I was proving that $$-\Delta_x G(x-y)=\delta_y \text { on } D^{\prime}\left(\mathbb{R}^3\right) $$
but during the calculation there is $$\frac{1}{4 \pi \varepsilon^2} \int_{\partial B_{\varepsilon}(y)} d \sigma(x) \varphi(x) \underset{\varepsilon \rightarrow 0}{\longrightarrow} \varphi(y) $$ my question is why?