Fatou Lemma extension for every measurable sequence (negative & non-negative) Let $(f_n)_{n\in \Bbb N}$ be a sequence of elements in $M(X,S)$, let $g\in M^+(X,S)$ such that $\int gd\mu<\infty$ and $f_n\ge-g\ $ $(a.e.- \mu)\ \forall n\in \Bbb N\ $ in $E\in S$. Then, it follows:
$$\int_{E} \liminf_{n\to \infty} (f_n)d\mu\le \liminf_{n\to \infty} \int_E f_nd\mu$$

I have a doubt in somepoint of my proof in the a.e.-$\mu$ part. It goes like this: 
Lets define $\forall n\in \Bbb N\ h_n:X\rightarrow \Bbb R\ $ as $$\ h_n(x)=f_n(x)+g(x)\ \forall\ x\in X$$ 
So clearly $h_n$ is measurable $\forall n\in \Bbb N$, because $f_n\in M(X,S)\ \forall n\in \Bbb N\ $ and $g\in M^+(X,S)$.
And we get that $h_n\ge 0\ (a.e.-\mu)\ \forall n\in \Bbb N$
Now let $$l_m=\inf_{k\ge m} \{ h_k \}\ \forall m\in \Bbb N\ $$
thus $$l_m\le l_{m+1}\ \forall m\in \Bbb N$$
then $$\lim_{m\to \infty} l_m = \sup_{m\in \Bbb N} \{ l_m \} = \liminf_{n\to \infty} (h_n)$$
So we apply Fatou's Lemma and get that:
$$\int_{E} \liminf_{m\to \infty} (l_m)d\mu\le \liminf_{m\to \infty}\int_E l_md\mu$$
but, since the limit of $l_m$ exists, $$\liminf_{m\to \infty} (l_m)=\lim_{m\to \infty} l_m$$
so
$$\int_{E} \liminf_{n\to \infty} (h_n)d\mu = \int_{E} \lim_{m\to \infty} (l_m)d\mu =\int_{E} \liminf_{m\to \infty} (l_m)d\mu \le \liminf_{m\to \infty}\int_E l_md\mu$$
thus
$$\int_{E} \liminf_{n\to \infty} (h_n)d\mu\le \liminf_{m\to \infty}\int_E l_md\mu$$
Where $l_m\le h_n\ \forall n\ge m \Rightarrow\ \int_E l_m d\mu \le \int_E h_n d\mu\ \forall n\ge m \Rightarrow\ \int_E l_m d\mu \le \inf_{n\ge m} \int_E h_n d\mu\ \forall m\in \Bbb N\ \\ \Rightarrow\ \int_E l_m d\mu \le \sup_{m\in \Bbb N} \inf_{n\ge m} \int_E h_n d\mu = \liminf_{n\to \infty} \int_E h_n d\mu\ \\ 
\Rightarrow\ \liminf_{m\to \infty} \int_E l_m d\mu \le \liminf_{n\to \infty} \int_E h_n d\mu\ \\$ thus
$$\int_{E} \liminf_{n\to \infty} (h_n)d\mu\le \liminf_{m\to \infty}\int_E l_md\mu\le \liminf_{n\to \infty} \int_E h_n d\mu$$
Where $h_n=f_n+g\ \forall n\in \Bbb N\ \Rightarrow\ \liminf_{n\to \infty} (h_n)=g+\liminf_{n\to \infty} (f_n)\ \\ \Rightarrow\ \int_E \liminf_{n\to \infty} (h_n)d\mu =\int_E gd\mu +\int_E \liminf_{n\to \infty} (f_n)d\mu$ 
so
$$\int_E gd\mu +\int_E \liminf_{n\to \infty} (f_n)d\mu = \int_{E} \liminf_{n\to \infty} (h_n)d\mu\le \liminf_{n\to \infty} \int_E h_n d\mu$$
where 
$$\liminf_{n\to \infty} \int_E h_n d\mu = \liminf_{n\to \infty} \int_E (f_n+g) d\mu = \liminf_{n\to \infty} \{ \int_E f_nd\mu +\int_E gd\mu \} =\int_E gd\mu + \liminf_{n\to \infty} \int_E f_nd\mu$$
so
$$\int_E gd\mu +\int_E \liminf_{n\to \infty} (f_n)d\mu \le \int_E gd\mu + \liminf_{n\to \infty} \int_E f_nd\mu$$
where $\int_E gd\mu< \infty$, so we can substract it from the inequality and thus
$$\int_E \liminf_{n\to \infty} (f_n)d\mu \le \liminf_{n\to \infty} \int_E f_nd\mu$$
I don't know where (or how) to use the fact that $h_n \ge 0\ \ (a.e.-\mu)\ \forall n\in \Bbb N$
 A: Set
$$h_n := f_n + g.$$
As $h_n \geq 0$ we may apply Fatou's lemma and obtain
$$L := \int_E \liminf_{n \to \infty} h_n \, d\mu \leq \liminf_{n \to \infty} \int_E h_n \, d\mu =: R. \tag{1}$$
By the very definition of $h_n$, we have
$$\begin{align*} R &= \liminf_{n \to \infty} \int_E h_n \, d\mu \\ &= \liminf_{n \to \infty} \left( \int_E g_n+ \int f_n \, d\mu \right) \\ &= \int_E g \, d\mu + \liminf_{n \to \infty} \int_E f_n \, d\mu \end{align*}$$
where we have used in the last step that
$$\liminf_{n \to \infty} (c+a_n) = c+ \liminf_{n \to \infty} a_n$$
for any sequence $(a_n)_n$ and constant $c$. On the other hand,
$$\begin{align*} L &= \int_E \liminf_{n \to \infty} h_n \, d\mu \\ &= \int_E \liminf_{n \to \infty} (g+f_n) \, d\mu \\ &= \int_E (g+ \liminf_{n \to \infty} f_n) \, d\mu \\ &= \int_E g \, d\mu + \int_E \liminf_{n \to \infty} f_n \, d\mu. \end{align*}$$
Plugging this into $(1)$, we get
$$\int_E g \, d\mu + \int_E \liminf_{n \to \infty} f_n \, d\mu \leq \int_E g \, d\mu + \liminf_{n \to \infty} \int_E \, d\mu.$$
Subtracting $\int_E g \, d\mu < \infty$ on both sides finishes the proof.
