Pointwise Maximum of Measurable Functions is Measurable
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Theorem
Let $\struct {X, \Sigma}$ be a measurable space.
Let $f, g: X \to \overline \R$ be $\Sigma$-measurable functions.
Then the pointwise maximum $\max \set {f, g}: X \to \overline \R$ is also $\Sigma$-measurable.
Proof
For all $x \in X$ and $a \in \R$, we have by Max Operation Yields Supremum of Parameters that:
- $\max \set {\map f x, \map g x} \le a$
if and only if both $\map f x \le a$ and $\map g x \le a$.
That is, for all $a \in \R$:
- $\set {x \in X: \max \set {\map f x, \map g x} \le a} = \set {x \in X: \map f x \le a} \cap \set {x \in X: \map g x \le a}$
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:
- $\set {x \in X: \max \set {\map f x, \map g x} \le a} \in \Sigma$
Hence $\max \set {f, g}$ is measurable, by Characterization of Measurable Functions: $(2) \implies (1)$.
$\blacksquare$
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $8.10$