Proof of Fatou-Lebesgue Theorem Good evening everyone,
how can I prove the following inequality?
Let $f,f_n\in L(X,\mu)$ on a measure space with $0\leq f_n(x)\leq f(x)$. If $f,f_n$ are $\mu$-almost everywhere in $X$, then is
$$\int_X\liminf_{n\to\infty} f_n d\mu \leq \liminf_{n\to\infty}\int_X f_n d\mu\leq\limsup_{n\to\infty}\int_X f_n d\mu\leq \int_X\limsup_{n\to\infty} f_nd\mu.$$
How can I prove it?
Firstable, I know that
$$\liminf_{n\to\infty} \int_X f_n d\mu \leq \limsup_{n\to\infty}\int_X f_n d\mu.$$
But I don't know how I can prove the inequality. Could someone help me please? 
 A: Okay, I've tried to prove the "big inequality", so:
We have the property that $f,f_n$ are measurable and $0\leq f_n(x)\leq f(x)$.
Let be
$$a_n:=f+f_n,\qquad b_n := f-f_n.$$
Then, $a_n,b_n$ are real and integrable because of $f$ and $f_n$. So, 
$$0\leq a_n\leq 2\cdot f,\ 0\leq b_n\leq a_n\leq 2\cdot f.$$
Now we consider $\liminf$ of the sequences $a_n,b_n$:
$$\begin{align}
\liminf_{n\to\infty} a_n &= f+\liminf_{n\to\infty} f_n,\\
\liminf_{n\to\infty} b_n &= f+\liminf_{n\to\infty} (-f_n) = f-\limsup_{n\to\infty} f_n.
\end{align}$$
By Fatou's lemma is
$$\int_X\liminf_{n\to\infty} f_n d\mu \leq \liminf_{n\to\infty} \int_X f_n d\mu.$$
So:
$$\begin{align}
\int_X\liminf_{n\to\infty} a_n d\mu &= \int_X(f+\liminf_{n\to\infty} f_n)d\mu\\
&=\int_X fd\mu+\int_X\liminf_{n\to\infty} f_n d\mu\\
\Leftrightarrow\int_X\liminf_{n\to\infty} f_n d\mu &=\int_X\liminf_{n\to\infty} a_n d\mu-\int_X fd\mu\\
&\leq\liminf_{n\to\infty}\int_X a_nd\mu -\int_X fd\mu\\
&=\liminf_{n\to\infty}\int_X (f+f_n) d\mu-\int_X fd\mu\\
&=\int_X fd\mu-\int_X fd\mu+\liminf_{n\to\infty}\int_X f_nd\mu\\
&=\liminf_{n\to\infty}\int_X f_nd\mu.
\end{align}$$
Now for $b_n$:
$$\begin{align}
\int_X\liminf_{n\to\infty} b_n d\mu &= \int_X(f-\limsup_{n\to\infty} f_n)d\mu\\
&=\int_X fd\mu-\int_X\limsup_{n\to\infty} f_n d\mu\\
\Leftrightarrow\int_X\limsup_{n\to\infty} f_n d\mu &=-\int_X\liminf_{n\to\infty} b_n d\mu+\int_X fd\mu\\
&\geq\int_X fd\mu-\liminf_{n\to\infty}\int_X b_nd\mu\\
&=\int_X fd\mu-\liminf_{n\to\infty}\int_X (f-f_n)d\mu\\
&=\int_X f d\mu-\int_X f d\mu - \liminf_{n\to\infty}\int_X (-f_n)d\mu\\
&=-\left(-\limsup_{n\to\infty}\int_X f_n d\mu\right)\\
&=\limsup_{n\to\infty}\int_X f_n d\mu.
\end{align}$$
Could anyone help me, how I can prove the following inequality?
$$\liminf_{n\to\infty}\int_X f_n d\mu \leq \limsup_{n\to\infty}\int_X f_n d\mu$$
A: The inequality you're asking about has nothing to do with sequences of integrals (as pointed out in the first comment). In general $$\liminf_{n} x_{n} \leq \limsup_{n} x_{n}$$ for any sequence $\{x_{n}\}$.
If you really need to prove the inequality, you should do it for a sequence of real numbers (or extended real numbers). It's a matter of checking the definitions of $\liminf$ and $\limsup$.
