Pointwise Product of Simple Functions is Simple Function

Theorem
Let $\left({X, \Sigma}\right)$ be a measurable space.

Let $f,g : X \to \R$ be simple functions.

Then $f \cdot g: X \to \R, \left({f \cdot g}\right) \left({x}\right) := f \left({x}\right) \cdot g \left({x}\right)$ is also a simple function.

Proof
From Measurable Function is Simple Function iff Finite Image Set, it follows that there exist $x_1, \ldots, x_n$ and $y_1, \ldots y_m$ comprising the image of $f$ and $g$, respectively.

But then it immediately follows that any value attained by $f \cdot g$ is of the form $x_i \cdot y_j$.

Hence, there are at most $n \times m$ distinct values $f \cdot g$ can take.

From Pointwise Product of Measurable Functions is Measurable, $f \cdot g$ is also measurable.

Hence by Measurable Function is Simple Function iff Finite Image Set, $f \cdot g$ is a simple function.

Also see

 * Operation Induced on Set of Mappings, of which the definition of $f \cdot g$ is an instantiation
 * Scalar Multiple of Simple Function is Simple Function
 * Pointwise Sum of Simple Functions is Simple Function
 * Space of Simple Functions forms Ring