Pointwise Product of Measurable Functions is Measurable

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Theorem

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

Let $f, g: X \to \overline{\R}$ be $\Sigma$-measurable functions.


Then the pointwise product $f \cdot g: X \to \overline{\R}$ is also $\Sigma$-measurable.


Proof

By Measurable Function Pointwise Limit of Simple Functions, we find sequences $\left({f_n}\right)_{n \in \N}, \left({g_n}\right)_{n \in \N}$ such that:

$\displaystyle f = \lim_{n \to \infty} f_n$
$\displaystyle g = \lim_{n \to \infty} g_n$

where the limits are pointwise.

It follows that for all $x \in X$:

$f \left({x}\right) \cdot g \left({x}\right) = \displaystyle \lim_{n \to \infty} f_n \left({x}\right) \cdot g_n \left({x}\right)$


so that we have the pointwise limit:

$\displaystyle f \cdot g = \lim_{n \to \infty} f_n \cdot g_n$

By Pointwise Product of Simple Functions is Simple Function, $f \cdot g$ is a pointwise limit of simple functions.

By Simple Function is Measurable, $f \cdot g$ is a pointwise limit of measurable functions.


Hence $f \cdot g$ is measurable, by Pointwise Limit of Measurable Functions is Measurable.

$\blacksquare$


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