[proofplan] Murray-von Neumann equivalence means that the two projections are the initial and final projections of a single partial isometry in $M$. We choose such a partial isometry $v$ with $v^*v=p$ and $vv^*=q$. The tracial property of $\tau$ gives $\tau(v^*v)=\tau(vv^*)$ as an equality in $[0,\infty]$, and substituting the two projection identities proves the result. [/proofplan]

[step:Choose a partial isometry implementing the equivalence]Since $p\sim q$ in $M$, by the definition of Murray-von Neumann equivalence there exists a partial isometry $v\in M$ such that \begin{align*} v^*v=p. \end{align*} and \begin{align*} vv^*=q. \end{align*} Here $v^*v$ is the initial projection of $v$, and $vv^*$ is the final projection of $v$.[/step]

[guided]The hypothesis $p\sim q$ is not merely an abstract [equivalence relation](/page/Equivalence%20Relation): in a von Neumann algebra it means that the equivalence is implemented inside the same algebra. Thus there is a partial isometry $v\in M$ whose initial projection is $p$ and whose final projection is $q$. Written algebraically, this means \begin{align*} v^*v=p. \end{align*} and \begin{align*} vv^*=q. \end{align*} This is the only point where Murray-von Neumann equivalence is used. Once this operator $v$ has been chosen, the rest of the proof is an application of the trace identity to the two positive elements $v^*v$ and $vv^*$.[/guided]

[step:Apply the trace identity to the implementing partial isometry] Because $\tau$ is a trace on $M$, it satisfies the tracial identity \begin{align*} \tau(x^*x)=\tau(xx^*) \end{align*} for every $x\in M$, with equality understood in the extended interval $[0,\infty]$. Applying this identity to $x=v$ gives \begin{align*} \tau(v^*v)=\tau(vv^*). \end{align*} Substituting $v^*v=p$ and $vv^*=q$ yields \begin{align*} \tau(p)=\tau(q). \end{align*} This includes the case in which both sides are equal to $\infty$. [/step]