Given $f : \Omega \to \mathbb{R}$ an integrable function defined on the domain $\Omega \subset \mathbb{R}^n$ ($n > 2$), its newtonian potential $w : \mathbb{R}^n \to \mathbb{R}$ is defined by $$w(x) = \int_\Omega \Gamma(x - y) f(y) dy,$$ where $\Gamma(x) = \frac{\|x\|^{2-n}}{n(2-n) \omega_n}$, $x \neq 0$, is the fundamental solution of Laplace equation (According to Gilbarg and Trudinger).
I am trying to prove $w$ is well defined, which amounts to prove that $$\lim_{\varepsilon \to 0} \int_{B_\varepsilon(x)} \Gamma(x - y) f(y) dy = 0,$$ where $B_\varepsilon(x)$ denotes the $\varepsilon$-ball around $x$.
Any suggestions?
In order to have $w$ well-defined everywhere, I don't think it is enough to have $f$ integrable. The way this is usually proved is using Holder's inequality. Ignoring constants and choosing $p \in [1,\infty]$ later, we see $$\left \lvert\int_{B_{\varepsilon}(x)} \frac{f(y)}{\| x-y \|^{n-2}} dy \right \rvert \le \left(\int_{B_\varepsilon(x)} \lvert f(y) \rvert^p dy\right)^{1/p} \left( \int_{B_\varepsilon(x)} \frac{dy}{\| x-y \|^{q(n-2)}}\right)^{1/q}$$ where $\tfrac 1 p + \tfrac 1 q = 1$. The first integral on the right hand side is a finite constant so long as $f \in L^p(\Omega)$. Switching to polars in the latter integral, we see $$\int_{B_\varepsilon(x)} \frac{dy}{\| x-y \|^{q(n-2)}} = \alpha_n \int^\varepsilon_0 \frac{\rho^{n-1}}{\rho^{q(n-2)}} d\rho = \alpha_n \int^\varepsilon_0 \rho^{n-1-q(n-2)} d\rho$$ where $\alpha_n$ is some dimensional constant. The integrand here is integrable near $0$ iff $$n-1-q(n-2) > -1$$ which happens iff $$q < \frac{n}{n-2} \,\,\,\,\,\,\,\,\,\, \Longleftrightarrow \,\,\,\,\,\,\,\,\,\,\, 1- \frac 1 p > \frac{n-2}{n} \,\,\,\,\,\,\,\,\,\,\, \Longleftrightarrow \,\,\,\,\,\,\,\,\,\, p > \frac n 2. $$
Thus for the Newton potential to be well-defined everywhere, we need $f \in L^p(\Omega)$ with $p > n/2.$
If you only need it to be well-defined almost everywhere (w.r.t. the Lebesgue measure), then it is enough to have $f$ integrable. To prove this, just integrate again. We see $$\int_{\Omega} \lvert w(x) \rvert dx\le C \int_{\Omega} \int_{\Omega} \frac{\lvert f(y)\rvert}{\| x-y \|^{n-2}} \,dy\, dx,$$ for some constant $C$ (again, I've supressed all of the constants in the definition of the Poisson kernel). By Tonelli's theorem, we can swap the order of integration yielding, $$\int_{\Omega} \lvert w(x) \rvert dx \le C \int_\Omega \lvert f(y) \rvert \int_\Omega \frac{1}{\| x-y \|^{n-2}} \,dx\, dy.$$ The inner integral converges since $n-2 < n$ (this is obvious just by using polars again). Say this integral is equal to $\beta_n$. Then we see $$\int_{\Omega} \lvert w(x) \rvert dx \le C \beta_n \int_{\Omega} \lvert f(y) \rvert dy < \infty$$ since $f$ is integrable. This shows that $w$ is integrable. But if $w$ is integrable, then it is finite almost everywhere. Hence $w$ is well-defined almost everywhere.
PS. Sorry my response is so long. I hope it helps.