Pointwise Minimum of Simple Functions is Simple/Proof 2

Proof
From Simple Function is Measurable, we have that:


 * $f$ and $g$ are $\Sigma$-measurable.

For brevity let:


 * $h = \min \set {f, g}$

From Pointwise Minimum 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:


 * $\map h x = \map f x$

so that:


 * $\map h x \in \map f X$

If $\map g x \le \map f x$, then:


 * $\map h x = \map g x$

so that:


 * $\map h x \in \map g X$

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.