Composition of Measurable Mappings is Measurable

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Theorem

Let $\left({X_1, \Sigma_1}\right)$, $\left({X_2, \Sigma_2}\right)$ and $\left({X_3, \Sigma_3}\right)$ be measurable spaces.

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

Let $g: X_2 \to X_3$ be a $\Sigma_2 \, / \, \Sigma_3$-measurable mapping.


Then their composition $g \circ f: X_1 \to X_3$ is $\Sigma_1 \, / \, \Sigma_3$-measurable.


Proof

Let $E_3 \in \Sigma_3$.

Then $g^{-1} \left({E_3}\right) \in \Sigma_2$, and $f^{-1} \left({g^{-1} \left({E_3}\right)}\right) \in \Sigma_1$ as $f,g$ are measurable.


That is, $\left({g \circ f}\right)^{-1} \left({E_3}\right) \in \Sigma_1$ for all $E_3 \in \Sigma_3$.

Hence, $g \circ f$ is $\Sigma_1 \, / \, \Sigma_3$-measurable.

$\blacksquare$


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