- #1

SqueeSpleen

- 141

- 5

Let [itex]E[/itex] be of finite measure and let [itex]\{ f_{n} \} _{n \geq 1} : E \rightarrow \overline{\mathbb{R}}[/itex] measurable functions, finites almost everywhere in [itex]E[/itex] such that [itex]f_{n} \rightarrow_{n \to \infty}[/itex] f almost everywhere in [itex]E[/itex]. Prove that exists a sequence [itex](E_{i})_{i \geq 1}[/itex] of measurable sets of [itex]E[/itex] such that:

1) [itex]| E- \displaystyle \bigcup_{i=1}^{\infty} E_{i} | = 0[/itex]

2) For every [itex]i \geq 1[/itex], [itex]f \rightrightarrows_{n \to \infty}[/itex] in [itex]E_{i}[/itex]

Notation:

[itex]f_{n} \rightarrow_{n \to \infty} f[/itex] means "[itex]f_{n}[/itex] converges to [itex]f[/itex] as [itex]n \to \infty[/itex]"

[itex]f \rightrightarrows_{n \to \infty}[/itex] means "[itex]f_{n}[/itex] converges uniformly to [itex]f[/itex] as [itex]n \to \infty[/itex]"

I don't know if there are multiple common definitions, but here is the mine:

A function [itex]f : \mathbb{R}^{n} \rightarrow \mathbb{R}[/itex] is measurable if for all [itex]a \in \mathbb{R}[/itex]:

[itex]\{ f > a \} = \{ x \in \mathbb{R}^{n} / f(x) > a \}[/itex] is measurable in the sense of Lebesgue.

[itex]E \in R^{n}[/itex] is measurable in the Lebesgue sense if for every [itex]\varepsilon > 0 \exists U \in \mathcal{U} / E \in U \wedge |U-E|_{e} < \varepsilon[/itex]

Where [itex]|[/itex] [itex]|_{e}[/itex] is the outer measure.

Attemp/idea:

I tried to divide [itex]E[/itex] in [itex]E_{k,n} = \{ x \in E / | f(x)-f_{n} | < \frac{1}{k} \}[/itex] but it will not help at all because something may "converge" similar in the first m terms but converge otherwise later.

1) [itex]| E- \displaystyle \bigcup_{i=1}^{\infty} E_{i} | = 0[/itex]

2) For every [itex]i \geq 1[/itex], [itex]f \rightrightarrows_{n \to \infty}[/itex] in [itex]E_{i}[/itex]

Notation:

[itex]f_{n} \rightarrow_{n \to \infty} f[/itex] means "[itex]f_{n}[/itex] converges to [itex]f[/itex] as [itex]n \to \infty[/itex]"

[itex]f \rightrightarrows_{n \to \infty}[/itex] means "[itex]f_{n}[/itex] converges uniformly to [itex]f[/itex] as [itex]n \to \infty[/itex]"

I don't know if there are multiple common definitions, but here is the mine:

A function [itex]f : \mathbb{R}^{n} \rightarrow \mathbb{R}[/itex] is measurable if for all [itex]a \in \mathbb{R}[/itex]:

[itex]\{ f > a \} = \{ x \in \mathbb{R}^{n} / f(x) > a \}[/itex] is measurable in the sense of Lebesgue.

[itex]E \in R^{n}[/itex] is measurable in the Lebesgue sense if for every [itex]\varepsilon > 0 \exists U \in \mathcal{U} / E \in U \wedge |U-E|_{e} < \varepsilon[/itex]

Where [itex]|[/itex] [itex]|_{e}[/itex] is the outer measure.

Attemp/idea:

I tried to divide [itex]E[/itex] in [itex]E_{k,n} = \{ x \in E / | f(x)-f_{n} | < \frac{1}{k} \}[/itex] but it will not help at all because something may "converge" similar in the first m terms but converge otherwise later.

Last edited: