Crystal Field Splitting of d-Orbitals in Octahedral and Tetrahedral Ligand Fields
The labels you are talking about refer to the orbitals’ irreducible representations. Which irreducible representations can occur is dictated by the point group of the entire compound. For octahedral compounds, the point group is the very symmetric $O_\mathrm{h}$, which basically includes every element of symmetry you can have in a cube. Tetrahedral complexes belong to the $T_\mathrm{d}$ point group with a considerably lower symmetry (i.e. a smaller number of symmetry operations).

Octahedral complexes are probably more present in the average non-coordination chemist’s brain. Since $O_\mathrm{h}$ includes a centre of symmetry, all irreducible representations have a subscripted g or u to indicate whether equal phases of an orbital are mapped onto each other upon inversion (g for gerade) or whether phases are mapped onto their opposites upon inversion (u for ungerade). Due to the d-orbitals having two nodal planes, they are automatically g, which is why they as a whole transform as $\mathrm{e_g + t_{2g}}$. Since, as I said, this is the more prevalent case in general, this is what a lot of chemists have in mind.

The tetrahedral point group $T_\mathrm{d}$ does not contain a centre of symmetry. This is easily seen when looking at a tetrahedron, which is not inversion-symmetric. Thus, the irreducible representations do not gain a subscripted g and u. The d-orbitals thus transform as $\mathrm{e + t_2}$. Adding a g to these is wrong, but common (see above).

Thus, you were correct in assuming: ‘just because everyone does it does not mean it is correct.’ The same thing can be said for $\ce{S=O}$ double bonds in sulfate.