Consider the QED action:
$$
S = \int d^4x \, \left[ - \frac{1}{4}F^{\mu\nu}F_{\mu\nu} + \bar\psi(i\gamma^\mu\partial_\mu - m)\psi + ej^\mu A_\mu \right]
$$
We need the action to be dimensionless, and $d^4x$ has dimension $-4$ (i.e. $[d^4x] = -4$). Thus, the terms in the Lagrangian density must have mass dimension $4$. Expanding the electromagnetic tensor as $F^{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ and $[\partial] = 1$, we see that $[F^{\mu\nu} F_{\mu\nu}] = 2[\partial] + 2[A_\mu] = 4$ implies that $[A_\mu] = 1$. Doing the same exercise for the term with the spinor field and derivative, we find $[\psi] = 3/2$. Then, from $[\bar\psi m \psi] = 2[\psi] + [m] = 4$, we get $[m] = 1$. Therefore, $m$ has mass dimension $1$ (thus, we call it the mass term). Finally, since $j^\mu = \bar\psi\gamma^\mu\psi$, we get $[e] = 0$.
Therefore, this theory has a dimensionless parameter $e$ and a dimensionful parameter $m$. Now, under the renormalization group (RG), the fact that $m$ has dimension $1$ and $[m] = 1 < 4 = d$ means that the coupling is relevant, i.e., it grows in the low-energy regime. Thus, at low-energies, particles become more massive and non-relativistic, while at high-energies they become massless and relativistic. Since $[e] = 0$, this coupling is marginal under RG, meaning that we need to compute its beta function to understand how it behaves at different energy scales. At one loop order, we have
$$
\beta(e) = \frac{e³}{12\pi^2}
$$
Which means that $e$ increases with the energy scale, introducing another scale dependence to the theory.
