cours/mesure image.md
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up:: mesure positive d'une application #maths/intégration

[!definition] mesure image Soit (E, \mathcal{A}, \mu) un espace mesuré Soit (F, \mathcal{B}) un espace mesurable Soit f: E \to F mesurable de (E, \mathcal{A}) dans (F, \mathcal{B}) Notons \nu l'application de \mathcal{B} dans \overline{\mathbb{R}}_{+} définie : \begin{align} \nu: \mathcal{B} &\to \overline{\mathbb{R}}_{+}\\ B &\mapsto \mu(f^{-1}(B)) \end{align} \nu est une mesure positive d'une application appelée mesure image de \mu par $f$ ^definition

$= "![[" + dv.current().file.name + ".svg|700]]"

Propriétés

[!proposition] La mesure image est bien une mesure Soit (E, \mathcal{A}, \mu) un espace mesuré Soit (F, \mathcal{B}) un espace mesurable Soit f: E \to F mesurable de (E, \mathcal{A}) dans (F, \mathcal{B}) \begin{align} \nu: \mathcal{B} &\to \overline{\mathbb{R}}_{+}\\ B &\mapsto \mu(f^{-1}(B)) \end{align} \nu est une mesure de (F, \mathcal{B})

[!démonstration]- Démonstration

  1. Soit (B_{n})_{n\in\mathbb{N}} une suite d'éléments 2 à 2 disjoints de \mathcal{B} \displaystyle\nu\left( \bigcup _{n=0}^{+\infty} B_{n} \right) = \mu\left( f^{-1}\left( \bigcup _{n=0}^{+\infty}B_{n} \right) \right) = \mu\left( \bigcup _{n=0}^{+\infty} \underbrace{f^{-1}(B_{n})}_{A_{n}} \right) \forall n \in \mathbb{N}, \quad A_{n} \in \mathcal{A} Si n \neq p alors f^{-1}(B_{n}) \cap f^{-1}(B_{p}) = \emptyset Si x \in f^{-1}(B_{n}) \cap f^{-1}(B_{p}), alors f(x) \in B_{n} et f(x) \in B_{p} or, B_{n} \cap B_{p} = \emptyset ainsi, \displaystyle\nu\left( \bigcup _{n=0}^{+\infty}B_{n} \right) = \sum\limits_{n=0}^{+\infty} \mu(f^{-1}(B_{n})) = \sum\limits_{n=0}^{+\infty}\nu(B_{n}) Et donc \nu est bien une mesure sur \mathcal{B}

Exemples

[!example] Exemple 1 Sur l'espace mesuré ([0; 1], \mathcal{B}([0; 1]), \lambda _{[0; 1]}) Soit la tribu \mathcal{B} = \{ \emptyset, \{ 0 \}, \{ 1 \}, \{ 0, 1 \} \} Soit 0 \leq a < b \leq 1 on a \lambda _{[0; 1]}(]\![ a, b ]\![) = b - a Soit p \in [0; 1] Soit l'application : \begin{align} \mathbb{1}_{[0, p]} : [0; 1] &\to \{ 0; 1 \}\\ x &\mapsto \begin{cases} 0 \text{ si } x \in [0; p]\\ 1 \text{ sinon} \end{cases} \end{align} (fonction indicatrice de [0; p]) On cherche la mesure image de \lambda _{[0; p]} par \mathbb{1}_{[0; p]} C'est une mesure sur \mathcal{B}. Notons la \nu. \nu(\{ 1 \}) = \lambda _{[0; 1]}((\mathbb{1}_{[0; p]})^{-1}(\{ 1 \})) = \lambda _{[0; p]}([0; p]) = p Pour tout x \in [0; 1] \mathbb{1}_{[0; p]} = 1 \iff x \in [0; p], donc : \nu(\{ 0 \}) = \lambda _{[0; 1]}((\mathbb{1}_{[0; 1]})^{-1}(\{ 0 \})) = \lambda _{[0; 1]}(]p; 1]) = 1-p \nu(\{ 0; 1 \}) = \nu(\{ 0 \} \cup \{ 1 \}) = p+1 - p = 1 Donc, \nu est la mesure de Bernoulli de paramètre p : \nu = p\delta_1+(1-p)\delta_0

[!example] Exemple 2 Soit \nu = \frac{1}{6}d_0 + \frac{1}{3}d_1 + \frac{1}{2}d_2 Soit f: [0; 1] \to \{ 0, 1, 2 \} telle que \nu soit l'image de \lambda _{[0; 1]} par f !mesure image 2024-09-18 15.26.58.excalidraw

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adb2e6d5f0fd60f27aefc6aea6ed09461088e5ee: E

e96807732150d61ddfa7c8a0c877a47cbc3b3236: F

ee89b59b1a1e5a14b6c025939a698bbac5f376a9: f^{-1}(B)

e30129f05a01a4a4a7b432e0291c53bf6dcbb40e: $$B



e2b8d6503e865684f3d10c742442f5b01f2337d6: $$\mu(B)$$

5da2f1c2ab6d8be668e628b41596152fe3d036e7: $$\nu(B) = \mu(f^{-1}(B))$$

## Drawing
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