# Proving Measurable Functions Convergence in Finite Measure Sets

• SqueeSpleen
In summary, the following sequence of measurable sets exists:E_{1}, E_{2}, E_{3}, E_{4}, E_{5}, E_{6}, E_{7}, E_{8}, E_{9}, E_{10}f

#### 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.

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medible functions

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

This was a mistake, I meant measurable.

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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.

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