Proving Measurable Functions Convergence in Finite Measure Sets

[SqueeSpleen]

Let E be of finite measure and let \{ f_{n} \} _{n \geq 1} : E \rightarrow \overline{\mathbb{R}} measurable functions, finites almost everywhere in E such that f_{n} \rightarrow_{n \to \infty} f almost everywhere in E. Prove that exists a sequence (E_{i})_{i \geq 1} of measurable sets of E such that:
1) | E- \displaystyle \bigcup_{i=1}^{\infty} E_{i} | = 0
2) For every i \geq 1, f \rightrightarrows_{n \to \infty} in E_{i}

Notation:
f_{n} \rightarrow_{n \to \infty} f means "f_{n} converges to f as n \to \infty"
f \rightrightarrows_{n \to \infty} means "f_{n} converges uniformly to f as n \to \infty"
I don't know if there are multiple common definitions, but here is the mine:
A function f : \mathbb{R}^{n} \rightarrow \mathbb{R} is measurable if for all a \in \mathbb{R}:
\{ f > a \} = \{ x \in \mathbb{R}^{n} / f(x) > a \} is measurable in the sense of Lebesgue.
E \in R^{n} is measurable in the Lebesgue sense if for every \varepsilon > 0 \exists U \in \mathcal{U} / E \in U \wedge |U-E|_{e} < \varepsilon
Where | |_{e} is the outer measure.Attemp/idea:
I tried to divide E in E_{k,n} = \{ x \in E / | f(x)-f_{n} | < \frac{1}{k} \} but it will not help at all because something may "converge" similar in the first m terms but converge otherwise later.

 

[brmath]

medible functions

Sorry, did you mistype this? I don't know what it means.

 

[SqueeSpleen]

This was a mistake, I meant measurable.

 

[SqueeSpleen]

In a previous exercise (I didn't realize it could be useful here, so I didn't post it) I had the following result:
Let E be a set of finite measure and \{ f_{k} \}_{k \in \mathbb{N}} : E \rightarrow \mathbb{R} a sequence of measurable functions such that for every x \in E, exists M_{x} \in \mathbb{R}_{>0}:
| f_{k} (x) | \leq M_{x}, \forall k \in \mathbb{N}
Then for every \varepsilon > 0 exists F \subset E closed and M \in \mathbb{R}_{>0} such that:
| E - F | < \varepsilon and | f_{k} (x) | \leq M, \forall k \in \mathbb{N}, \forall x \in F

If there's a set where \forall M \in \mathbb{R}_{>0} \exists k \in \mathbb{N} / | f - f_{k} | > M, then this set has to have null measure, because either the functions doesn't converge here or they're not finite.

Let's be E = H \cup Z where \{ f_{k} \}_{k \in \mathbb{N}} are finite and converges to f in H and | Z | = 0 (E-Z=H) (They may be not finite in differents sets, but all are of zero measure and the countable union of sets of null measure has null measure).

As H is measurable, for every \varepsilon > 0 exists C closed such that C \in H and | H - C | < \varepsilon.

I'll try with E_{k} = \{ x \in H / f_{n} (x) < k \forall n \in \mathbb{N}

Edit: I had a non-trivial mistake so I was trying to fix it but I failed and I have a lecture in 6 hours, so I'd better to sleep now.

 

[brmath]

Getting late here -- will look at this tomorrow if someone doesn't help you sooner.