Bounded Dirichlet problem for hitting probabilities

Fix $x\in E$ and a bounded real $q:E\to \mathbb{R}$, with $\mathrm{\mid }q\mathrm{\mid }\le M$. Take an increasing sequence of finite sets $E_j\uparrow E$ (eventually $E_j=E$ if $E$ is finite) and put $q_j=q{\mathbf{1}}_{E_j}$. The positive and negative parts of $q_j$ are finite-range nonnegative simple functions. By [F1], [F7] and [F8], $K{q}_j(x)={\int }_E{q}_{j}dK(x,\cdot )={\sum }_{y\in E_j}p(x,y)q(y)\mathrm{.}$ The constant $M$ is integrable for the probability measure $K(x,\cdot )$, so [F9] gives $K{q}_j(x)\to Kq(x)$. Also ${\sum }_{y}p(x,y)\mathrm{\mid }q(y)\mathrm{\mid }\le M{\sum }_{y}p(x,y)=M$ by [F1], so the finite sums converge to the absolutely convergent row sum $Pq(x)$ from [F6]. Therefore $Kq(x)=Pq(x)\mathrm{.}$ This identity holds for every bounded real $q$ and each row; no positivity of the individual entries beyond being transition weights is required.

Choose $M<\mathrm{\infty }$ with $\mathrm{\mid }f(a)\mathrm{\mid }\le M$ on $A$, extend $f$ by zero off $A$, and define $H(\omega )$ to be $f({\omega }_{T_A(\omega )})$ when the first-hit time of $A$ is finite, and zero otherwise. Each event that the first hit is at time $n$ is cylinder-measurable; $H$ is the pointwise limit of its finite sums over these disjoint events, so it is product-measurable and $\mathrm{\mid }H\mathrm{\mid }\le M$. By [F5], $h(x):={\mathbb{E}}_{x}H(X_0,X_1,\dots )$ is measurable; [F10] gives $\mathrm{\mid }h(x)\mathrm{\mid }\le M$. The payoff in the Statement equals $H(X_0,X_1,\dots )$ pathwise, including its zero value on nonhit paths, so $h=u$. If $x\in A$, [F2] and [F3] give $T_A=0$ and $h(x)=f(x)$. If $x\mathrm{\notin }A$, deleting the first coordinate does not change $H$, including when the path never hits $A$. Thus [F5] at time $1$ and [F10] give $h(x)={\mathbb{E}}_{x}H(X_1,X_2,\dots )={\mathbb{E}}_{x}h(X_1)\mathrm{.}$ Apply [F4] at time $0$ to the bounded function $h$, use [F3] and [F10] to take expectations, and then use step 1.1 to obtain $u(x)=h(x)=Kh(x)=Ph(x)=Pu(x)\mathrm{.}$ This proves existence and both equations.

Let $v:E\to \mathbb{R}$ be any bounded solution of the stated boundary and harmonic equations, fix $x\in E$, set $T=T_A$, and define $Y_n=v(X_{n\wedge T})$. By [F2] this is adapted, and it is bounded by $\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}$. On $\{T\le n\}$, $Y_{n+1}=Y_n$; on $\{T>n\}$, $X_n\in A^c$ and $Y_{n+1}=v(X_{n+1})$. The one-step identity [F4], the row identity in step 1.1, and $v=Pv$ on $A^c$ imply ${\mathbb{E}}_x[Y_{n+1}\mid {\mathcal{F}}_n]=Y_n\text{a.s.}$ Consequently $Y_n$ is a bounded martingale and [F3], [F10] give ${\mathbb{E}}_x{Y}_n=v(x)$ for every $n$. The hypothesis makes $T<\mathrm{\infty }$ ${\mathbb{P}}_x$-almost surely, so eventually $Y_n=v(X_T)=f(X_T)=F_A$ almost surely. Since $\mathrm{\mid }{Y}_n\mathrm{\mid }\le \mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}$ is an integrable majorant, [F9] yields $v(x)={\lim }_n{\mathbb{E}}_x{Y}_n={\mathbb{E}}_x{F}_A=u(x)\mathrm{.}$ As $x$ was arbitrary, every bounded solution equals $u$; this proves uniqueness.

If $E$ is finite, $p$ is irreducible and $A\mathrm{\ne }\mathrm{\varnothing }$, [F11] gives ${\mathbb{P}}_x(T_A=\mathrm{\infty })\le {\mathbb{P}}_x(T_A>km)\le (1-\epsilon {)}^k⟶0$, so the almost-sure hitting hypothesis holds for every start. The preceding steps give the finite irreducible instance. When $f\equiv 1$ on $A$, the pathwise payoff is ${\mathbf{1}}_{\{T_A<\mathrm{\infty }\}}$, hence its expectation is the hitting probability; under the theorem's hypothesis it equals $1$.

If $E=\mathrm{\varnothing }$, there is no state or deterministic-start law and all assertions are vacuous. If $E\mathrm{\ne }\mathrm{\varnothing }$ and $A=\mathrm{\varnothing }$, then $T_A=\mathrm{\infty }$ everywhere, contradicting the hypothesis; thus every nonvacuous instance has $A\mathrm{\ne }\mathrm{\varnothing }$. If $A=E$, then $T_A=0$, the boundary equation determines $u$ and every solution directly. For a one-state chain these are the only admissible cases. If $f=0$, then $H=u=0$ and the uniqueness proof still applies. Deterministic rows are covered by the row identity and the same martingale calculation; a finite irreducible deterministic chain reaches every nonempty $A$ by the finite-tail argument. The endpoint $T_A=0$ is handled in step 2.1, while a first hit at time $1$ is included in the shifted-path and stopped-process identities in steps 2.1 and 2.2. AC [A1] is used through the canonical chain laws and conditional-expectation Markov identities [F4], [F5] and [F10]; choosing one enumeration of a countable $E$ adds no family-wise choice. The equations and uniqueness claim are not an iff statement.