If the Poisson formula is valid when the boundary condition is $ L^2 $? 
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 $?
 A: 
Can I say that $u(x)$ actually solve the Dirichlet problem?

Yes or no, depending on what you mean by solve.

In other words, can I show that for a.e. $\xi\in\partial B_{r}(0)$, $u(x) \to g(\xi)$ if $x \to \xi$?

No, this isn't possible in this generality. So let me divide my answer into two parts.

Negative. This boils down to the following:

Lemma. Let $C \subseteq \partial B_r$ be the set of all $\xi \in\partial B_{r}$ with the following property:
$$
x_n \in B_r, \ x_n \to \xi 
\quad \Longrightarrow \quad 
u(x_n) \to g(\xi).
$$
Then $g|_C$ is continuous.

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:

Claim. Let $u_s \colon \partial B_r \to \mathbb{R}$ by defined as $u_s(x) := u(sx)$ (for $0<s<1$). Then $u_s \xrightarrow{s \to 1} g$ in $L^2(\partial B_r)$.

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.
