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.
Best Answer
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 $\int g\,d\mu < \infty$