Truncation of BG? Let $G$ be a topological group. In some cases, e.g. when $G$ is discrete or when the spaces $G^n$ are locally contractible and the coefficients are discrete, the cohomology of the classifying space $BG$ is the group cohomology of $G$. So, for simplicity, let us assume that $G$ is discrete.
My question: is there a nice explicit space $B_{\leq k}G$ that is functorial in $G$, such that $H^n(B_{\leq k} G, M) = H^n (BG, M)$ for $n \leq k$ and $0$ for $n>k$? Here $M$ is a $G$-module.
For example, for $G = S^1$, $BG = CP^\infty$ and a possible choice for $B_{\leq 2} S^1$ is $CP^1$. (If it simplifies the question, feel free to assume $M = \mathbf Z$.)
Thank you.
(Edit: In response to Ralph's comment, let us assume $G$ is discrete to simplify, but an answer for non-discrete groups would be interesting, too.)
 A: There are several functorial models for $BG$, see for example [Adem, Milgram: Cohomology of Finite Groups, Chapter II] where 
$$BG = \coprod_{i=0}^\infty \sigma^i \times G^i/(\text{relations})$$
with $\sigma^i=\lbrace (x_1,...,x_i)\mid 0\le x_1 \le \cdots \le x_i \le 1\rbrace$ the standard $i$-simplex.  
Now assume $G$ is discrete. Then $$B_nG := \coprod_{i=0}^n \sigma^i \times G^i/(\text{relations})$$ 
is the n-skeleton of $BG$ and depends functorially on $G$. Since the cohomology in degree $\lt n$ of a CW complex is determined by the $n$-skeleton, we obtain $H^k(B_nG;M)=H^k(BG;M)$ for $0 \le k < n$ and $H^k(B_nG;M)=0$ for $k> n$ and all (local) coefficients $M$. 
In general $H^n(B_nG;M)\neq H^n(BG;M)$, but there is an exact sequence: Write $BG=EG/G$ $(EG$ is described explicitely in Adem-Milgram) and let $E_nG$ be the $n$-skeleton of $EG$. Then the following sequence is exact (Cartan, Eilenberg: Homologica Algebra, XVI, §9, Appl. 1): 
$$0 \to H^n(BG;M)\to H^n(B_nG;M)\to H^n(E_nG;M)^G \to H^{n+1}(BG;M)\to 0$$
Remark: In case that $G$ is not discrete, the above has to be adjusted accordingly. For example, if $G$ is (topologically) a $m$-dimensional CW complex then $B_nG$ is the $n(m+1)$-skeleton of $BG$. 
