In the book "Regularity Theory for Elliptic PDE", here is a theorem as follows
Theorem(Harnack's inequality). Assume $ u\in H^1(B_1) $ is a non-negative, is the weak solution for the laplace equation $ \Delta{u}=0\text{ in }B_1 $ where $ B_1 $ is the ball with center $ 0 $ and radius $ 1 $. Then the infimum and the supermum of $ u $ are comparable in $ B_{1/2} $. That is, there exists $ C $ depending only on $ n $, such that \begin{eqnarray} \sup_{B_{1/2}}u\leq C\inf_{B_{1/2}}u. \end{eqnarray}
The proof of this theorem in this book is by using Poisson kernel representation as follows \begin{eqnarray} u(x)=C_n\int_{\partial B_{1}}\frac{1-|x|^2}{|x-y|^n}u(y)dS(y),(*) \end{eqnarray} where $ C_n $ is a constant depending only on $ n $. I know that the proof of the final result is trivial by this representation formula. However I doubt the correctness of this Poisson kernel representation. By using the Weyl Lemma, I see that $ u\in C^{\infty}(B_1) $, but on $ \partial B_1 $, I cannot get that $ u(x)|_{\partial B_1} $ is continuous. Therefore, my question is the formula (*) correct for the weak solution $ u $ and how to derive this formula? Moreover, I know that if we use the Poisson integration on $ \partial B_{3/4} $, we can also have the Harnack inequality, but I still want to know if we can use the Poisson integration on $ \partial B_{1} $