A Hilbert module defined in "L^2-invariants: theory and applications to geometry and K-theory", Springer-Verlag, 2002, by W. Lück:

A Hilbert $\mathcal N(G)$-module $V$ is a Hilbert space $V$ together with a linear isometric $G$-action such that there exists a Hilbert space $H$ and an isometric linear $G$-embedding of $V$ into the tensor product of Hilbert spaces $H\bar\otimes\ell^2(G)$ with the obvious $G$-action.

($\mathcal N(G)$ is the group von Neumann algebra of a group $G$).

After that the notion of von Neumann trace defined as:

Let $f:V\rightarrow V$ be a positive endomorphism of a Hilbert $\mathcal N(G)$-module. Choose a Hilbert space $H$, a Hilbert basis $\{b_i: i\in I\}$ for $H$, a $G$-equivariant projection $\text{pr}:H\otimes\ell^2(G)\rightarrow H\otimes\ell^2(G)$ and an isometric $G$-isomorphism $u:\text{im(pr)}\xrightarrow{\cong}V$. Let $f:H\otimes\ell^2(G)\rightarrow H\otimes\ell^2(G)$ be the positive operator given by the composition $$ \bar f:H\otimes\ell^2(G)\xrightarrow{\text{pr}}\text{im(pr)}\xrightarrow{u}V\xrightarrow{f}V\xrightarrow{u^{-1}}\text{im(pr)}\hookrightarrow H\otimes\ell^2(G)$$ Define the von Neumann trace of $f:V\rightarrow V$ by

$$ \text{tr}_{\mathcal N(G)}(f):=\sum_{i\in I}\langle f(b_i\otimes e), b_i\otimes e\rangle \quad\in[0,\infty]; $$ where $e\in G\subset\ell^2(G)$ is the unit element.

The "positive" means "positive operator" in the sense of Hilbert spaces, and "endomorphism" mean that $f$ needs to commute with the $\mathcal N(G)$-action.

The author claims after this definition that

This definition is independent of the choices of $H$, $\{b_i: i\in I\}$, $\text{pr}$ and $u$.

But only the independence of the choice $\{b_i: i\in I\}$ was proved. I could not prove the independence of the choice of $H$ in the definition above. What is the main idea to prove this claim?