Poisson integral Given a bounded harmonic function $u(z)$ on the open unit disk, and given radial limits $\lim_{r\rightarrow 1^{-1}}u(re^{i\theta})$ being some constant $a$ for $0<\theta<\pi$ and being some other constant $b$ for $\pi<\theta<2\pi$, can I use the Poisson integral formula $u(z_{0})=\dfrac{1}{2\pi}\int_{0}^{2\pi}u(e^{i\theta})\dfrac{1-|z_{0}|^{2}}{|e^{i\theta}-z_{0}|^{2}}d\theta$ to get values of $u$ inside the disk?
I'm unsure about the conditions under which the formula can be applied.
 A: Boundedness is  enough. To justify the integral representation of $u$, one usually considers the functions $u_r(z)=u(rz)$ with $0<r<1$. These are smooth on the boundary, so the formula is not an issue: $$u_r(z) = \frac{1}{2\pi} \int_0^{2\pi} u_r(e^{i\theta})\frac{1-|z|^2}{|e^{i\theta}-z|^2}\,d\theta \tag1$$
As $r\to 1$, we have $u_r(z)\to u(z)$ for every $z$ in the open unit disk $\mathbb D$, because $u$ is continuous there. So the issue is how to pass to the limit in the right hand side of (1). Since we do it with $z$ fixed,  the kernel $\frac{1}{2\pi}\frac{1-|z|^2}{|e^{i\theta}-z|^2}$ is a nice bounded function of $\theta$; it creates no problems. (One could even wrap it together with $d\theta$ into a probability measure $\omega_z$ on the boundary, called the harmonic measure with respect to $z$.) 
Suppose that as $r\to 1$,  $u_r(e^{i\theta})$ converges for almost every $\theta$ to some function, which we can denote $u^*(e^{i\theta})$. (The asterisk is optional but it helps to keep things in order, emphasizing the distinction between a function and its limit  values on the boundary.) In the presence of a uniform bound $|u_r|\le M$ the dominated convergence theorem applies; hence,   $$\frac{1}{2\pi} \int_0^{2\pi} u_r(e^{i\theta})\frac{1-|z|^2}{|e^{i\theta}-z|^2}\,d\theta \to \frac{1}{2\pi} \int_0^{2\pi} u^*(e^{i\theta})\frac{1-|z|^2}{|e^{i\theta}-z|^2}\,d\theta \tag2$$  which answers your question.
Sometimes we don't have a dominating function handy. Instead, it suffices to assume that there exists $p>1$ and $M$ such that $\int |u_r(e^{i\theta})|^p\,d\theta\le M$ for all $r$. Indeed, a bounded sequence in $L^p$ with $1<p<\infty$ has a weakly convergent subsequence. Let  $u^*$ be this weak limit. Since  the Poisson kernel is  a bounded function of $\theta $ (for a fixed $z$), integration against it a continuous linear functional on $L^p$. Thus, (2) holds  for a subsequence, which  is enough to get  the integral representation of $u$. (Using the representation, one can show that the integrals indeed converge  as $r\to 1$.) 
With $p=1$ the above does not work since $L^1$ is not reflexive. But if $\int |u_r(e^{i\theta})|\,d\theta$ is bounded, then we can find a subsequence that converges in the sense of weak* convergence of measures. Then the integral representation of $u$ involves not the radial limits $u^*$, but some finite signed measure   on the boundary, which need not be absolutely continuous.
