In the book Quantum Field Theory, Ryder defines the vacuum-to-vacuum transition amplitude in the presence of the source $J(x)$ as $$Z[J] = \int \mathcal{D}\phi \, \exp\left\lbrace i\int \, d^4x \, \left[\mathcal{L}(\phi) + J(x) \phi(x) + \frac{i}{2} \epsilon\phi^2 \right]\right\rbrace \propto \langle0, \infty|0,-\infty\rangle.\tag{6.1}$$
This is Eq. (6.1) of the book.
In discrete form, $$Z[J] = \lim_{N \to \infty} \int \prod_{n=1}^{N^4} d\phi_n \, \exp\left\lbrace i\sum_{n=1}^{N^4} \delta^4 \left( \mathcal{L}_n + \phi_nJ_n + \frac{i}{2} \epsilon \phi_n^2 \right) \right\rbrace,\tag{6.2}$$ where $n$ stands for the four indices $(i, j, k, l)$.
Question
What does $d\phi_n$ mean?
I think that $$d\phi_n = \frac{\partial\phi_n}{\partial x_i}dx_i + \frac{\partial\phi_n}{\partial y_j}dy_j + \frac{\partial\phi_n}{\partial z_k}dz_k + \frac{\partial\phi_n}{\partial t_l}dt_l, $$ where $\phi(x_i, y_j, z_k, t_l) \equiv \phi_n$ and, the other symbols have their usual meanings. Is this correct?