be a Følner sequence of $G$. Then $$\dim^{{\operatorname{Elek}}}_{kG}(M)=\lim_{n\to\infty}\frac{\dim_k\left(M[F_n]\right)}{|F_n|}.$$

The map $f$ is given by right multiplication with a matrix $A\in
  M_{r,s}(kG)$. Viewing $A$ as a map $G\to k^{r\times s}$ it is clear what we mean by the support ${\operatorname{supp}}(A)$ of $A$. Let $R >0$ be the diameter of ${\operatorname{supp}}(A)\cup{\operatorname{supp}}(A)^{-1}$. Since $$\lim_{n \to \infty} \frac{\bigl|\partial_R
    F_n\bigr|}{\bigl|F_n\bigr|} = 0,$$ it is enough to show that for every $n\ge 1$ $$\label{eq:estimate_1}
    \bigl|\dim_k(M[F_n])-\dim_k\bigl({\operatorname{im}}(p^\ast)\vert_{F_n}\bigr)\bigr| \le s\cdot \bigl|\partial_R F_n\bigr|.$$

For the definition of inner products $(\_,\_)$ and $\langle\_,\_\rangle$ we refer to Remark \[rem:Pontrjagin\_duality\]. Define the following $k$-linear subspaces of ${\operatorname{map}}(F_n,k^s)$: $$\begin{aligned}
    W_n &= \bigl\{\phi\colon F_n\to k^s\mid
    \langle{\operatorname{pr}}_n\circ f \circ j_n(x),\phi\rangle=0~\forall x\in k[F_n]^r\bigr\};\\
    V_n &= \bigl\{\phi\colon F_n\to k^s \mid \exists \bar\phi:G\to k^s
    \; \text{satisfying}\;
    \bar\phi\vert_{F_n}=\phi, \langle f(y), \bar\phi\rangle=0~\forall y\in kG^r\bigr\};\\
    Z_n &=\bigl\{\phi\colon F_n\to k^s \mid \phi\vert_{\partial_R
      F_n}=0\bigr\}.
  \end{aligned}$$

Since $\dim_k(M[F_n]) = \dim_k(W_n)$ and $\dim_k\bigl({\operatorname{im}}(p^\ast)\vert_{F_n}\bigr) = \dim_k(V_n)$, the desired estimate  is equivalent to $$\label{eq:estimate_2}
    \bigl|\dim_k(W_n)-\dim_k(V_n)\bigr| \le s\cdot \bigl|\partial_R F_n\bigr|.$$

By additivity of $\dim_k$ we obtain that $$\begin{aligned}
    \dim_k (W_n\cap Z_n)&\ge
    \dim_k(W_n)-\dim_k({\operatorname{map}}(F_n,k^s))+\dim_k(Z_n)\\
    &\ge\dim_k(W_n)-s \cdot |F_n| + s \cdot (|F_n|-|\partial_R F_n|)\\
    &=\dim_k(W_n)-s \cdot |\partial_R F_n|.
  \end{aligned}$$ Similarly, we get $$\dim_k (V_n\cap Z_n)\ge \dim_k(V_n)-s \cdot |\partial_R F_n|.$$ To prove  it hence suffices to show that $$\begin{aligned}
    W_n\cap Z_n&\subset V_n; \label{eq:first_inclusion}\\
    V_n\cap 