For $V \subset \mathbb{Z}^d$ finite, consider $(h_x^V)_{x\in \mathbb{Z}^d}$, a stochastic process indexed by $\mathbb{Z}^d$. $h$ is a discrete Gaussian Free Field with zero boundary condition outside $V$ if its law is given by
$$P(h \in A) = \frac{1}{Z}\int_A e^{-\frac{1}{4d}\sum_{xy \in E}(h_x-h_y)^2} \prod_{x\in V}dh_x \prod_{x \notin V}\delta_0(dh_x)$$
for any measurable $A \subset \mathbb{R}^{\mathbb{Z}^d}$. Here $\delta_0$ is the Dirac point-mass at $0$. $E$ is the edge set consisting edges between any pair of vertices at unit Euclidean distance. $Z$ is the normalisation factor.
It was pointed out in the reading that without the zero boundary condition, the measure won't be well defined since the measure of the whole space evaluates to infinity. ie, we have $$\int_{R^{\mathbb{Z}^d}} e^{-\frac{1}{4d}\sum_{xy \in E}(h_x-h_y)^2} \prod_{x\in \mathbb{Z}^d}dh_x = \infty$$ and $$\int_{R^{\mathbb{Z}^d}} e^{-\frac{1}{4d}\sum_{xy \in E}(h_x-h_y)^2} \prod_{x\in V}dh_x \prod_{x \notin V}\delta_0(dh_x) < \infty$$.
I'm struggling to show that the first integral is infinite and the second is finite. My intuition is that without the zero boundary condition, we can translate $h$ without making an impact on the measure, therefore this makes the measure of the whole space infinite. On the other hand with the zero boundary condition, we are fixing all points outside $V$ to be $0$ and hence we can no longer make the translation freely.