# Pointwise Maximum of Simple Functions is Simple

## 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 1

$f + g$ is simple.
$-g$ is simple.

Then, from Pointwise Sum of Simple Functions is Simple Function, we have:

$f - g$ is simple.
$\size {f - g}$ is simple.

From Pointwise Sum of Simple Functions is Simple Function, we then have:

$\paren {f + g} + \size {f - g}$ is simple.

Finally, from Scalar Multiple of Simple Function is Simple Function, we have:

$\dfrac 1 2 \paren {\paren {f + g} + \size {f - g} }$ is simple.

By Maximum Function in terms of Absolute Value, we have:

$\ds \max \set {f, g} = \frac 1 2 \paren {\paren {f + g} + \size {f - g} }$

so:

$\max \set {f, g}$ is simple.

$\blacksquare$

## Proof 2

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.
$\map f X$ 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.

$\blacksquare$