Pointwise Maximum of Measurable Functions is Measurable

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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 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: $(2) \implies (1)$.

$\blacksquare$


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