[proofplan]
We argue by contradiction. A faithful semifinite trace would produce a nonzero positive finite-trace contraction, and the spectral theorem then gives a nonzero projection of finite trace. Such a projection is finite, because [trace invariance under Murray-von Neumann equivalence](/theorems/9317) prevents it from being equivalent to a proper subprojection. This contradicts the characterization of type III factors as factors with no nonzero finite projections.
[/proofplan]
[step:Assume a faithful normal semifinite trace and extract a nonzero finite-trace positive contraction]
Suppose, for contradiction, that $\tau:M_+\to[0,\infty]$ is a faithful normal semifinite trace on $M$.
Since $1\in M_+$ is nonzero and $\tau$ is faithful, $\tau(1)>0$. By semifiniteness of $\tau$ applied to $1$, there exists an element $y\in M_+$ such that $0\le y\le 1$, $\tau(y)<\infty$, and $y\ne 0$. Since $\tau$ is faithful and $y\ne 0$, we have $\tau(y)>0$. Set $x:=y$. Then $0\le x\le 1$, $x\ne 0$, and $0<\tau(x)<\infty$.
[guided]
We begin by assuming the negation of the theorem: there is a faithful normal semifinite trace
$\tau:M_+\to[0,\infty]$ on $M$.
The first goal is not yet to find a projection. Semifiniteness gives finite-trace positive elements below a given positive element, so we apply it to the identity $1\in M_+$. Because $M$ is a factor, it is nonzero, and therefore $1\ne 0$. Faithfulness of $\tau$ says that a positive element has trace zero only when it is zero, so $\tau(1)>0$.
Semifiniteness means that the trace of a positive element is detected from below by finite-trace positive elements. Applied to $1$, it gives a positive element $y\in M_+$ with $0\le y\le 1$ and $\tau(y)<\infty$, and with $y\ne 0$; otherwise all finite-trace positive elements below $1$ would be zero and could not detect the positive value of $\tau(1)$. Since $y\ne 0$ and $\tau$ is faithful, $\tau(y)>0$. Defining $x:=y$, we have obtained a nonzero finite-trace positive contraction: $0\le x\le 1$, $x\ne 0$, and $0<\tau(x)<\infty$.
This is the bridge from semifiniteness to projection theory.
[/guided]
[/step]
[step:Use a spectral projection of $x$ to obtain a nonzero finite-trace projection]
Let $E_x$ denote the spectral measure of the positive operator $x$. For each $n\in\mathbb N$, define $e_n:=E_x([1/n,1])\in M$. Since $0\le x\le 1$, the projection $E_x((0,1])$ is the support projection of $x$. If $e_n=0$ for every $n\in\mathbb N$, then $E_x((0,1])=\bigvee_{n=1}^{\infty} E_x([1/n,1])=0$, and hence $x=0$, contradicting the choice of $x$. Therefore there exists $n_0\in\mathbb N$ such that $e:=e_{n_0}\ne 0$.
Set $\varepsilon:=1/n_0$. By the functional calculus for $x$, $\varepsilon e\le x$. Positivity and homogeneity of the trace give $\varepsilon\,\tau(e)=\tau(\varepsilon e)\le \tau(x)<\infty$. Since $\varepsilon>0$, it follows that $0<\tau(e)\le \varepsilon^{-1}\tau(x)<\infty$. Thus $e\in M$ is a nonzero projection with finite trace.
[/step]
[step:Show that the finite-trace projection is finite]
We prove that $e$ is finite in $M$. Suppose instead that $e$ is infinite. Then there is a projection $f\in M$ with $f<e$ and $f\sim e$ in the Murray-von Neumann sense. Thus there exists a partial isometry $v\in M$ such that
\begin{align*}
v^*v=e,\qquad vv^*=f.
\end{align*}
Trace invariance under Murray-von Neumann equivalence, as in [citetheorem:9317], gives
\begin{align*}
\tau(e)=\tau(f).
\end{align*}
Because $f\le e$, the projection $e-f$ belongs to $M$, is orthogonal to $f$, and is nonzero since $f<e$. Finite additivity of the trace on orthogonal positive elements gives
\begin{align*}
\tau(e)=\tau(f)+\tau(e-f).
\end{align*}
Since $\tau(e)<\infty$ and $\tau(e)=\tau(f)$, subtraction in $[0,\infty)$ gives
\begin{align*}
\tau(e-f)=0.
\end{align*}
Faithfulness of $\tau$ then implies $e-f=0$, contradicting $f<e$. Hence $e$ is a finite projection.
[/step]
[step:Contradict the type III characterization]
We have produced a nonzero finite projection $e\in M$. But by the characterization of type III factors in [citetheorem:9301], every nonzero projection in a type III factor is infinite. This contradicts the finiteness of $e$.
Therefore the assumed faithful normal semifinite trace $\tau$ cannot exist. Hence a type III factor admits no faithful normal semifinite trace.
[/step]
