has $$\int_0^\infty G(s)^{1+h} \mathrm{d}s \le \int_0^\infty (C \psi'(s))^{1+h} \mathrm{d}s.$$ Since the function $\psi$ satisfies , it follows that $$\int_0^\infty G(s)^{1+h} \mathrm{d}s \le C^{1+h} (\psi(\mathrm{e}^{\frac{1}{h}}))^{1+h},$$ or, equivalently, $$\label{G}
\limsup_{h\searrow0} \frac{\|G\|_{1+h}}{\psi(\mathrm{e}^{\frac{1}{h}})} \le C.$$

First, $$\frac{1}{\psi(\mathrm{e}^{\frac{1}{h}})} \left(\int_0^{t_0} G(s)^{1+h} \mathrm{d}s\right)^{\frac{1}{1+h}} \le \frac{1}{\psi(\mathrm{e}^{\frac{1}{h}})} \left(\int_0^{t_0} (C \psi'(s))^{1+h} \mathrm{d}s\right)^{\frac{1}{1+h}} \mathop{\longrightarrow}\limits_{h\searrow0} 0,$$ since $\psi(\infty)=\infty$ and $\psi'\in L^{1+h}(0,\infty)$ for every $h>0$.

Second, by Lebesque Monotone Convergence Theorem and  we obtain $$\left(\int_0^{t_0} H(s)^{1+h} \mathrm{d}s\right)^{\frac{1}{1+h}} \mathop{\longrightarrow}\limits_{h\searrow0}\int_0^{t_0} H(s) \mathrm{d}s \le C \psi(t_0).$$

Therefore, $$\limsup_{h\searrow0} \frac{1}{\psi(\mathrm{e}^{\frac{1}{h}})} \left(\int_0^{t_0} G(s)^{1+h} \mathrm{d}s\right)^{\frac{1}{1+h}} = \limsup_{h\searrow0}\frac{1}{\psi(\mathrm{e}^{\frac{1}{h}})} \left(\int_0^{t_0} H(s)^{1+h} \mathrm{d}s\right)^{\frac{1}{1+h}}=0.$$

Since $H(t)=G(t)$ for $t\ge t_0$, it follows from  that $$\limsup_{h\searrow0} \frac{\|H\|_{1+h}}{\psi(\mathrm{e}^{\frac{1}{h}})}\le C.$$ This proves that $$\|H\|_{1,\limsup} \le  \|H\|_{1,\lim \psi}.$$

The following result is well-known at least in the commutative setting (see e.g. [@DPSSS Proposition 2.1]. For the convenience of the reader we provide a short proof.

\[prop7.5\] Let $\psi: [0,\infty)\to [0,\infty)$ be an increasing concave function satisfying the conditions and and moreover that $\psi(0)=0$, $\lim_{t\to\infty} \psi(t)=\infty$. Assume that $\mathcal{N}$ is atomic. For any $T\in \mathcal{M}_\psi^{(p)}$, we have that $$\mathrm{d}_{\mathcal{M}_\psi}(|T|^p,\mathcal{M}_{\psi,0})= \limsup_{t\to \infty}  \frac1{\psi(t)}\int_0^t \mu_T(s)^p \mathrm{d}s.$$

Let $\mathcal{M}'_{\psi,0}$ denote the norm closure of the space of elements $T\in \mathcal{M}_\psi$ with compactly supported singular value function. The assumption that $\mathcal{N}$ is atomic ensures that $\mathcal{M}_{\psi,0}=\mathcal{M}_{\psi,0}'$. Our proof will in fact consist of 