It is a cautionary remark that is often made that solutions to the Dirichlet problem (with continuous boundary conditions) are not unique when the domain in question is the upper half plane. Yes, you have the Poisson integral formula if $f$ is in certain Lebesgue classes. But one can always add $y$ to it, which satisfies the $0$ boundary condition Dirichlet problem without being itself the $0$ harmonic function. In fact, it's not even just one degree of freedom. One can always add also multiples of $e^x\sin(y)$. And anyway, we have to relax the sense in which we mean a solution to the Dirichlet problem. The boundary condition may not be approached in the sense that one can extend the solution continuously to the boundary. The approach may only be in $L^p$. In short, the situation for both existence and uniqueness, as well as the types of allowable boundary conditions, become significantly more complicated than the classical theory for the unit disk. When I refer to $L^p$, I will not allow $p=\infty$.
To be precise, let us say that $f:\mathbb{R}^{d-1}\rightarrow \mathbb{R}$ is an $L^p$ boundary condition, and that we are seeking harmonic functions $F: \mathbb{H}^d\rightarrow \mathbb{R}$ for which $\lim_{\epsilon\downarrow0}F(\epsilon, .)=f$ where the limit is in $L^p$. (Here, $\mathbb{H}^d$ denotes the upper half plane.) Suppose that $F_1$ and $F_2$ are like this and also bounded. Does $F_1=F_2$?
Can $F_1=F_2$ be deduced even if the sense in which the boundary conditions are satisfied derives from a mixture of limits? For instance, $F_1$ may have $f$ as its boundary values in the sense of continuous extension to the boundary being possible, while $F_2$ may have it in the sense of approach in $L^p$ ($p \in (1, \infty)$) with neither of theses conditions implying the other.