In Lawrence C Evans' Partial Differential Equations, Chapter 2, he proves by construction that the representation formula for the solution to Poisson's equation with initial values:
$$\begin{cases} -\Delta u = f & \text{in }U,\\ \phantom{-\Delta}u = g & \text{on }\partial U.\end{cases}$$
using Green's functions is given by
$$u(x) = -\int_{\partial U} g(y) \frac{\partial}{\partial \nu} G(x,y)dS(y) - \int_{U}f(y)G(x,y)dy.$$
He then proceeds to say that by fixing $x \in U$, we may symbolically write
$$\begin{cases} -\Delta G = \delta_x & \text{in } U,\\ \phantom{-\Delta}G = 0& \text{on }\partial U. \end{cases}$$
Doesn't plugging this set of equations into the representation make the first term identically zero? As in, since the first integral is over the boundary of $U$, isn't $G$ zero there, and hence its normal derivative as well?