First let me define $\bar{L^{\Phi}}$ space Let $\Phi:\mathbb{R}\to\mathbb{R^+}$ be a convex function such that
$$\Phi(0)=0,\Phi(-x)=\Phi(x) \quad\text{ and}\quad \lim_{x\to \infty}\Phi(x)=\infty$$
$\bar{L^{\Phi}}$=$\{f:\Omega\to\mathbb{R} \text{ measurable such that } \int_{\Omega}\Phi(|f|)d\mu<\infty \}$
In general it is not a vector space, since it is not closed under scalar multiplication. For example let $\Omega=(0,1)$, $\Phi(t)=e^t$ and take $f(x)=\frac{-1}{2}\log x$ then $2f\notin \bar{L^\Phi}$
It is given in the book that $\bar{L^\Phi}$ is a vector space if $\Phi\in \Delta_2$ globally when $\mu(\Omega)=\infty$ and locally when $\mu(\Omega)<\infty$
Let me define $\Delta_2$: $\Phi$ is said to be $\Delta_2$ if $\Phi(2x)\leq K\Phi(x)$ for some $K>0$ and $x\geq x_0 \geq 0$. But I do not understand how this $\Delta_2$ condition matters.
$\textbf{Proof:}$ The proof goes like this, first we see a result that if $h\in\bar{L^{\Phi}}$ and $|f|\leq |h|$,f is measurable then $f\in \bar{L^{\Phi}}...(1)$ so if I take $f\in \bar{L^{\Phi}}$ and if we show that $2f\in \bar{L^{\Phi}}$ then this will imply that $nf\in \bar{L^{\Phi}}$ for any integer $n$.and from (1) we can say that for any $\alpha$ such that $|\alpha f|\leq |nf|$ implies $\alpha f\in \bar{L^{\Phi}}$.
$\textbf{My Question:}$ This is not the complete proof but the only part I do not understand in the proof is why $2f\in\bar{L^{\Phi}}$ implies that $nf \in \bar{L^{\Phi}}$ for any integer $n?$