showing that for a general $\mathcal{N}$, it holds that $$\label{disisfifit}
\mathrm{d}_{\mathcal{M}_\psi}(|T|^p,\mathcal{M}_{\psi,0}')= \limsup_{t\to \infty}  \frac1{\psi(t)}\int_0^t \mu_T(s)^p \mathrm{d}s, \quad\forall T\in \mathcal{M}_\psi^{(p)}$$ for any function $\psi$ additionally satisfying $\lim_{t\to0} \frac{t}{\psi(t)}=0$. Since the original statement is for atomic $\mathcal{N}$, we can always guarantee that this condition holds.

It follows from [@Chi_Suk] that for every $T\in \mathcal{M}_\psi^{(p)}$ there exists a rearrangement-preserving (and thus, isometric) embedding $i_T$ of $\mathcal{M}_\psi^{(p)}(0,\infty)$ into $\mathcal{M}_\psi^{(p)}$ such that $i_T(\mu(T))=T$. Thus, following the argument in [@CRSS Page 267], it is sufficient to prove the formula for every $x=\mu(x) \in \mathcal{M}_\psi^{(p)}(0,\infty)$.

For every $x=\mu(x) \in \mathcal{M}_\psi^{(p)}(0,\infty)$ and every $n\in\mathbb N$ the function $x^p\chi_{(0,n)} \in \mathcal{M}'_{\psi,0}(0,\infty)$. Hence, for every $n\in\mathbb N$ we have $$\mathrm{d}_{\mathcal{M}_\psi(0,\infty)}(x^p,\mathcal{M}'_{\psi,0}(0,\infty))=\mathrm{d}_{\mathcal{M}_\psi(0,\infty)}(x^p\chi_{[n,\infty)},\mathcal{M}'_{\psi,0}(0,\infty)).$$

Therefore, $$\begin{aligned}
\mathrm{d}_{\mathcal{M}_\psi(0,\infty)}(x^p,\mathcal{M}'_{\psi,0}(0,\infty))
&\le\lim_{n\to\infty}\|x^p\chi_{[n,\infty)}\|_{\mathcal{M}_\psi(0,\infty)}\\
&=\lim_{n\to\infty}\sup_{t>0} \frac{1}{\psi(t)} \int_0^t \mu(x^p\chi_{[n,\infty)})(s) \ \mathrm{d}s\\
&=\lim_{n\to\infty}\sup_{t>0} \frac{1}{\psi(t)} \int_0^t (x(s+n))^p \ \mathrm{d}s\\
&=\lim_{n\to\infty}\sup_{t>0} \frac{1}{\psi(t)} \int_n^{t+n} (x(s))^p \ \mathrm{d}s.\end{aligned}$$

By the definition of supremum for every $n\in {{\mathbb N}}$ there exists $t_n>0$ such that $$\sup_{t>0} \frac{1}{\psi(t)} \int_n^{t+n} (x(s))^p \ \mathrm{d}s \le \frac{1}{\psi(t_n)} \int_n^{t_n+n} (x(s))^p \ \mathrm{d}s +\frac1n.$$

Denote for brevity $$a:=\limsup_{t\to \infty}  \frac1{\psi(t)}\int_0^t (x(s))^p \mathrm{d}s.$$

1\. If $\limsup_{n\to\infty} t_n=\infty$, then $$\mathrm{d}_{\mathcal{M}_\psi(0,\infty)}(x^p,\mathcal{M}'_{\psi,0}(0,\infty))\le \lim_{n\to\infty}\sup_{t>n} \frac{1}{\psi(t)} \int_n^{t+n} (x(s))^p \ \mathrm{d}s\le a,$$ since $x=\mu(x)$.

2\. If $0 < \liminf_{n\to\infty} t_n \le \limsup_{n\to\infty} t_n<\infty$, then $$\frac{1}{\psi(t_n)} \int_n^{t_n+n} (x(s))^p \ \mathrm{d}s \le \frac{t_n 