Pointwise Maximum of Simple Functions is Simple/Proof 2
Theorem
Let $\struct {X, \Sigma}$ be a measurable space.
Let $f, g : X \to \R$ be simple functions.
Then the pointwise maximum $\max \set {f, g}: X \to \R$ is also simple function.
Proof
From Simple Function is Measurable, we have that:
- $f$ and $g$ are $\Sigma$-measurable.
For brevity let:
- $h = \max \set {f, g}$
From Pointwise Maximum of Measurable Functions is Measurable, we have that:
- $h$ is $\Sigma$-measurable.
From Measurable Function is Simple Function iff Finite Image Set, we aim to show that:
- $\map h X$ is a finite set.
From Measurable Function is Simple Function iff Finite Image Set, we have:
- $\map f X$ and $\map g X$ are finite sets.
Let $x \in X$.
If $\map f x < \map g x$, then:
- $\map h x = \map g x$
so that:
- $\map h x \in \map g X$
If $\map g x \le \map f x$, then:
- $\map h x = \map f x$
so that:
- $\map h x \in \map f X$
Since for $x \in X$ we have either $\map g x \le \map f x$ or $\map f x < \map g x$, we obtain:
- $\map h X \subseteq \map f X \cup \map g X$
From Union of Finite Sets is Finite, we have that:
- $\map f X \cup \map g X$ is finite.
Then, from Subset of Finite Set is Finite:
- $\map h X$ is finite.
So:
- $h$ is simple.
$\blacksquare$