When we apply Gauss's law should we consider also the charge over the gaussian surface?

Lets say I have a coaxial cable, the internal material cylinder has radius $a$ and uniform volumetric charge density $\rho$, the external shield cylinder has radius $b$ and uniform superficial charge density $\sigma$, and the total charge of the cable is zero. I want to find the electric field at any point is space.Using Gauss's law I found $$E(r)= \begin{cases} \frac{r \rho}{2\epsilon_0} \quad \mbox{ if $r\leq a$} \\ \frac{a^2\rho}{2\epsilon_0 r} \quad \mbox{if $a\leq r<b$} \\ 0 \quad \mbox{ . .if $b<r$} \end{cases}$$ but what happen at $r=b$? Should I understand "the charge enclosed by the gaussian surface" as the charge in the interior of the volume having as frontier that surface or it does incude the charge in the surface itself? In the first case it would be $E(b)=\frac{a^2\rho}{2\epsilon_0b}=\frac{-\sigma}{\epsilon_0}$ while in the second one it would be just zero. Which one is the right interpretation?