Pointwise Product of Measurable Functions is Measurable

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Theorem

Let $\struct {X, \Sigma}$ 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 is Pointwise Limit of Simple Functions, we find sequences $\sequence {f_n}_{n \mathop \in \N}, \sequence {g_n}_{n \mathop \in \N}$ such that:

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

where the limits are pointwise.

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

$\map f x \cdot \map g x = \ds \lim_{n \mathop \to \infty} \map {f_n} x \cdot \map {g_n} x$


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so that we have the pointwise limit:

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

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.

$\blacksquare$


Sources

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