Grand-canonical Gibbs measure for continuous systems

Looking at your formula (1), it appears that $\mu$ must be a measure defined on a $\sigma$-algebra $\mathscr F$ over the finite set $\Lambda$. The natural $\sigma$-algebra over the finite set $\Lambda$ is the largest $\sigma$-algebra over $\Lambda$, which is the (power) set $2^\Lambda$ of all subsets of $\Lambda$. By definition, the product measure $\mu^{\otimes N}$ is defined on the product $\sigma$-algebra $\mathscr F^{\otimes N}$. If $\mathscr F=2^\Lambda$, then $\mathscr F^{\otimes N}=(2^\Lambda)^{\otimes N}=2^{\Lambda^N}$, the set of all subsets of $\Lambda^N$.

Your formula (1) can then be rewritten simply as $$ \Xi_{\Lambda}(\beta, z) := \sum_{N=0}^{\infty}\frac{z^{N}}{N!} \int_{\Lambda^N}e^{-\beta U_{N}(x_{1},...,x_{N})}\,\mu^{\otimes N}\Big(\prod_{j=1}^N dx_j\Big) \\ =\sum_{N=0}^{\infty}\frac{z^{N}}{N!} \sum_{(x_{1},...,x_{N})\in\Lambda^N}e^{-\beta U_{N}(x_{1},...,x_{N})}\,\prod_{j=1}^N \mu(\{x_j\}). $$

The configuration space is the disjoint union of $\Lambda^N$ for each nonnegative integer $N$. You can take the Borel $\sigma$-algebra on each of these (or Lebesgue if you prefer, but you're unlikely to encounter non-Borel sets in real life).