Pointwise Multiplication preserves A.E. Equality

Theorem

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

Let $f, g, h : X \to \overline \R$ be functions with:

$f = g$ $\mu$-almost everywhere.

Then:

$f \times h = g \times h$ $\mu$-almost everywhere

where $f \times h$ and $g \times h$ are the pointwise products of $f$ and $h$, and $g$ and $h$ respectively.

Proof

Since:

$f = g$ $\mu$-almost everywhere

there exists a $\mu$-null set $N \subseteq X$ such that:

if $x \in X$ has $\map f x \ne \map g x$ then $x \in N$.

Note that if $x \in X$ is such that:

$\map f x = \map g x$

then:

$\map f x \map h x = \map g x \map h x$

By the Rule of Transposition, we therefore have:

if $x \in X$ has $\map f x \map h x \ne \map g x \map h x$ then $\map f x \ne \map g x$.

So:

if $x \in X$ has $\map f x \map h x \ne \map g x \map h x$ then $x \in N$.

Since $N$ is $\mu$-null, we therefore have:

$f \times h = g \times h$ $\mu$-almost everywhere.

$\blacksquare$