TITLE: Two inequalities about using Fatou Lemma
QUESTION [10 upvotes]: 1) Let $\{f_n\}$ be a sequence of nonnegative measurable functions of $\mathbb R$ that converges pointwise on $\mathbb R$ to $f$ integrable. Show that
$$\int_{\mathbb R} f = \lim_{n\to \infty}\int_{\mathbb R}f_n  \Rightarrow \int_{E} f = \lim_{n\to \infty}\int_{E}f_n $$
for any measurable set $E$
I know that $\int_{\mathbb R} f = \int_{\mathbb R \setminus E} f + \int_{E} f$ and $\int_{\mathbb R \setminus E} f \le \liminf_{n\to {\infty}}\biggr(\int_{\mathbb R \setminus E} f_n \biggr)$ from Fatau's Lemma.
I couldn't obtain $\int_{E} f = \liminf_{n\to \infty}\int_{E}f_n = \limsup_{n\to \infty}\int_{E}f_n$ and I have seen that inequality below for obtaining it but I couldn't understand. Could someone explain me please?

$$\liminf_{n\to \infty}\int_{\mathbb R \setminus E}f_n = \int_{\mathbb R}f-\limsup_{n\to \infty}\int_{E}f_n$$

2) It has been written "since $\int_Ef_n \le \int_Ef$ (this inequality from monotonicity I have understood) thus 

$$\limsup\int_Ef_n \le \int_Ef$$

in proof of The Monotone Convergence Theorem in Royden's Real Analysis. I couldn't see why that inequality obtains.
Thanks for any help
Regards

REPLY [4 votes]: Good question. Here is another possible approach.
You already get $\int_{\mathbb R} f = \int_{\mathbb R \setminus E} f + \int_{E} f$ and $\int_{\mathbb R \setminus E} f \le \liminf_{n\to {\infty}}\biggr(\int_{\mathbb R \setminus E} f_n \biggr)$. So we have
$$
\lim_{n\to {\infty}}\int_{\mathbb R} f_n =\int_{\mathbb R} f \leq\liminf_{n\to {\infty}}\biggr(\int_{\mathbb R \setminus E} f_n \biggr)+ \int_{E} f.
$$
Notice that for real sequences $a_n$, $\liminf_{n\to \infty}a_n = -\limsup_{n\to \infty}(-a_n)$ (see the "property" section in wiki), and $\lim_{n\to {\infty}}\int_{\mathbb R} f_n = \limsup_{n\to \infty}\int_{\mathbb R} f_n$. Thus the inequality above can be rewritten as
$$
\limsup_{n\to {\infty}}\int_{\mathbb R} f_n \leq-\limsup_{n\to {\infty}}\biggr(-\int_{\mathbb R \setminus E} f_n \biggr)+ \int_{E} f,
$$
or equivalently,
$$
\limsup_{n\to {\infty}}\int_{\mathbb R} f_n +\limsup_{n\to {\infty}}\biggr(-\int_{\mathbb R \setminus E} f_n \biggr)\leq \int_{E} f,
$$
Next we will use another property for limit superior, which says that for two sequences $a_n$ and $b_n$, we have
$$
\limsup_{n\to \infty}(a_n+b_n)\leq \limsup_{n\to \infty}a_n + \limsup_{n\to \infty}b_n.
$$
Therefore we have
$$
\limsup_{n\to {\infty}}\biggr(\int_{\mathbb R} f_n-\int_{\mathbb R \setminus E} f_n \biggr)\leq \limsup_{n\to {\infty}}\int_{\mathbb R} f_n +\limsup_{n\to {\infty}}\biggr(-\int_{\mathbb R \setminus E} f_n \biggr)\leq \int_{E} f,
$$
that is, $$\int_{E} f\geq \limsup_{n\to {\infty}}\int_{ E} f_n.$$
For the last step, by Fatou's lemma we have $\int_{E} f\leq \liminf_{n\to {\infty}}\int_{ E} f_n$, thus
$$
\lim_{n\to {\infty}}\int_{ E} f_n = \liminf_{n\to {\infty}}\int_{ E} f_n = \limsup_{n\to {\infty}}\int_{ E} f_n = \int_{E} f.
$$