[proofplan] We use the definition of an infinite projection to find a proper subprojection $q<p$ with $q\sim p$. [Trace invariance under Murray-von Neumann equivalence](/theorems/9317) gives $\tau(q)=\tau(p)$, while finite additivity of the trace decomposes $\tau(p)$ as $\tau(q)+\tau(p-q)$. Since $p-q$ is a nonzero projection, faithfulness forces $\tau(p-q)>0$, so the equality cannot hold if $\tau(p)$ is finite. [/proofplan] [step:Choose a proper equivalent subprojection of $p$] Since $p$ is infinite, there exists a projection $q\in\mathcal P(M)$ such that $q<p$ and $q\sim p$ in the Murray-von Neumann sense. Here $q<p$ means $q\le p$ and $q\ne p$. Define \begin{align*} r:=p-q. \end{align*} Because $q\le p$, the operator $r$ is a projection in $M$, and $q r=0=r q$. Since $q\ne p$, we have $r\ne 0$. [/step] [step:Use trace invariance and additivity on the orthogonal decomposition] The trace $\tau$ is faithful, normal, and semifinite by hypothesis, and $p,q\in\mathcal P(M)$ satisfy $q\sim p$. Therefore [citetheorem:9317] gives \begin{align*} \tau(q)=\tau(p). \end{align*} Also $p=q+r$ with $q,r\in M_+$ and $q r=0=r q$. By finite additivity of the trace on positive elements, \begin{align*} \tau(p)=\tau(q)+\tau(r). \end{align*} Combining the two equalities gives \begin{align*} \tau(p)=\tau(p)+\tau(r). \end{align*} [guided] The point of choosing $q$ is that it gives two incompatible pieces of information about the trace of $p$. First, because $q\sim p$, the projection $q$ has the same trace as $p$. The hypotheses of [citetheorem:9317] are satisfied: $M$ is equipped with a faithful normal semifinite trace $\tau$, and $p,q$ are projections in $M$ with $q\sim p$. Hence \begin{align*} \tau(q)=\tau(p). \end{align*} Second, because $q$ is a subprojection of $p$, the difference \begin{align*} r:=p-q \end{align*} is a projection in $M$. The equality $q\le p$ implies $p q=q=q p$, so \begin{align*} q r=q(p-q)=q-q=0 \end{align*} and similarly $r q=0$. Thus $p$ is the orthogonal sum of the two positive elements $q$ and $r$: \begin{align*} p=q+r. \end{align*} Finite additivity in the definition of a trace gives \begin{align*} \tau(p)=\tau(q)+\tau(r). \end{align*} Substituting the trace-invariance identity $\tau(q)=\tau(p)$ yields \begin{align*} \tau(p)=\tau(p)+\tau(r). \end{align*} This is the equality that will force $\tau(p)$ to be infinite, because $r$ is a nonzero positive projection. [/guided] [/step] [step:Apply faithfulness to rule out finite trace] Since $r\ne 0$ and $r\in M_+$, faithfulness of $\tau$ gives \begin{align*} \tau(r)>0. \end{align*} Suppose for contradiction that $\tau(p)<\infty$. Then the equality \begin{align*} \tau(p)=\tau(p)+\tau(r) \end{align*} is impossible in $[0,\infty]$, because adding the strictly positive extended-real number $\tau(r)$ to the finite number $\tau(p)$ gives a strictly larger value, or gives $\infty$. Therefore $\tau(p)$ cannot be finite. Since $\tau(p)\in[0,\infty]$, it follows that \begin{align*} \tau(p)=\infty. \end{align*} [/step]