For such an $M$, set $\lambda^{\prime}(M) := \dim_{\mathbf{F}_p} M/pM - \dim_{\mathbf{F}_p} M[p]$; since $\mu(M) = 0$, the dimensions are actually finite---this can be extracted from the argument below. Also, if $M$ is itself finite, then $\lambda^{\prime}(M) = 0$.

Since $\mu(M) = 0$, due to the structure theorem, there is a $p$-torsion-free $\mathbf{Z}_p[[T]]$-module $X$ and a $\mathbf{Z}_p[[T]]$-homomorphism $f\colon M \rightarrow X$ with finite kernel and cokernel. Since $\lambda(M) = \lambda(X) = \lambda^{\prime}(X)$ (the latter because $X$ is $\mathbf{Z}_p$-free), to show the desired $\lambda(M) = \lambda^{\prime}(M)$, it remains to see observe that a repeated application of the snake lemma gives $\lambda^{\prime}(\mathrm{Ker} f) + \lambda^{\prime}(X) = \lambda^{\prime}(M) + \lambda^{\prime}(\mathrm{Coker} f)$, whereas $\lambda^{\prime}(\mathrm{Ker} f) = \lambda^{\prime}(\mathrm{Coker} f) = 0$ due to the finiteness of $\mathrm{Ker} f$ and $\mathrm{Coker} f$.

For such an $M$, set $\lambda^{\prime}(M) := \dim_{\mathbf{F}_p} M/pM - \dim_{\mathbf{F}_p} M[p]$; since $\mu(M) = 0$, the dimensions are actually finite---this can be extracted from the argument below. Also, if $M$ is itself finite, then $\lambda^{\prime}(M) = 0$.

Since $\mu(M) = 0$, due to the structure theorem, there is a $p$-torsion-free $\mathbf{Z}_p[[T]]$-module $X$ and a $\mathbf{Z}_p[[T]]$-homomorphism $f\colon M \rightarrow X$ with finite kernel and cokernel. Since $\lambda(M) = \lambda(X) = \lambda^{\prime}(X)$ (the latter because $X$ is $\mathbf{Z}_p$-free), to show the desired $\lambda(M) = \lambda^{\prime}(M)$, it remains to observe that a repeated application of the snake lemma gives $\lambda^{\prime}(\mathrm{Ker} f) + \lambda^{\prime}(X) = \lambda^{\prime}(M) + \lambda^{\prime}(\mathrm{Coker} f)$, whereas $\lambda^{\prime}(\mathrm{Ker} f) = \lambda^{\prime}(\mathrm{Coker} f) = 0$ due to the finiteness of $\mathrm{Ker} f$ and $\mathrm{Coker} f$.