I want to prove that Orlicz space is complete. Here is my attempt
Let $(X,\mu)$ be a measurable space and suppose $\phi(x)$ is a young function. Define $$L^{\phi}=\{f\hspace{0.2cm} \text{measurable:}\int_{X}\phi(|f(x)|/M)d\mu<\infty, \text{for some}\hspace{0.2cm} M>0\}$$
And $||f||_{\phi}=\inf_{M>0}\int_{X}\phi(|f(x)|/M)d\mu\le1$.
I want to prove that $L_{\phi}$ is complete in this norm.
I thought of the following:
step one let $\{ f_n \}$ be a Cauchy sequence in $L^{\phi}$, then $\forall \epsilon > 0$ $\exists$ a positive integer $N$ s.t $\forall$ $n>m>N$ we have $\|f_n-f_m\| < \epsilon$.
Now, for all $n>m>N$ the following is true $\int_{X}\phi(\frac{|f_n(x)-f_m(x)|}{\|f_n-f_m\|})d\mu\le1$. Can I deduce from here that the quantity $|f_n(x)-f_m(x)|<\epsilon$ for all $n>m>N$? I thought I could since everything is fixed ($\phi$, $X$, the bound 1) but the $\|f_n-f_m\|<\epsilon$ and here that epsilon is arbitrary. If this deduction is correct. I then want to proceed as follows:
step two construct a pointwise limit $f:X \to \mathbb{R}$ of $f_n$ on $X$ s.t $\lim_\limits{n \to \infty}f_n(x)=f(x)$
step three prove that $f$ is indeed in $L^{\phi}$ and that $\|f_n-f\|\to 0$
I appreciate any help or hints.