Suppose $\Omega \subset \mathbb{R}^3$ is closed, bounded, with smooth boundary and the density $\rho: \Omega \to \mathbb{R}$ is continuous. The potential function is
$$\phi(x) = \int_\Omega \frac{\rho(x')}{|x - x'|} dx',$$
I already showed it continuous for $x \notin \Omega$ and am trying to show it continuous for $x \in \Omega$.
My effort: I take a ball $B_\delta(x_0)$ around a fixed point $x_0$ with $|x-x_0| < \delta$ and look at
$$|\phi(x) - \phi(x_0)| \leq \left|\int_{\Omega- B_\delta(x_0)}\rho(x') \left(\frac{1}{|x-x'|} - \frac{1}{|x_0-x'|}\right) dx'\right|+ \left|\int_{ B_\delta(x_0)}\frac{\rho(x') }{|x_0-x'|} dx' \right| + \left|\int_{ B_\delta(x_0)} \frac{\rho(x')}{|x-x'|} dx' \right|$$
Since the integrand in the first integral is continuous there exists $\delta_0$ such that if $|x - x_0| < \delta_0$ then
$$\left|\int_{\Omega- B_\delta(x_0)}\rho(x') \left(\frac{1}{|x-x'|} - \frac{1}{|x_0-x'|}\right) dx'\right| < \frac{\epsilon}{3}$$
Also convergence of the potential integral means if $\delta$ is small enough then
$$\left|\int_{ B_\delta(x_0)}\frac{\rho(x') }{|x_0-x'|} dx' \right| < \frac{\epsilon}{3}$$
My question is how to show the third integral can be made $< \frac{\epsilon}{3}$ since $B_\delta(x_0)$ is not centered at $x$.
The Absolute continuity of the Lebesgue integral can be used here. Since the function $\rho(x)/|x-x_0|$ is integrable on the ball $B(x_0, 1)$, it follows that for every $\epsilon$ there exists $\delta>0$ such that if $E\subset B(x_0, 1)$ has measure less than $\delta$, then $$ \int_E \frac{|\rho(x)|}{|x'-x_0|} dx' < \epsilon $$
This argument works for both 2nd and 3rd integrals, they don't have to be treated separately.