To prove that $L^{\infty}$ is a subspace of $L^{\Phi}$(Orlicz space) Let me define it first
Definition 1: A mapping $\Phi:\mathbb{R}\to\mathbb{R^+}\cup{\infty}$ is said to be an N function if
(1) $\Phi$ is convex
(2) $\Phi(0)=0$
(3) $\lim_{x\to \infty}\Phi(x)=\infty$
Orlicz space: Let $(\Omega,F,\mu)$ be arbitrary measure space. The $\Phi$-Orlicz space is defined as
$$\tag{1}
L^{\Phi}(\mu)=\Big\{f:\Omega \to \mathbb{R} \hspace{0.2cm} \text{such that} \int_{\Omega}\Phi(|\alpha f|)\,\mathrm d\mu<\infty \mbox{ for some } \alpha>0\Big\}
$$
What we have is that $L^{\Phi}$ is a normed linear space with the norm
$$
\|f\|_{\Phi} = \inf\Big\{k>0:\int_{\Omega}\Phi(\tfrac{f}{k})\,\mathrm d\mu\leq 1\Big\}
$$
$L^{\infty}$ space: Let $(\Omega,F,\mu)$ be a measure space then $L^{\infty}=\{f:\Omega\to \mathbb{R}\text{ such that } ||f||_{\infty}<\infty\}$
where $||f||_{\infty}=inf\{K\geq 0\text{ such that } |f(x)|\leq K \text{ a.e }\}$
I have to show that $L^{\infty}$ is a subspace of $L^{\Phi}$. What I understand is that in $L^{\infty}$ we have functions that are bounded almost everywhere. To show that it is a subspace of $L^{\Phi}$ we have to first show that $L^{\infty}\subseteq L^{\Phi}$ and in $L^{\Phi}$ we have all the measurable real-valued functions which satisfy the given integral condition. so if I pick $f\in L^{\infty}$ then how we can say that it is in $L^{\Phi}$ since bounded almost everywhere does not imply measurable? So I am stuck here and also I have no idea how do I proceed to prove other properties like scalar multiplication and how do we induce norm on $L^{\infty}$ space? Can someone help to prove this.That will be great help thanks.
 A: When $\mu(\Omega)<\infty$ the statement is true: since $|f|\leq\|f\|_\infty$ $\mu$-almost surely and $\Phi$ is monotone increasing (why?)
$\Phi(|f|)\leq \Phi(\|f\|_\infty)$. Hence
$$\int\Phi(|f|)\,d\mu\leq \Phi(\|f\|_\infty)\mu(\Omega)<\infty$$
Thus, $L_\infty(\mu)\subset L_\Phi(\mu)$
Furthermore, in such a case, by choosing $a>0$ such that $0<\Phi(a)<1/\mu(\Omega)$ we have that
$\Phi\big(a\tfrac{|f|}{\|f\|_\infty}\big)\leq\Phi(a)$ $\mu$-almost surely ($\Phi$ is monotone nondecreasing) and so,
$\int\Phi\big(a\tfrac{|f|}{\|f\|_\infty}\big)\,d\mu\leq \Phi(a)\mu(\Omega)<1$ which means that
$$\|f\|_{\Phi}\leq \frac{\|f\|_\infty}{a}$$
In other words, the inclusion map $f\stackrel{\iota}{\hookrightarrow} f$ from $L_\infty(\mu)$ into $L_\Phi(\mu)$ is a continuous bounded operator.
If $\mu(\Omega)=\infty$ the inclusion $L_\infty(\mu)\subset L_\Phi(\mu)$ does not hold. For example take a function $f\in L_\infty(\mu)$ that is bounded away from zero, that is for some $\varepsilon>0$, $|f|>\varepsilon$. The $$\int\Phi(\alpha^{-1}|f|)\,d\mu\geq\int\Phi(\alpha^{-1}\varepsilon)\,d\mu=\Phi(\alpha^{-1}\varepsilon)\mu(\Omega)=\infty, \qquad \alpha>0$$
