# A strong version of Dominated Convergence Theorem - Real Analysis

1. Nov 24, 2013

### SqueeSpleen

1. The problem statement, all variables and given/known data

Let $(g_{n})_{n \in \mathbb{N}}$ a sequence functions integrable over $\mathbb{R}^{p}$ such that:
$g_{n} (x) \longrightarrow g(x)$ almost everywhere in $\mathbb{R}^{p}$, where $g$ is a function integrable over $\mathbb{R}^{p}$.
Given $(f_{n})_{n \in \mathbb{N}}$ a sequence of functions measurable over $\mathbb{R}^{p}$ such that:
$| f_{n} | \leq g_{n}$, for all $n \in \mathbb{N}$ and $f_{n}(x) \longrightarrow f(x)$ a.e in $\mathbb{R}^{p}$, prove that:
$\displaystyle \int_{R^{p}} g(x) dx = \lim_{n \to \infty} \displaystyle \int_{\mathbb{R}^{p}} g_{n} (x) dx \Longrightarrow \displaystyle \int_{R^{p}} f(x) dx = \lim_{n \to \infty} \displaystyle \int_{\mathbb{R}^{p}} f_{n} (x) dx$

2. The attempt at a solution

I tried to $g_{n} - | f_{n} |$ as a sequence of nonnegative functions, but I need the sequence to be an increasing sequence to be able to apply Beppo-Levi's theorem.

Fatou's lemma didn't help me neither:

$\displaystyle \int g-|f|\leq \liminf_{n \to \infty} \displaystyle \int g_{n} - |f_{n}|$
$\displaystyle \int - | f | \leq \liminf_{n \to \infty} \displaystyle \int - | f_{n} |$
$\limsup_{n \to \infty} \displaystyle \int |f_{n}| \leq \displaystyle \int |f |$
And by Fatou's lemma:
$\displaystyle \int | f | \leq \liminf_{n \to \infty} \displaystyle \int |f_{n}|$
As $\liminf \leq \limsup$ both limits are the same and
$\lim_{n \to \infty} \displaystyle \int |f_{n}| = \displaystyle \int |f |$

Instead of using $| f |$ I could use $f^{+}$ and $f^{-}$ and prove the theorem?

I couldn't solve this problem in weeks, I was going to post it to get some help and I only tried to do it once more because I don't like to have to read my handwrite to transcribe it to the PC and I wanted to show that I really tried but it appears that I finally did it, it's not the first time that the exercise of writing it in this forum made me solve a problem, but this time I'm posting it anyway because I'm not sure if my proof is correct.

Last edited: Nov 24, 2013