Let $U\subset \mathbb{R}^d$ be a bounded domain and $g: U\to \mathbb{R}$ be continuous. A continuous function $u:\overline{U}\to \mathbb{R}$, $u\in \mathcal{C}^2(U)$ is said to be a solution of Poisson's problem for $g$ if : \begin{cases}u(x)=0 \text{ for all } x \in \partial U\\ \Delta u(x)=-2g(x)\text{ for all } x \in U. \end{cases}
If $u$ is a solution of Poisson's problem it can be shown by appliying Itô's formula and taking the expectation that $$u(x)= \mathbb{E}_x(\int_0^Tg(B_t)dt) \text{ for all } x \in U, $$ where $T:=\inf\{t>0 : B_t \notin U \}$.
I want to show the converse, that is to say that $u(x):=\mathbb{E}_x(\int_0^Tg(B_t)dt) $ is a solution to the Poisson's problem associated to $g$ (if every point of $\partial U$ is regular). I don't see how to prove that $\Delta u=-2g$. I know that we wan write $u(x)$ as $\int G(x,y)g(y)dy$ where $G$ denote Green's function and that $x \to G(x,y)$ is harmonic on $U\setminus \{y\}$.