I am bearing seeking for a sequential decomposition of the norm in Orlicz space. 
Let me state what is known in the particular case of Lebesgue space $L^p(\Bbb R^d)$. 

Given $u\in L^p(\Bbb R^d)$ let
 $$n\in \Bbb Z,\qquad  D_n= \{x\in \Bbb R^d: 2^n \leq |u(x)|<2^{n+1}\},\quad \text{put}\quad d_n= |D_n|= \int_{D_n}d x $$

Then it is not difficult to show that 
\begin{align}\tag{1}\label{eq1}\|u\|_{L^p(\Bbb R^d)}=  \Big(\int_{\Bbb R^d} |u(x)|^pd x\Big)^{1/p}\sim \Big(\sum_{n\in \Bbb Z} 2^{np}d_n\Big)^{1/p}\end{align}


Indeed, we have 
 $$\frac12 \|u\|_{L^p(\Bbb R^d)}\leq \Big(\sum_{n\in \Bbb Z} 2^{np}d_n\Big)^{1/p}\leq \|u\|_{L^p(\Bbb R^d)}.$$

**Question** is there a way to get a decomposition similar to \eqref{eq1} for an Orlicz space $L^\phi(\Bbb R^d)$? Where $\phi$ is  a sufficiently nice Young function, e.g., $\phi$ is continuous, increasing, convex and in addition the mapping $x\mapsto \frac{\phi(x)}{x}$, $x>0$ is increasing and satisfies 

\begin{align*}
&\lim_{x\to 0^+}\frac{\phi(x)}{x}= \lim_{x\to \infty}\frac{x}{\phi(x)}= 0.
\end{align*} 

Recall that: 
 \begin{align*}
 L^\phi(\Bbb R^d)&= \Big\{u: \Bbb R^d\to \Bbb R\text{ meas.}:~ \int_{\Bbb R^d}  \phi\Big(\frac{|u(x)|}{\lambda}\Big)d x<\infty ~~\text{for some $\lambda>0$}\Big\}.
 \end{align*}
The space $L^\phi(\Bbb R^d)$ is equipped with the Luxemburg norm $\|\cdot\|_{L^\phi(\Bbb R^d)}$ by 
\begin{align}
 \|u\|_{L^\phi(\Bbb R^d)}=\inf \Big\{ \lambda>0~: \int_{\Bbb R^d} \phi\Big(\frac{|u(x)|}{\lambda}\Big)d x\leq 1\Big\}.
 \end{align}