About the completeness of orlicz space Let $(X,\mu)$ be a measurable space and suppose $\phi(t)$ is continuous,convex,and increasing function on$[0,\infty)$,with $\phi(0)=0$. 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. But i have no idea how to start.
 A: Suppose that $\{f_n\}$ is a Cauchy sequence in $L^{\phi}$. This means that in norm terms for every $\varepsilon>0$ there exists some $n_0>0$ such that $$||f_n-f_m||_{\phi}<\varepsilon\hspace{0.2cm}\text{whenever}\hspace{0.2cm}n,m\geq n_0$$
Since $\{f_n\}$ is a Cauchy sequence we can find an increasing sequence of positive integers $n_k$ and form a subsequence $\{f_{n_k}\}$ with the following property $$||f_n-f_{n_k}||_{\phi}<\frac{1}{2^k}\leq 1\hspace{0.2cm}\text{whenever}\hspace{0.2cm}n\geq n_k$$
Define the following function $$g_N=\sum^{N}_{k=1}|f_{n_{k+1}}-f_{n_k}|$$
First notice that $\{g_N\}$ is a sequence of functions that is nonnegative (since absolute value of differences is taken) and nondecreasing (as $N$ increases more and more nonnegative terms are included). Lets see whether $g_N\in L^{\phi}$ or not. Using the triangle inequality (since $||\cdot||_{\phi}$ is a norm} 
$$||g_N||_{\phi}=\Big|\Big|\sum^{N}_{k=1}|f_{n_{k+1}}-f_{n_k}|\Big|\Big|_{\phi}\leq \sum^{N}_{k=1} ||f_{n_{k+1}}-f_{n_k}||_{\phi}<\sum^{N}_{k=1}\frac{1}{2^k}<1$$ 
So we conclude that $g_N\in L^{\phi}$. Let $\lim_{N\to\infty}g_N=g$ then $g\in L^{\phi}$ since $$||g||_{\phi}=\Big|\Big|\sum^{\infty}_{k=1}|f_{n_{k+1}}-f_{n_k}|\Big|\Big|_{\phi}\leq \sum^{\infty}_{k=1} ||f_{n_{k+1}}-f_{n_k}||_{\phi}<\sum^{\infty}_{k=1}\frac{1}{2^k}\leq1$$
On the other side $$\phi\Big(\frac{g_N(x)}{M}\Big)\equiv \phi_N(x)$$
is also a nondecreasing sequence of functions because $\phi$ is increasing and $\{g_N\}$ is nondecreasing. Additionally $\int_X\phi_N(x)d \mu<\infty$ because $g_N\in L^{\phi}$. By Monotone Convergence Theorem then $\phi_N$ converges almost everywhere on $X$. Let $\lim_{N\to\infty}\phi_N=\phi$ then by continuity of $\phi$ we obtain $$\lim_{N\to\infty}\phi_N=\lim_{N\to\infty}\phi\Big(\frac{g_N(x)}{M}\Big)=\phi\Big(\lim_{N\to\infty}\frac{g_N(x)}{M}\Big)=\phi\Big(\frac{g(x)}{M}\Big)$$
This means that $\sum^{\infty}_{k=1}|f_{n_{k+1}}-f_{n_k}|$ is convergent almost everywhere on $X$. In particular $\sum^{\infty}_{k=1}(f_{n_{k+1}}-f_{n_k})$ is convergent almost everywhere on $X$. On the other side 
$$\sum^{N}_{k=1}(f_{n_{k+1}}-f_{n_k})=f_{n_N}-f_{n_1}$$
then $f_{n_N}$ converges almost everywhere on $X$. Denote by $\lim_{N\to\infty}f_{n_N}=f$. For any fixed $i$ the sequence $|f_{n_j}-f_{n_i}|$ satisfies Fatou's Lemma, that is $|f_{n_j}-f_{n_i}|\geq 0$ and $||f_{n_j}-f_{n_i}||<1/2^{i}<\infty$ whenever $j\geq i$. Therefore $\lim\inf_j|f_{n_j}-f_{n_i}|=|f-f_{n_i}|$ almost everywhere on $X$. We have $g\in L^{\phi}$ so $|f-f_{n_i}|\in L^{\phi}$ implying $f-f_{n_i}\in L^{\phi}$ and thus $f\in L^{\phi}$ (since $L^{\phi}$ is a linear space and $f_{n_i}\in L^{\phi}$ by assumption). 
A: First, it is worth pointing out that the problem, which is probably from Exercise 23, Chapter 1, in Stein and Shakarchi's Functional Analysis, has an error in its current form. We can simply let $\Phi(x) = 0$, which is continuous, convex, and increasing on $[0, \infty)$. Now every non-negative measurable function is in $L^\Phi$ with $\|f\|_\Phi = 0$. Every sequence {$f_n$} is a Cauchy sequency. Yet it does not have a well-defined limit.
A sufficient modification is to require that $\Phi$ is not the constant function 0.
Also note that the answer provided by Arian is not completely correct. (Pun intended.) It is implicitly assumed that the limit $\lim_{N\rightarrow\infty}g_N$ converges in $L^\Phi$. But this is not yet proved, and the inequality that immiedately follows is therefore invalid.
Below is a proof that closely follows that of Theorem 1.3 in Stein and Shakarchi's Functional Analysis.
Let $\{f_n\}_{n=1}^\infty$ be a Cauchy sequence in $L^\Phi$, and consider a subsequence $\{f_{n_k}\}_{k=1}^\infty$ of $\{f_n\}$ with the following property $\| f_{n_{k+1}} − f_{n_k} \|_\Phi \leq 2^{−k}$ for all $k \ge 1$. We now consider the series whose convergence will be seen below
$$f(x) = f_{n_1}(x) + \sum_{k=1}^\infty \left(f_{n_{k+1}}(x) - f_{n_k}(x)\right)$$ and
$$g(x) = |f_{n_1}(x)| + \sum_{k=1}^\infty \left|f_{n_{k+1}}(x) - f_{n_k}(x)\right|,$$
and the corresponding partial sums
$$S_K(f)(x) = f_{n_1}(x) + \sum_{k=1}^K\left(f_{n_{k+1}}(x) - f_{n_k}(x)\right)$$ and
$$S_K(g)(x) = |f_{n_1}(x)| + \sum_{k=1}^K \left|f_{n_{k+1}}(x) - f_{n_k}(x)\right|.$$
Note that so far we have not yet shown that either $f$ or $g$ actually converges, either point-wise or in the norm $L^\Phi$.
The triangle inequality for $L^\Phi$ implies
$$\|S_K(g)\|_\Phi \leq \|f_{n_1}\|_\Phi + \sum_{k=1}^K \|f_{n_{k+1}} - f_{n_k}\|_\Phi \leq \|f_{n_1}\|_\Phi + \sum_{k=1}^K 2^{-k} \leq \|f_{n_1}\|_\Phi + 1 \equiv M.$$
By the definition of the $L^\Phi$ norm, we have
$$\int_X \Phi(S_K(g)(x)/M) d\mu \le 1.$$
As $\Phi(S_K(g)/M)$ is increasing with regard to $K$, we can let $K$ tend to infinity and apply the monotone convergence theorem. This proves that $$\int_X \Phi(g(x)/M) d\mu \le 1.$$
This in turn implies that $\Phi(g(x)/M)$ is finite almost everywhere. Since $\Phi$ is increasing, convex, and not the constant function 0, $g(x)$ is finite almost everywhere. Therefore the series defining $g$, and hence the series defining $f$ converges almost everywhere, and $f \in L^\Phi$.
We now show that $f$ is the desired limit of the sequence $\{ f_n \}$ in the norm $L^\Phi$. Since (by construction of the telescopic series) the $(K − 1)$th partial sum of this series is precisely $f_{n_K}$, we ﬁnd that
$$f_{n_K}(x) \rightarrow f(x) \; \mathrm{a.e.} \;x.$$
We observe that for any $\epsilon > 0$, there exists a large enough integer N such that
$$\int_X \Phi(|f_m - f_n|/\epsilon) d\mu \leq 1 $$
for all $m,n > N$.
Note that $\liminf_{K\rightarrow\infty} \Phi(|f_m(x)-f_{n_K}(x)|/\epsilon) = \Phi(|f_m(x)-f(x)|/\epsilon)$ almost everywhere, as $\Phi$ is continuous. With Fatou’s lemma this gives
$$\int_X \Phi(|f_m-f|/\epsilon) d\mu \leq \liminf_{K\rightarrow\infty} \int_X \Phi(|f_m-f_{n_K}|/\epsilon) d\mu \leq 1$$
for any $m > N$.
This proves that $f_n \rightarrow f$ in the norm $L^\Phi$ as $n$ tends to infinity.
(A key step in the proof is the point-wise convergence of $f$. This could be greatly simplified if we assume $X$ is $\sigma$-finite. For more details, see this paper.)
