# Pointwise Product of Simple Functions is Simple Function

## Theorem

Let $\struct {X, \Sigma}$ be a measurable space.

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

Then $f \cdot g: X \to \R, \map {\paren {f \cdot g} } x := \map f x \cdot \map g x$ 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.

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

$\blacksquare$

## Also see

- Definition:Pointwise Operation, 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

## Sources

- 2005: René L. Schilling:
*Measures, Integrals and Martingales*... (previous) ... (next): $8.7 \ \text{(iv)}$