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Draeden писал(а):Source of the post
$\mu B = 0 \Rightarrow \forall \varepsilon \exists \{ E_n \}_{n=1}^{\infty},E_n \in E,E\in G,B \subset \cup_{n=1}^{\infty}E_n,\sum_{n=1}^{\infty}\mu E_n < \varepsilon$

$B \subset \cup_{n=1}^{\infty}E_n$

$B \in M$ ? M - измеримое мн-во

Since $B = \bigcup_{n=1}^{\infty}(B \cap E_n)$ assume $A_n = B \cap E_n$ and prove that $A_n \in M_F$ .
As $A_n \subset E_n$ and $E_n \in {\mathbb G} \cap {\mathbb E}$ we can write, that $A_n \bigtriangleup E_n \subset E_n$ , but the last one has small measure: $\mu ^* E_n < \varepsilon$ .
That results in $\forall \varepsilon \mu^*(A_n \bigtriangleup E_n) < \varepsilon$ .
Therefore, it's proved that $A_n \in M_F$ , but $B = \bigcup_{n=1}^{\infty} A_n$ i.e. $B \in {\mathbb M}$ .
Q.E.D.

Draeden писал(а):Source of the post $\forall \varepsilon \mu^*(A_n \bigtriangleup E_n) < \varepsilon$

As I understand, $$A_n$$

depends on $\varepsilon$ .

You're right. You'd like a proof of lebesgue measurability without that ?

Draeden писал(а):Source of the post
You're right. You'd like a proof of lebesgue measurability without that ?

I think it's not correct. You look over only one $\epsilon$

$G \in \mathbb{G}(\mathbb{R}^m) \\ I:G \to \mathbb{E \cap \mathbb{G}} \\ a \in I(a) \subset G$

therefore

$G \subset \bigcup_{a \in \mathbb{Q}^m \cap G} I(a)$

$\bigcup_{a \in \mathbb{Q}^m \cap G} I(a) \subset G$

it's clearly, that $\mathbb{Q}^m \cap G$ is countable.

Theorem

$A \in \mathbb{M}_F \Rightarrow \forall \varepsilon \exists G \in \mathbb{G}, A \subset G, \mu ( G \setminus A ) < \varepsilon$

Original proof

$A \in \mathbb{M}_F \Rightarrow \forall \varepsilon \exists E \in \mathbb{E}, \mu ( A \bigtriangleup E ) < \varepsilon$

$\mu ( A \bigtriangleup E ) < \varepsilon \Rightarrow \exists \{ E_n \}_{n=1}^{\infty}, E_n \in \mathbb{G \cap E}, A \bigtriangleup E \subset \bigcup_{n=1}^{\infty}E_n, \sum_{n=1}^{\infty}m(E_n) < \varepsilon$

$\exists E' \in \mathbb{G \cap E}, E \subset E', m(E'\setminus E) < \varepsilon$

$G = \bigcup_{n=1}^{\infty}E_n \cup E' \in \mathbb{G} \Rightarrow A \subset G, \mu ( G \setminus A ) \le \mu ( E' \setminus E ) + \mu \bigcup_{n=1}^{\infty}E_n < 2\varepsilon$

My proof

$A \in \mathbb{M}_F \Rightarrow \mu(A) < \infty \Rightarrow \forall \varepsilon \exists \{ E_n \}_{n=1}^{\infty}, E_n \in \mathbb{G \cap E}, A \subset \bigcup_{n=1}^{\infty}E_n, \sum_{n=1}^{\infty}m(E_n) < \mu(A) + \varepsilon$

$G = \bigcup_{n=1}^{\infty}E_n \in \mathbb{G} \Rightarrow \mu(G \setminus A ) < \varepsilon$

where's mistake ?

Integral

$\int_{-\infty}^{+\infty} \frac{sin x}x dx = Im \left( \int_{-\infty}^{+\infty} \frac{e^{ix}}x dx \right) = Im \left( \lim_{a \to 0} \int_{-\infty}^{+\infty} \frac{e^{ix}}{x-a} dx \right)$

then, if Im a > 0, the integral equals to

$2\pi i \cdot res_a \left( \frac{e^{ix}}{x-a} \right ) = 2\pi i \cdot e^{ia} \to^{a \to 0} 2\pi i$

$\int_{-\infty}^{+\infty} \frac{sin x}x dx = 2 \pi$

if Im a < 0, the integral is zero ( since $e^{ix} = e^{i \alpha x}, \alpha > 0$ )

and, at last, if Im a changes its sign while approaching to 0, the limit doesn't exist.

where's mistake ?

Док-во вроде правильное.
Ha счет интеграла - поясни 2-e равенство.

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