Lat $G$ be a locally compact group and $\mathcal{L}(G)$ its von Neumann algebra equipped with the Plancherel weight $\omega_G$.
If $G$ is discrete then $\omega_G$ is a finite trace.
Suppose that $\omega_G$ satisfies $\omega_G(1)<\infty$. Is $G$ discrete ?
I know I'm a bit late, but maybe someone will be interested in this question.
The answer is yes, here is an argument.
If $\omega_G(1)<+\infty$ then we can rescale it so that it is a normal state on $\mathcal{L}(G)$. Since $\mathcal{L}(G)\subseteq B(L^2(G))$ is represented in the standard way, there is a unit vector $\xi\in L^2(G)$ such that $\omega_G(x)=\langle \xi | x \xi \rangle$. Left invariance implies that $\omega_G(\lambda_g)=\delta_{g,e}$. Indeed, we have $\omega_G(\lambda_g)\lambda_g=(id\otimes \omega_G)\Delta(\lambda_g)=\omega_G(\lambda_g)1$ for all $g\in G$, where $\Delta$ is the comultiplication on $\mathcal{L}(G)$. Consequently $\langle \xi | \lambda_g \xi\rangle =\delta_{g,e}$. As the function $G\ni g \mapsto \langle \xi | \lambda_g \xi \rangle\in \mathbb{C}$ is continuous, set $A=\{g\in G\,|\,|1-\langle \xi|\lambda_g \xi\rangle|<1/2\}=\{e\}$ is open, hence $G$ is discrete.