Dealing with the electrostatic boundary condition

In Griffiths, it is noted that there is a discontinuity in the electric field for a material with a surface charge density.

What is the significance of this boundary condition in practicality when calcuating the electric fields of say, a conductor, where all the charge is located at the surface (because there can be no electric field in the “meat” of the conductor)?

For instance, say I want to find the electric field everywhere for a metal conducting sphere of radius $a$.

I can deduce that the only existent charge density is that of a surface charge density $sigma$ as the object is conducting.

By application of Gauss’ Law, I can show that for $r<a$, $mathbf E = 0$ as there is no enclosed charge until $r=a$.

For $r ge a$, I can envelop the sphere in a Gaussian surface with radius $r$ such that $r ge a$.

$$implies oint mathbf E cdot hat n dS = frac{Q_{enc}}{epsilon_0} = frac{sigma A}{epsilon_0}$$

$$implies |mathbf E| 4 pi r^2 = frac{sigma}{epsilon_0} 4 pi a^2$$

$$implies mathbf E = frac{sigma}{epsilon_0} frac{a^2}{r^2} hat r$$

Perhaps I’ve just found the answer to my question by considering $mathbf E (r=a)$? However, $$frac{sigma}{epsilon_0} hat n$$ seems to be a change in $mathbf E$ as it is $mathbf E_{above} – mathbf E_{below}$, so why is it a value at $r =a$ here, and not, for instance, the change in $mathbf E$ for $mathbf E (a – dr) to mathbf E (a + dr)$?