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 $?
Yes or no, depending on what you mean by solve.
No, this isn't possible in this generality. So let me divide my answer into two parts.
Negative. This boils down to the following:
Applying it to our case, $C$ would have full measure, so $g$ would be continuous when restricted to a set of full measure. This is not true for an arbitrary function $g \in L^2(\partial B_r)$ (otherwise Lusin's theorem would work with $\varepsilon=0$). As an example, the characteristic function of a fat Cantor set doesn't have this property.
Proof of the lemma. Assume (for the sake of contradiction) that $g|_C$ is discontinuous at $\xi \in C$. Then there is a sequence $\xi_n \in C$ for which $\xi_n \to \xi$ but $|g(\xi_n)-g(\xi)| \ge \varepsilon > 0$. For each $n$, consider the sequence $y_m := (1-\frac{1}{m}) \xi_n$. For large $m$, we have $|y_m-\xi_n| \le |\xi_n-\xi|$, and by our assumption also $|u(y_m)-g(\xi_n))| \le \varepsilon/2$. Take $x_n$ as one of these points, i.e. some $y_m$ for a large value of $m$. Then $$ |x_n - \xi| \le |x_n-\xi_n| + |\xi_n-\xi| \le 2 |\xi_n-\xi| \to 0, $$ while at the same time $$ |u(x_n)-g(\xi)| \ge |g(\xi_n)-g(\xi)| - |u(x_n)-g(\xi_n)| \ge \varepsilon - \frac{\varepsilon}{2} = \frac{\varepsilon}{2}, $$ which leads to a contradiction with our assumption.
Positive. I expect (although I haven't checked it) that $u(x) \to g(\xi)$ in an $L^2$ sense. More precisely:
In principle, I'd expect it to be similar to the classical case ($g$ continuous, $u$ continuous up to the boundary). One has $$ u_s(x)-g(x) = c \int_{\partial B_r} \frac{r^2-s^2r^2}{|sx-y|^n} (g(y)-g(x)) \, dy, $$ then after applying Jensen's inequality: $$ \int_{\partial B_r} |u_s(x)-g(x)|^2 \, dx \le c \int_{\partial B_r} \int_{\partial B_r} \frac{r^2-s^2r^2}{|sx-y|^n} |g(y)-g(x)|^2 \, dy \, dx, $$ and hopefully this tends to zero.