at successive integer times there is a fixed positive probability that the next independent unit-time Brownian motion increment has length greater than the ball's diameter, forcing an exit. The survival probability is therefore bounded by a geometric sequence. Path continuity gives $B_{{\tau }_D}\in \partial D$. Thus the bounded Dirichlet boundary data give a bounded Borel function $u$ on $D$.

For interior harmonicity, fix a ball with closure in $D$, centred at $x$, and let $\sigma$ be its Brownian exit time. The Strong Markov property, followed by the tower property of conditional expectation, givesThe orthogonal invariance of Brownian motion makes $B_{\sigma }$ uniform on the boundary sphere. Hence $u$ has the spherical mean value property for harmonic functions for every such ball. A locally bounded Borel function with this property is smooth and harmonic: integrating the spherical averages against any smooth radial mollifier gives $u=u\ast \rho$ locally, which first proves smoothness; the mean value property for harmonic functions then implies $\Delta u=0$.

For the boundary limit at $\xi$, choose the harmonic barrier for the Dirichlet problem $h_{\xi }$ from (i). The Itô formula, localized inside $D$, makes $h_{\xi }(B_{t\wedge {\tau }_D})$ a bounded martingale. Compact localization and the bounded convergence theorem, first to the exit and then as $t\to \infty$, yieldFor $\epsilon >0$, if $\partial D\setminus B(\xi ,\epsilon )$ is nonempty, compactness and barrier positivity giveIf that boundary subset is empty, the probability is already zero. ThereforeFirst let $x\to \xi$ and then $\epsilon \downarrow 0$. The continuity of the Dirichlet boundary data proves $u(x)\to f(\xi )$.

Finally, the difference of two continuous solutions is a harmonic function vanishing on the boundary. The maximum principle for harmonic functions on the bounded domain gives that it is zero. Thus the Kakutani solution of the Dirichlet problem exists, attains every prescribed boundary value, and is unique.

A standard form of the Kakutani solution of the Dirichlet problem uses a bounded domain $D\subset {\mathbb{R}}^d$, continuous Dirichlet boundary data $f\in C(\partial D)$, and regularity of every boundary point for the Dirichlet problem. The boundedness of $D$ makes $\partial D$ a compact set, so $f$ is a bounded function and a uniformly continuous function. A bounded domain with $C^2$ boundary is a sufficient geometric case; one must not omit boundary regularity for an arbitrary bounded domain.

One standard analytic characterization of a regular boundary point on a bounded domain is the existence of a positive harmonic barrier: for each $\xi \in \partial D$ there is a harmonic function $h_{\xi }\in C(\overline{D})$ with $h_{\xi }(\xi )=0$ and $h_{\xi }(x)>0$ for $x\in \overline{D}\setminus \{\xi \}$. This is the harmonic form of a barrier for the Dirichlet problem; the barrier characterization of regularity is a standard fact of potential theory.

Let $B$ be $d$-dimensional Brownian motion started at $x\in D$ and let ${\tau }_D=\inf \{t\ge 0:B_t\notin D\}$ be its Brownian exit time. Then the unique solution in $C(\overline{D})\cap C^2(D)$ isEquivalently, $u(x)={\int }_{\partial D}f(\xi ){\omega }_D(x,d\xi )$, where ${\omega }_D$ is harmonic measure, the exit probability distribution of Brownian motion. The harmonic function equation is $\Delta u=0$, with the Laplace operator convention $\Delta ={\sum }_j{\partial }_j^2$.

First take finite endpoints $\ell <r$ and a starting point $z\in (\ell ,r)$. For the Brownian exit time $H=\inf \{t:B_t\notin (\ell ,r)\}$, the stopped path stays in $[\ell ,r]$. The process $(B_t-z{)}^2-t$ is a martingale. Bounded-time optional stopping theorem givesBy the monotone convergence theorem, $\mathbb{E}H<\infty$, so exit from every bounded interval is almost sure:For completeness, a proper unbounded interval also has almost sure exit. Applying optional stopping theorem to $B_{H\wedge t}$ and then bounded convergence theorem gives the endpoint probabilitiesFor a fixed lower level $\ell$, let $r\to \infty$. Hitting $\ell$ before $r$ is a subset of ever hitting $\ell$, and its probability tends to one. Similarly every fixed higher level is hit almost surely. After a level is reached, the same argument from that level shows that a level strictly outside a given proper interval is eventually reached. This proves exit even when endpoints are included. A starting point already outside needs no argument. The whole real line has no exit, so the phrase “every interval” is understood to exclude that trivial exception. Countably many rational levels make the exit conclusion simultaneous for all proper intervals. The bounded-interval assertion is the one used below.

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.

Put $u(z)=\mathrm{Im}(z-g_A(z))$ on $D_A$. This is a harmonic function, tends to zero at infinity, and is bounded. To justify the boundary behavior without requiring a locally connected space as its boundary, let $f_A=g_A^{-1}$. Its expansion at infinity implies that bounded $z$ cannot have $|g_A(z)|\to \infty$. If $z_n$ approaches a finite point of $\partial D_A$ and $g_A(z_n)$ had a subsequential limit inside $\mathbb{H}$, continuity of $f_A$ would force that point to lie inside $D_A$, a contradiction. Thus boundary degeneration under a mapping-out function gives $\mathrm{Im}{g}_A(z_n)\to 0$. Consequently $u$ extends continuously to the finite boundary with value $\mathrm{Im}z$.

Let $B$ be planar Brownian motion started at $z\in D_A$, and $\tau$ its Brownian exit time from $D_A$. This time is finite almost surely: it is at most the first time the imaginary coordinate, a one-dimensional Brownian motion, hits zero. By the Itô formula, $u(B_{t\wedge \tau })$ is a bounded martingale. The optional stopping theorem and dominated convergence theorem therefore giveBoundedness is important here: directly stopping the unbounded imaginary-coordinate martingale would require an unjustified uniform integrability assertion.

At $z=iy$, the Laurent series gives $u(iy)=a/y+O(y^{-2})$. HenceThe sign follows because the exit point lies on $\partial D_A\subseteq \overline{\mathbb{H}}$. This proves the Brownian representation of half-plane capacity.

The implication is false. Take the thin rectanglesEach is a compact H-hull, and $\mathrm{diam}(A_{\epsilon })=\sqrt{4+{\epsilon }^2}\to 2$. We verify that its half-plane capacity nevertheless tends to zero.

Let $K=\{z\in \mathbb{H}:|z|\le 2\}$, which contains all these rectangles. At the Brownian exit time $\tau$ from $\mathbb{H}\setminus A_{\epsilon }$, the exit height is at most $\epsilon$, and a positive exit height requires hitting $K$ before $\mathbb{R}$. ThereforeFor $y>2$, conformal invariance of planar Brownian motion and $g_K(z)=z+4/z$ identify the probability on the right with the harmonic measure of $[-4,4]$ from $i(y-4/y)$ in $\mathbb{H}$. Integrating the Poisson kernel for the upper half-plane givesThe Brownian representation of half-plane capacity now proves the explicit estimateChoose $\epsilon =1/n$ to obtain the required sequence. This is an instance of half-plane capacity of a low rectangle: shrinking height can make half-plane capacity small while horizontal extent stays fixed.

Put $\Delta B_h=B_{t+h}-B_t$ and $R_h={\int }_t^{t+h}(H_s-H_t)d{B}_s$. The constant-in-time coefficient $H_t$ on $(t,t+h]$ is previsible, so the ratio minus $H_t$ is $R_h/\Delta B_h$. Since $|H|\le C$, continuity and the dominated convergence theorem implyThe Itô isometry gives $\mathbb{E}{R}_h^2\le h{\eta }_h$. For $\epsilon ,\delta >0$, use the Chebyshev inequality and the normal distribution of $\Delta B_h/\sqrt{h}$ to obtainFirst let $h\downarrow 0$, then $\delta \downarrow 0$; the normal distribution has no atom at zero. This provesNo independence between $R_h$ and the denominator is needed.

The first estimate suggested by the hint is valid: Brownian scaling and the Gamma function give $\mathbb{E}|\Delta B_h{|}^{-1/2}=h^{-1/4}{2}^{-1/4}\Gamma (1/4)/\sqrt{\pi }$. The second hint requires care about the placement of its expectation: with $V_h={\int }_t^{t+h}(H_s-H_t{)}^{2}ds$, the interpretation $\mathbb{E}|R_h{|}^{1/2}\le (\mathbb{E}{V}_h{)}^{1/4}$ is correct by the Jensen inequality and the Itô isometry, whereas the interpretation $\mathbb{E}|R_h{|}^{1/2}\le \mathbb{E}(V_h^{1/4})$ is false with constant $1$. For the latter interpretation, the Burkholder-Davis-Gundy inequality supplies a constant $C_{1/2}$ instead. One can see the obstruction with a Brownian motion stopped on exiting $(-1,1)$: if $\tau$ is that Brownian exit time, then $|B_{\tau }|=1$, $\mathbb{E}\tau =1$, and strict Jensen inequality gives $\mathbb{E}{\tau }^{1/4}<1=\mathbb{E}|B_{\tau }{|}^{1/2}$.

This obstruction also survives the continuity condition on $H$. On a long finite interval approximate $1_{\{s<\tau \}}$ by a continuous adapted process equal to $1$ until $\tau$ and decreasing linearly to $0$ during the next $ϵ$ units of time, and take its initial value to be $0$ by inserting a short continuous ramp before starting the exit experiment. The Itô isometry controls these approximations, and letting the interval length tend to infinity recovers the strict inequality above; hence some bounded continuous adapted example also violates the latter interpretation with constant $1$. The direct proof avoids relying on this ambiguous hint.

Let $T={\tau }_b\wedge {\tau }_{-a}$ and $A=\{{\tau }_b<{\tau }_{-a}\}$. The Brownian exit time $T$ is finite almost surely. Optional stopping of the bounded martingale $B_{t\wedge T}$ givesFor $T_n=T\wedge n$, optional stopping of $B_t^2-t$ gives $\mathbb{E}{T}_n=\mathbb{E}[B_{T_n}^2]\le \max (a^2,b^2)$. Letting $n\to \infty$ by monotone and bounded convergence proves $\mathbb{E}T=\mathbb{E}[B_T^2]=ab$.

The third derivative in part b at $\lambda =0$ is the cubic martingale $B_t^3-3t{B}_t$. Optional stopping at $T_n$ is valid because $T_n$ is bounded. Since $B_{T_n}$ is bounded and $T_n\to T$ in $L^1$, its stopped identity passes to the limit and givesPut $x=\mathbb{E}[T{1}_A]$ and $y=\mathbb{E}[T{1}_{A^c}]$. Then $x+y=ab$, whileSolving gives $x=ab(2a+b)/(3(a+b))$. Dividing by $\mathbb{P}(A)=a/(a+b)$ proves the conditional Brownian interval-exit time formula

Let $D\subset {\mathbb{R}}^d$ be bounded, choose $R$ with $D\subset B(0,R)$, and let $T$ and ${\tau }_R$ be the respective exit times. Then $T\le {\tau }_R$. Sinceis a martingale, the optional sampling theorem for a supermartingale at ${\tau }_R\wedge n$ givesMonotone convergence theorem now gives

For a nonnegative random variable, $Y={\int }_0^{\infty }{1}_{\{Y>t\}}dt$ pointwise, including when $Y=\infty$. Tonelli's theorem permits taking expectation under the integral and proves the formula in the extended nonnegative reals. This relates finiteness of a Brownian exit time expectation to integrability of its survival probability.