[proofplan]
We argue by contradiction. If a faithful normal semifinite trace $\tau$ existed, semifiniteness applied to the unit would produce a nonzero positive contraction $x$ with finite trace. Spectral calculus then extracts a nonzero projection $p$ below a scalar multiple of $x$, so positivity gives $\tau(p)<\infty$. Since $p$ is infinite by hypothesis, this contradicts the theorem that infinite projections have infinite faithful normal semifinite trace.
[/proofplan]
[step:Assume a faithful normal semifinite trace and extract a finite trace positive element]
Suppose, for contradiction, that $\tau:M_+\to[0,\infty]$ is a faithful normal semifinite trace on $M$, where $M_+:=\{a\in M:a\ge 0\}$ is the positive cone of $M$. Since $M$ is nonzero, its identity projection $1_M$ is nonzero. By semifiniteness of $\tau$, applied to the nonzero positive element $1_M\in M_+$, there exists an element $x\in M_+$ such that
\begin{align*}
0\le x\le 1_M
\end{align*}
and
\begin{align*}
x\ne 0,\qquad \tau(x)<\infty.
\end{align*}
[/step]
[step:Use spectral calculus to find a nonzero spectral projection below $x$]
Let $\mathcal{B}([0,1])$ denote the Borel $\sigma$-algebra of $[0,1]$, and let $E_x:\mathcal{B}([0,1])\to \mathcal{P}(M)$ denote the spectral measure of the positive contraction $x$, where $\mathcal{P}(M)$ is the projection lattice of $M$. For each $n\in\mathbb{N}$, define the spectral projection
\begin{align*}
p_n:=E_x([1/n,1])=\mathbb{1}_{[1/n,1]}(x)\in \mathcal{P}(M).
\end{align*}
The projections $(p_n)_{n\in\mathbb{N}}$ increase to the support projection $s(x)$. Since $x\ne 0$, its support projection is nonzero, so there exists $n_0\in\mathbb{N}$ such that $p_{n_0}\ne 0$. Define
\begin{align*}
\varepsilon:=1/n_0,\qquad p:=p_{n_0}.
\end{align*}
Then $p$ is a nonzero projection in $M$. By functional calculus for the scalar inequality $\varepsilon\mathbb{1}_{[\varepsilon,1]}(t)\le t$ on $[0,1]$, we have
\begin{align*}
\varepsilon p\le x.
\end{align*}
[guided]
The purpose of this step is to pass from a finite trace positive element to a finite trace projection. A trace on positive elements is easiest to compare on projections, so we use the spectral theorem for the positive contraction $x$.
Let $\mathcal{B}([0,1])$ denote the Borel $\sigma$-algebra of $[0,1]$, and let $\mathcal{P}(M)$ denote the projection lattice of $M$. Let
\begin{align*}
E_x:\mathcal{B}([0,1])\to \mathcal{P}(M)
\end{align*}
be the spectral measure of $x$. This is defined on the Borel subsets of $[0,1]$ because $0\le x\le 1_M$, so the spectrum of $x$ is contained in $[0,1]$. For each $n\in\mathbb{N}$, set
\begin{align*}
p_n:=E_x([1/n,1])=\mathbb{1}_{[1/n,1]}(x).
\end{align*}
Each $p_n$ is a projection in $M$ by the [Borel functional calculus](/theorems/2696) for von Neumann algebras.
We now check that at least one of these projections is nonzero. The projections $p_n$ increase to the support projection $s(x)=\mathbb{1}_{(0,1]}(x)$. If every $p_n$ were zero, then their supremum would be zero, so $s(x)=0$. The equality $s(x)=0$ would force $x=xs(x)=0$, contradicting the choice of $x\ne 0$. Hence there is some $n_0\in\mathbb{N}$ such that $p_{n_0}\ne 0$.
Define
\begin{align*}
\varepsilon:=1/n_0,\qquad p:=p_{n_0}.
\end{align*}
Then $p$ is a nonzero projection in $M$. Finally, the scalar inequality
\begin{align*}
\varepsilon\mathbb{1}_{[\varepsilon,1]}(t)\le t
\end{align*}
holds for every $t\in[0,1]$: it is equality-bounded by $t$ on $[\varepsilon,1]$ and is $0\le t$ outside that interval. Applying functional calculus to this scalar inequality gives
\begin{align*}
\varepsilon p\le x.
\end{align*}
[/guided]
[/step]
[step:Bound the trace of the spectral projection]
Since $\tau$ is positive and $\varepsilon p\le x$, monotonicity of the positive trace gives
\begin{align*}
\tau(\varepsilon p)\le \tau(x).
\end{align*}
By positive homogeneity of $\tau$ on $M_+$,
\begin{align*}
\varepsilon\tau(p)\le \tau(x)<\infty.
\end{align*}
Because $\varepsilon>0$, it follows that
\begin{align*}
\tau(p)\le \varepsilon^{-1}\tau(x)<\infty.
\end{align*}
[/step]
[step:Contradict infiniteness of the nonzero projection]
The projection $p$ is nonzero, so by hypothesis $p$ is infinite in $M$. Since $\tau$ is a faithful normal semifinite trace on $M$, the theorem that infinite projections have infinite faithful trace [citetheorem:9299] implies
\begin{align*}
\tau(p)=\infty.
\end{align*}
This contradicts the finite bound $\tau(p)<\infty$ obtained above. Therefore no faithful normal semifinite trace on $M$ exists.
[/step]
