Lemma 44.3.7.label Let $H$ be a complex Hilbert space, $(M, \tau)$ be a von Neumann algebra on $H$, and $x: D(x) \to H$ be a densely defined closable operator affiliated with $M$, then the closure of $x$ is also affiliated with $M$.
Proof. Let $\Gamma(x) = \bracsn{(\xi, x\xi)|\xi \in D(x)}$ be the graph of $x$, $\ol{x}$ be the closure of $x$, and $\Gamma(\ol{x})$ be its graph. For each $u \in U(M')$,
\begin{align*}\Gamma(u \ol{x})&= \bracsn{(\xi, u\ol x\xi)|\xi \in D(\ol x)}= \ol{\bracsn{(\xi, ux\xi)|\xi \in D(x)}}\\&= \ol{\bracsn{(\xi, xu\xi)|\xi \in D(x)}}={\bracsn{(\xi, \ol xu\xi)|\xi \in D(\ol x)}}= \Gamma(\ol x u)\end{align*}
$\square$
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