Pointwise Product of Simple Functions is Simple Function

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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.

$\blacksquare$


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