Do irreducible characters form a closed set? A character on a discrete group $\Gamma$ is a conjugation-invariant function $\tau$ which is of positive
type, and is normalized so that $\tau(e) = 1$, where $e$ is the identity element of $\Gamma$. A character $\tau$ is  irreducible if it cannot be represented as $\tau=a\tau_1+b\tau_2$ for some $a,b>0$ and some characters $\tau_1\ne \tau_2$. 

Is it true that irreducible characters of a discrete group form a closed set with respect to pointwise
  convergence?

 A: No. Here's an argument that shows the answer is no for non-abelian, torsion free, finitely generated  two-step nilpotent groups (this argument could easily be pushed to nilpotent groups of steps longer than 2).  
Let $G$ be such a group. Let $\lambda\in \widehat{Z(G)}$ (the dual group of $Z(G)$) be a group homomorphism
$\lambda:Z(G)\rightarrow \mathbb{T}=\{ z\in \mathbb{C}:|z|=1 \}$  with trivial kernel. Define $t_\lambda:G\rightarrow \mathbb{C}$ by $t_\lambda(x)=\lambda(x)$ if $x\in Z(G)$ and $t_\lambda(x)=0$ if $x\not\in Z(G).$
It's clear that $t_\lambda$ is a character.  It's not clear that $t_\lambda$ is actually an irreducible character.  But, it follows from  Howe's paper [On representations of discrete, finitely generated, torsion-free, nilpotent groups. Pacific J. Math. 73 (1977), no. 2, 281–305.] that $t_\lambda$ actually is an irreducible character on $G.$
Since $\widehat{Z(G)}\cong \mathbb{T}^d$ for some $d\geq1$, there is a sequence of homomorphisms $\lambda_n:Z(G)\rightarrow \mathbb{T}$ with trivial kernel that converge in $\widehat{Z(G)}$ to the trivial character  on $Z(G).$ Again by Howe's result each $t_{\lambda_n}$ is an irreducible character on $G.$
It follows that $t_{\lambda_n}$ converge pointwise to the character $t:G \rightarrow \mathbb{C}$ defined by $t(x)=1$ if $x\in Z(G)$ and $t(x)=0$ if $x\not\in Z(G).$ The function $t$ is not an irreducible character on $G.$  One way to see this is that the GNS representation associated with $t$ is just the left regular representation of $G$ on $\ell^2(G/Z(G)).$  Since $G$ is non-abelian and two step, the group $G/Z(G)$ is a non-trivial free abelian group.  Therefore the fact that $t$ is not irreducible  boils down to the fact that Haar measure on $\widehat{G/Z(G)}$ is not a point mass.
