It's related to a homework and exercise. The homework was more complicated, but I needed this to figure out the convention that was used.
Consider the integral $$\int D\phi\exp[-\frac{1}{2} \int d^4 x'\int d^4 x\phi(x')M(x',x)\phi(x)]=\frac{1}{\sqrt{\det M}}$$ in the standard QFT where $\phi$ were fields and $D\phi$ represented the functional integral over the fields, and $M(x',x)$ being the operator of the green's function and with the inverse being the denominator of the propagator. On the right hand side there might possible be a multiple of an infinite number.
The intuition was to use the Gaussian integral formula to integrate the matrix of the form $$\exp[-\frac{1}{2}X^TMX]$$
However, the functional integral was different because there were double integral inside the exponential, and no matter which approach to take, such as the taylor expansion, it was not easily removed.
There was an integration appendix where on page 6, where they wrote $$\int d^4 x\rightarrow \sum_i \Delta v_i$$ with $N=(L/\epsilon)^4$. In this way, $$\int d^4 x'\int d^4 x\phi(x')M(x',x)\phi(x)=\sum_i^N \Delta v_i \sum_j^N \Delta v_j \phi(x_i) M(x_i,x_j) \phi(x_j)$$ could be viewed as a matrix, which then the Gaussian integral apply.
But this also rised issues when taking a closer look. For example, if one view the integral to be of one field, excluded all the possible configuration and in contrast to the definition of $D\phi$, then the degree of freedom of the "vector" $\phi(x_i)$ was hard to explain.
If one adapt the $\phi_i=\phi(x_i)$, the integral was in fact over the field, then there would be a infinite cut off dependent number $$\Pi_{ij}(\Delta v_i \Delta v_j)$$ over the denominator, in contrast to that of a standard gaussian integral.
How to justify $\int D\phi\exp[-\frac{1}{2} \int d^4 x'\int d^4 x\phi(x')M(x',x)\phi(x)]$?
