Generalization of Fatou's lemma for nonpositive but bounded measurable functions. Let $(f_n)^{\infty}_{n=1}$ be a sequence of measurable (not-necessarily $\ge 0$). Let $g \gt 0$ be a measurable function with $\int g d\mu < \infty$ (integrable) such that $f_n\ge -g$ a.e. relative to $\mu$ in $E\in S$ ($S$ $\sigma$-algebra).
I want to show that $$\int_E \liminf f_n d\mu \le \liminf \int_E f_n d\mu.$$
My attempt:
Let $N_n=\{x\in E; f_n(x) < g(x)\}$. Then $\mu(N_n)=0$ for all $n \in \mathbb{N}$ by hypothesis. Let $N= \cup N_n$, it follows that $N$ is also $\mu$-null. 
With $N$ as above, one can see that $$(f_n+g)\chi_{E \setminus N}\ge 0,$$ everywhere, so applying Fatou's Lemma one gets:
$$ \int_{E\setminus N} \liminf (f_n+g) d\mu \le \liminf \int_{E\setminus N} (f_n +g) d\mu.$$
Since $\liminf g =g, \liminf \int g= \int g,$ and $\int g < \infty$, if I could "open" both integrals as the sum of the integrals of each function, I could cancel out the terms involving $g$, yet I can't seem to make it work. I have a semi-linearity theorem for positive measurable functions and a linearity theorem for integrable functions, but the $f_n$'s are neither positive nor integrable.
I was told that I should prove a linearity theorem concerning non positive measurable functions whose integrals can be extended real numbers, i.e., we might define $$\int f = \int f^+ -\int f^-,$$ given that said difference is well-defined (we do not get $\infty - \infty$).
I can't seem so see the light in the proof of said linearity. Any insight on the issue would be greatly appreciated.
 A: You can apply Fatou's lemma to $h_n = f_n + g \geq 0$. Since $\liminf h_n = \liminf f_n + g$, it yields,
$$
\int (\liminf f_n + g)\,d\mu \leq \liminf \int (f_n + g)\,d\mu
$$
Conclude using the condition $\int g\,d\mu < \infty$
A: Here is the detail solution for your question:
Apply Fatou's Lemma for $f_n + g \geq 0,$ we obtain
$$\int_{E}\liminf_{n\longrightarrow+\infty}( f_n +g)\leq \liminf_{n\longrightarrow+\infty} \int_E (f_n+g) $$
The idea is to expand both sides.
For the LHS
Let $E^+ =\{x \in E\,:\, \liminf\limits_{n \longrightarrow +\infty}f_n(x) \geq 0\}$ and $E^- =\{x \in E\,:\, \liminf\limits_{n \longrightarrow +\infty}f_n(x) <0\}$, we rewrite the LHS of first inequality
$$
\int_{E}\liminf_{n\longrightarrow+\infty}( f_n +g) = \int_E \left[(\chi_{E^+})\liminf_{n\longrightarrow+\infty}( f_n +g) + (\chi_{E^-})\liminf_{n\longrightarrow+\infty}( f_n +g) \right]\\
= \int_E (\chi_{E^+})\liminf_{n\longrightarrow+\infty}( f_n +g) + \int_E(\chi_{E^-})\liminf_{n\longrightarrow+\infty}( f_n +g) 
\\
= \int_{E^+}\liminf_{n\longrightarrow+\infty}( f_n +g) + \int_{E^-}\liminf_{n\longrightarrow+\infty}( f_n +g)
\\
= \int_{E^+}(\liminf_{n\longrightarrow+\infty} f_n +g) + \int_{E^-}(\liminf_{n\longrightarrow+\infty}f_n +g)\qquad(A)
\\
= \int_{E^+}\liminf_{n\longrightarrow+\infty} f_n +\int_{E^+}g + \int_{E^-}\liminf_{n\longrightarrow+\infty}f_n +\int_{E^-}g \qquad (B)
\\
= \int_{E}\liminf_{n\longrightarrow+\infty} f_n +\int_{E}g.$$
We have $(A)=(B)$ because $\liminf\limits_{n\longrightarrow+\infty} f_n $ and $g$ are nonnegative on $E^+$ and they are integrable on $E^-$ (from $f_n \geq -g$, we have $\liminf\limits_{n\longrightarrow+\infty} f_n \geq -g$, hence on $E^-$, $0 \geq\liminf\limits_{n\longrightarrow+\infty} f_n \geq -g$).
For the RHS
The main point is to prove 
$$
\int_E (f_{n} + g) = \int_{E} f_{n} + \int_E g,\qquad \forall n \in \mathbb{N}. \qquad (C)
$$
(both sides can be equal to infinity).
First, if $\int_E |f_n| < \infty$, then the identity is true because both functions are integrable. In the other case, that means $\int_E |f_n| = \infty$, put $f_n^+ = \max\{f_n,0\}$ and $f_n^-=-\min\{f_n,0\}$, then
$$
\int_E |f_n |= \int_E f_n^+ + \int_E f_n^-
$$
(because $|f_n|=f_n^+ +f_n^-$ and $f_n^+$, $f_n^-$ are nonnegative, the above identity follows from Theorem 1.27 p.22, Rudin, Real and Complex Analysis.)
Moreover, $ 0 \leq f_n^- \leq g$ so $f_n^-$ is integrable. It follows that $\int_E f_n^+ = \infty.$ Therefore,
$$
\int_E f_n = \int_E f_n^+ - \int_E f_n^- = \infty.
$$
(see Definition 1.31, identity (2), p.25, Rudin, Real and Complex Analysis.)
Hence
$$\int_{E} f_{n} + \int_E g = \infty.$$
We need to prove
$$\int_E (f_{n} + g) = \infty.$$
Assume by contradiction that $\int_E (f_{n} + g) < \infty.$
Since $|f_n| \leq |g| + |f_n +g| = |g| + (f_n +g)$ (because $f_n + g \geq 0$), we have
$$\int_E |f_n| \leq \int_{E} \left[ |g| + (f_n +g) \right]
= \int_{E} |g| + \int_{E} (f_n+g)  
< +\infty
$$
which is a contradiction and thus $(C)$ is proved.
The RHS of first inequality can be rewritten as
$$
\liminf_{n\longrightarrow+\infty} \int_E (f_n+g) = \liminf_{n\longrightarrow+\infty} \left(\int_E f_n+\int_E g\right) = \liminf_{n\longrightarrow+\infty} \int_E f_n + \int_E g
$$
The result follows.
