Measurable Mappings from Product Measurable Space

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Theorem

Let $\left({X, \Sigma}\right)$, $\left({X_1, \Sigma_1}\right)$ and $\left({X_2, \Sigma_2}\right)$ be measurable spaces.

Let $\Sigma_1 \otimes \Sigma_2$ be the product $\sigma$-algebra on $X_1 \times X_2$.

Let $f: X_1 \times X_2 \to X$ be a $\Sigma_1 \otimes \Sigma_2 \, / \, \Sigma$-measurable mapping.


Then:

$\forall x_1 \in X_1: f \left({x_1, \cdot}\right): X_2 \to X$ is $\Sigma_2 \, / \, \Sigma$-measurable
$\forall x_2 \in X_2: f \left({\cdot, x_2}\right): X_1 \to X$ is $\Sigma_1 \, / \, \Sigma$-measurable


Proof


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