# Lebesgue integral on measurable succesion

Let $(\Omega,\Sigma,\mu)$ a measure space and $f_n:\Omega \rightarrow [-\infty;+\infty]$ measurable functions. Supposing $g\leq f_n, \qquad \forall n \in N$, a measurable function such that his negative part $g^-$ in integrable.
Show that: $\int_\Omega \lim \inf_{n \rightarrow \infty} f_n d\mu \leq \lim \inf_{n \rightarrow \infty} \int_\Omega f_n d\mu$ and that the integral in the formula bove have sense.
This is an exercise I'm trying to solve, I started working on the demonstration that the integral have sense and then I'll try to prove the formula.
I know it's similar to the Fatou Lemma and his reverse but $f_n$ is not nonnegative and g is not dominating so I'm trying to work in that direction.

So first of all I tried to prove the existence of the limit on the right. Let $g=g^+ - g^-,\qquad g^+ - g^-\leq f_n$ so $g^-\geq g^+-f_n$ since $g^-$ is integrable I can assume that $g^+ - f_n$ is measurable too.Now can I tell that also $f_n$ is integrable as a consequence of it?

For the second integral $f_n$ in measurable,the liminf of the sucesion is measurable too but I don't know how to get to the result.

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We have $g\geqslant -g^-$ so $f_n\geqslant -g^-$ and $f_n+g^-\geqslant 0$. We apply Fatou's lemma to the sequence $\{f_n+g^-\}$.
How do you get that $f_n$ are integrable from what you wrote? – Davide Giraudo Nov 15 '12 at 21:23
I'm sorry,I have a problem with the connection, I wanted to write: If I apply the Fatou Lemma to ${f_n+g^-}$ and to $g^-$ I got the final result as you told but I'm wondering if the things I wrote above are right to justify that the integral have sense or if I'm missing something. If I use the fact that $f_N\geq -g^-$ I get that $g^-\geq -f_n$ so since $\int g^-$ is finished also $\int (-f_n)$ is finished and $\int f_n$ too.Is it right?What about the liminf? – Laura Nov 15 '12 at 21:26
It depends: on $\Bbb R_+$, $-1\leqslant e^{—x}$ but the map in the LHS is not integrable. – Davide Giraudo Nov 15 '12 at 21:28
I meant that the fact that $g^-\geqslant -f_n$ doesn't imply that $f_n$ integrable. If we had had $g^-\geqslant -f_n\geqslant 0$, it would be ok. The integral of $f_n+g^-$ is possibly infinite, but makes sense in $\overline{\Bbb R}$ (for example when $g(x)=-e^{-x^2}$ and $f_n=n$). "Make sense here is that not both $f_n^+$ and $f_n^-$ have infinite integral. – Davide Giraudo Nov 15 '12 at 21:41