Let f be continuous and let Br be the ball of radius r > 0 centered at $(x_0, y_0, z_0)$. Let V (Br) be the volume of Br. Prove that $$\lim_ {r\to0} \frac{1}{V(Br)}\ \iiint_{Br} \ f(x,y,z) dV = f(x_0, y_0, z_0).$$
I can visualise the image of $x^2+y^2+z^2\to0$ is a very small ball that the volume can be neglegible so the integral will be apprx = the real value of the function at the center of the ball.
My question is how to represent my idea in a more formal way.
$$ \Bigl|\frac{1}{V(Br)}\iiint_{Br} f(x,y,z)\,dV-f(x_0,y_0,z_0)\Bigr|\le\frac{1}{V(Br)}\iiint_{Br}\bigl|f(x,y,z)-f(x_0,y_0,z_0)\bigr|\,dV. $$ Now IF $f$ is continuous at $(x_0,y_0,z_0)$...