Use Fatou Lemma to show that $f$ takes real values almost everywhere. 
Let $(f_n)$ be a sequence in $L^p$ such that for each positive integer $n$,
  $
\| f_{n+1}-f_n\|_{p}
<\frac 1{2^n} 
$.
  Define $f: X \to [0,\infty]$ with
  $$
f(x)= \sum_{n=1}^\infty| f_{n+1}(x)-f_n(x)|.
$$
  Use Fatou Lemma to show that $f$ takes real values almost everywhere.


My thoughts:

We must show the existence of $A$ such that $\mu(A)=0$ and for all $x\in A^C$, $f\in\mathbb R$ using Fatous Lemma which states 
$$
\int_X\liminf_nf_n
\le\liminf_n\int_X f_n.
$$
My idea taking pieces of the completeness proof of $L^p$ is that we have that the series $f(x)=\displaystyle\sum_{n=1}^\infty\vert f_{n+1}(x)-f_n(x)\vert$ converges so $f(x)=\displaystyle\sum_{n=1}^\infty\vert f_{n+1}(x)-f_n(x)\vert<\infty$ and this holds for all $x\in X$ and then how will I define the set $A$?
I am not using at all Fatou Lemma because I don't know how.


[Added later:]
I would also like to understand steps of the proof in Alex R.'s answer below.
It is shown that

$$\int_A|f(x)|^pd\mu\leq \liminf_m\int_A|F_m(x)|^p dx\leq\liminf_m\sum_{n=1}^m\frac{1}{2^{pn}}d\mu<\infty.$$

There are a lot of missing steps. I tried to fill.
$\displaystyle\int_A\vert f(x)\vert^pdx
=\int_A|\liminf_mF_m(x)|^pdx=\int_A\liminf_m|F_m(x)|^pdx\le\liminf_m\int_A|F_m(x)|^pdx
=\liminf_m\int|\sum^m_{n=1}|f_{n+1}(x)-f_n(x)||^pdx
\fbox{=}\liminf_m\sum^m_{n=1}\int_A|f_{n+1}(x)-f_n(x)|^pdx
=\liminf_m\sum^m_{n=1}||f_{n+1}(x)-f_n(x)||_p^p\le\liminf_m\sum^m_{n=1}\frac{1}{2^{np}}\fbox=?$
Question1: Assuming the equalities and inequalities are correct, Why is the second equality, the boxed equality and what is equal in the last boxed equality?
It is also shown that

Taking $A$ to be $\mathbb{R}$, it follows that $\int_A|f(x)|^pd\mu<\infty$ which implies $\mu(\{|f|=\infty\})=0$. 

Question2: Should not have been written as
Take $A$ to be $\mathbb R.$
From all the arguments above (and not only $A=\mathbb R$), it follows that $\int_A|f(x)|^pd\mu<\infty$ which implies $\mu(\{|f|=\infty\})=0$ ??
Seems like as it originally stands, the affirmation depends only  on $A=\mathbb R$. But it is independent of taking $A$ as $\mathbb R$. Am I correct?
 A: Let $F_m(x)=\sum_{i=1}^m|f_{n+1}(x)-f_n(x)|$. Since $F_m(x)$ is increasing for each $x$, its limit in $m$ necessarily exists in the extended reals. So $\lim_{m\rightarrow\infty}F_m(x)=\liminf_{m\rightarrow\infty}F_m(x)=f(x)$
Then by Fatou:
$$\int_A|f(x)|^pd\mu\leq \liminf_m\int_A|F_m(x)|^p dx\leq\liminf_m\sum_{n=1}^m\frac{1}{2^{pn}}d\mu<\infty.$$
Taking $A$ to be $\mathbb{R}$, it follows that $\int_A|f(x)|^pd\mu<\infty$ which implies $\mu(\{|f|=\infty\})=0$. 
A: Hint: Since $\Vert f_{n+1}-f_n\Vert_p<\frac 1{2^n}$, if $g_k(x)=\displaystyle\sum_{n=1}^k\vert f_{n+1}(x)-f_n(x)\vert$ , then Minkowski's inequality implies that $\|g_k\|_p\le 1.$ And now an application of Fatou to $(g^p_k)$  shows that  $\|f\|_p\le 1$, for the $f$ as in your question, and this implies that $f(x)<\infty$ almost everywhere. 
A: I would like to directly answer your original question (for $1\le p< \infty$): 

Let $(f_n)$ be a sequence in $L^p$ such that
  $$
\| f_{n+1}-f_n\|_{p}
<\frac 1{2^n} \quad \forall n\in\mathbb{N}.
$$
  and define $f: X \to [0,\infty]$ with
  $$
f(x):= \sum_{n=1}^\infty| f_{n+1}(x)-f_n(x)|.\tag{1}
$$
  Use Fatou Lemma to show that $f$ takes real values almost everywhere.

For each positive integer $N$, define  $F_N:X\to[0,\infty]$ with $F_N(x)=\sum_{n=1}^N|f_{n+1}(x)-f_n(x)|.$ Then for each $x\in X$,
$$
f(x)=\lim_{N\to\infty}F_N(x)
$$ by the definition of $f$, and thus by continuity of the function $z\mapsto z^p$, for each $x\in X$,
$$
|f(x)|^p=\big(\lim_{N\to\infty}F_N(x)\big)^p=\lim_N |F_N(x)|^p.
$$
Now by Fatou lemma, 
$$
\int |f|^p\,d\mu\le \liminf_N \int |F_N|^p.\tag{2}
$$
But by the Minkowski inequality, for each $N$,
$$
\int |F_N|^p=
\|F_N\|_p^p\le (\sum_{n=1}^N\|f_{n+1}-f_n\|_p)^p\le(\sum_{n=1}^N\frac{1}{2^{n}})^p
\tag{3}
$$
Taking $\liminf_{N\to\infty}$ in (3) and then applying (2), one gets
$$
\int |f|^p<\infty
$$
and thus $\mu\{x\in X:|f(x)|^p=\infty\}=0$ and hence1 $\mu\{x\in X:|f(x)|=\infty\}=0$. 

1 See, e.g. this standard result: An integrable Functions is almost everywhere finite
