Pointwise Maximum of Measurable Functions is Measurable

Theorem
Let $\left({X, \Sigma}\right)$ be a measurable space.

Let $f, g: X \to \overline{\R}$ be $\Sigma$-measurable functions.

Then the pointwise maximum $\max \left({f, g}\right): X \to \overline{\R}$ is also $\Sigma$-measurable.

Proof
For all $x \in X$ and $a \in \R$, we have by Max Operation Yields Supremum of Parameters that:


 * $\max \left({f \left({x}\right), g \left({x}\right)}\right) \le a$

iff both $f \left({x}\right) \le a$ and $g \left({x}\right) \le a$.

That is, for all $a \in \R$:


 * $\left\{{x \in X: \max \left({f \left({x}\right), g \left({x}\right)}\right) \le a}\right\} = \left\{{x \in X: f \left({x}\right) \le a}\right\} \cap \left\{{x \in X: g \left({x}\right) \le a}\right\}$

By Characterization of Measurable Functions: $(1) \implies (2)$, the two sets on the RHS are elements of $\Sigma$, i.e. measurable.

By Sigma-Algebra Closed under Intersection, it follows that:


 * $\left\{{x \in X: \max \left({f \left({x}\right), g \left({x}\right)}\right) \le a}\right\} \in \Sigma$

Hence $\max \left({f, g}\right)$ is measurable, by Characterization of Measurable Functions: $(2) \implies (1)$.