Question about $L^\infty$ ,Lebesgue space and Orlicz space Let $(\Omega,\Sigma,\mu)$ be an abstract measure space. Let $\Phi:\mathbb{R}\to\mathbb{R^+}\cup\{+\infty\}$ be a convex function such that $\Phi(0)=0,\Phi(-x)=\Phi(x)$ and $lim_{x\to\infty}\Phi(x)=\infty$.
Define a set , $$L^{\Phi}=\{f:\Omega\to\mathbb{R}\text{/ f is measurable,}\int_{\Omega}\Phi(|\alpha f|)d\mu<\infty\text{ for some $\alpha>0$ }\}$$
My first question is, how we can say that $\Phi(f)$ is measurable.\
second question is, in the book it is written that if $\Phi(x)=0 : 0\leq x<1$ and $=\infty : x>1$ then $L^{\Phi}=L^{\infty}?$ How we got this one??
I know the definition of $L^{\infty }=\{f:\Omega\to\mathbb{R}\text{/ f is measurable},||f||_{\infty}<\infty\}$ where, $||f||_{\infty}=ess\hspace{0.2mm}sup|f(x)|=inf\{M\geq 0/ |f(x)|\leq M\text{ a.e }\}$
 A: First we say that since $\Phi$ is convex, it is continuous. Prove that every convex function is continuous
Then we say that $\Phi\circ f$ is measurable if $f$ is. Composition of measurable & continuous functions, is it measurable?
This should give measurability.
Now suppose that $\Phi(x)=0$ for $|x|<1$ and $\Phi(x)=\infty$ for $|x|\geqslant 1$.  If $f:\mathbb{R}\to \mathbb{R}_\infty$ is essentially bounded, then there exists $\alpha>0$ such that $\|\alpha f\|_\infty<1$. Then $\Phi(|\alpha f|)=0$ a.e. and $\int \Phi(|\alpha f|) dx=0<\infty$. On the other hand, if $\|\alpha f\|_\infty >1$, then $\mu(\{x: |\alpha f(x)|\geqslant 1\})>0$, and $\Phi(|alpha f|)=\infty$ on a set of positive measure. Therefore $\int \Phi(|\alpha f|)dx=\infty$ if $\|\alpha f\|_\infty>1$. In particular, if $f$ is not essentially bounded, then for every $\alpha>0$, $\int \Phi(|\alpha f|)dx=\infty$. Therefore there exists some $\alpha>0$ such that $\int \Phi(|\alpha f|)dx<\infty$ if and only if $f$ is essentially bounded.
