Dirichlet problem for Laplace equation as follows \begin{eqnarray} \Delta{u}&=&0\text{ in }B_r(0)\\ u&=&g\text{ on }\partial B_{r}(0), \end{eqnarray} where $ g $ is continuous.

It is already known that $ u(x)=C_n\int_{\partial B_{r}(0)}\frac{r^2-|x|^2}{|x-y|^n}g(y)dS(y) $ by the construction of Green function in balls, where $ C_n $ is a constant depending only on $ n $. From this formula, we can see that $ u(x)\in C^{\infty}(B_r(0)) $ and $ u(x)\rightarrow g(\xi) $ if $ x\rightarrow \xi $ with $ \xi\in\partial B_r(0) $. I am thinking the problem that when $ g $ is not continuous but $ g\in L^2(\partial B_{r}(0)) $. In that case the integration $ u(x)=C_n\int_{\partial B_{r}(0)}\frac{r^2-|x|^2}{|x-y|^n}g(y)dS(y) $ still makes sence and $ u(x) $ is still smooth. Can I say that $ u(x) $ actually solve the Dirichlet problem? In other words, can I show that for a.e. $ \xi\in\partial B_{r}(0) $, $ u(x)\rightarrow g(\xi) $ if $ x\rightarrow \xi $?

Dirichlet problem for Laplace equation as follows \begin{eqnarray} \Delta{u}&=&0\text{ in }B_r(0)\\ u&=&g\text{ on }\partial B_{r}(0), \end{eqnarray} where $ g $ is continuous.

It is already known that $ u(x)=C_n\int_{\partial B_{r}(0)}\frac{r^2-\lvert x\rvert^2}{\lvert x-y\rvert^n}g(y)dS(y) $ by the construction of Green function in balls, where $ C_n $ is a constant depending only on $ n $. From this formula, we can see that $ u(x)\in C^{\infty}(B_r(0)) $ and $ u(x)\rightarrow g(\xi) $ if $ x\rightarrow \xi $ with $ \xi\in\partial B_r(0) $. I am thinking about the problem when $ g $ is not continuous but $ g\in L^2(\partial B_{r}(0)) $. In that case the integration $ u(x)=C_n\int_{\partial B_{r}(0)}\frac{r^2-\lvert x\rvert^2}{\lvert x-y\rvert^n}g(y)dS(y) $ still makes sense and $ u(x) $ is still smooth. Can I say that $ u(x) $ actually solves the Dirichlet problem? In other words, can I show that for a.e. $ \xi\in\partial B_{r}(0) $, $ u(x)\rightarrow g(\xi) $ if $ x\rightarrow \xi $?