Composition of Measurable Mappings is Measurable

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