Prove that $\limsup_{n\to\infty}\int f_n\leq \int\limsup_{n\to\infty }f_n$ Les $f_n$ a sequence of measurable function such that $|f_n|\leq 1$ for all $n$.
1) Prove that $$\limsup_{n\to\infty}\int f_n\leq  \int\limsup_{n\to\infty }f_n$$
2) Does this inequality hold if supposed $f_n\geq 0$ but no necessarily bounded ?
My try
I set $f_n=f_n^++f_n^-$ where $f_n^+(x)=\max\{0,f_n(x)\}$ and $f_n^-=\min\{0, -f(x)\}$. Then 
\begin{align*}
\limsup_{n\to\infty }\int f_n&=\limsup_{n\to\infty }\int f_n^+-f_n^-\\
&\leq\limsup_{n\to\infty }\int f_n^++\limsup_{n\to\infty }\int(-f_n^-)\\
&=\limsup_{n\to\infty }\int f_n^+-\liminf_{n\to\infty }\int f_n^-\\
&\underset{Fatou}{\leq} \limsup_{n\to\infty }\int f_n^+-\int\liminf_{n\to\infty }f_n^-\\
&= \limsup_{n\to\infty }\int f_n^++\int\limsup_{n\to\infty }(-f_n^-)
\end{align*}
but I can't conclude.
For 2) I'm sure it's wrong, but I can't find a counter example.
 A: Question 1: First of all: If you consider the Lebesgue measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$, then your claim is not correct. Just consider
$$f_n(x) := 1_{[n,n+1]}(x),$$
then
$$\limsup_{n \to \infty} \int f_n(x) \, dx= 1 > 0 = \int \limsup_{n \to \infty} f_n(x) \, dx.$$
Now if you restrict the Lebesgue measure to $[0,1]$, i.e.
$$\limsup_{n \to \infty} \int_{[0,1]} f_n(x) \, dx \leq \int_{[0,1]} \limsup_{n \to \infty} f_n(x) \, dx,$$
then this inequality follows from Fatou's lemma applied to the non-negative sequence $g_n(x) := 1-f_n(x)$.
Question 2: No, this inequality is, in general, not satisfied if $f_n$ is not bounded. Consider
$$f_n(x) := n 1_{[0,1/n]}(x).$$
Remark: More generally, one can show the following statement

Let $(X,\mathcal{A},\mu)$ be a measure space. If $(f_n)_{n \in \mathbb{N}}$ is a sequence of measurable functions such that $f_n \leq g$ for some $g \in L^1(\mu)$, then $$\limsup_{n \to \infty} \int f_n \, d\mu \leq \int \limsup_{n \to \infty} f_n \, d \mu.$$

