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 sur (F, \mathcal{B}), que l'on appelle mesure image de \mu par $f$

[!démonstration]- Démonstration

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

[!example] Exemple 3 (E, \mathcal{A}, \mu) = ([0; 1], \mathcal{B}([0; 1]), \lambda _{[0; 1]}) \begin{align} f : ]0; 1[ &\to \mathbb{R}^{+} \\ x &\mapsto -\ln(x) \end{align} Alors, soit \nu la mesure image de \mu par f : $$\begin{align} \nu(]-\infty, t]) &= \lambda _{[0; 1]} ({ x \in ]0; 1] \mid -\ln(x) \leq t }) \ &= \lambda _{[0; 1]} ({ x \in ]0; 1] : x \geq e^{ -t } }) \ &= \lambda _{[0; 1]} ([e^{ -t }; +\infty[ \cap ]0; 1]) \ &= \lambda _{[0; 1]} ([e^{ -t }; +\infty[) \ &= \begin{cases} 0 \text{ si } t < 0 \ 1-e^{ -t } \text{ si } t \geq 0 \end{cases} \end{align}$$ !mesure image 2024-11-20 15.35.39.excalidraw

Excalidraw Data

Text Elements

Embedded Files

834f4f03db2630a85af1da27414acaf4a96e2d6b: (E, \mathcal{A}, \mu)

2a25a46d8b55d25eb04e97f32205978c7cfc4a56: (F, \mathcal{B})

93dd76f01859baaa41c85298a66215822c042d56: f

6ebd9d18c3e73ee6fa41fa7fe0561392117d6789: A = f^{-1}(B)

811c6101906f566736a7299004cd26f5fd73c338: B = f(A)

c788d57471a52a30e34e72a1c31c096d236baaca: \begin{align}\nu(B) &= \mu(f^{-1}(B)) \\&= \mu(A) \end{align}

e28573ea7722cca49ea84581133c4a77451f8cea: \mu(A)

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