On Beppo Levi’s Theorem

integrationlebesgue-integrallebesgue-measuremeasure-theoryprobability

Beppo Levi Theorem says that, if $f_n\uparrow f$ and $f_n$'s are integrable and also $\sup_n\int f_n d\mu < \infty$. Then $f$ is integrable and $\int f_n d\mu \uparrow \int f d\mu$.
Here, how do I show that $f$ is integrable?
I proposed a approach for doing that. Now
$f$ is integrable then it is enough to show $\int | f|d\mu<\infty$.
$\int|f|d\mu = \int |f-f_n + f_n|d\mu \leq \int|f_n – f|d\mu + \int|f_n|d\mu \forall n.$
Since $f_n$ is integrable then second term of the right hand side is finite say $A$. Now for fixed $\epsilon>0$ then there exists a $k \in \mathbb{N}$ such that $|f_n-f|<\epsilon$ for $n\geq k$.Then just replace the first term of right hand side by $\epsilon$ we get $\int|f|d\mu = \epsilon \mu(E) + A<\infty$. But for this conclution we need condition $\mu(\Omega)<\infty$, that is not given. So how do I manage to show integrability of $f$. Different approaches are appreciated. Thank you.

Best Answer

Apply Fatou's Lemma to $f_n-f_1$. This gives $\int (f-f_1) d\mu \leq \lim \inf \int (f_n-f_1) d\mu \leq \sup \int f_n d \mu -\int f_1 d\mu$. Hence $\int f d\mu <\infty$. On the other hand $f \geq f_1$ so $\int f d\mu >-\infty$.