Pointwise Multiplication on Space of Real-Valued Measurable Functions Identified by A.E. Equality is Well-Defined
Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\map {\mathcal M} {X, \Sigma, \R}$ be the set of real-valued $\Sigma$-measurable functions on $X$.
Let $\sim$ be the almost-everywhere equality equivalence relation on $\map {\mathcal M} {X, \Sigma, \R}$.
Let $\map {\mathcal M} {X, \Sigma, \R}/\sim$ be the space of real-valued $\Sigma$-measurable functions identified by $\sim$.
Then pointwise scalar multiplication on $\map {\mathcal M} {X, \Sigma, \R}/\sim$ is well-defined.
Proof
Let $E_1, E_2 \in \map {\mathcal M} {X, \Sigma, \R}/\sim$.
We need to show that $E_1 \cdot E_2$ is independent of the choice of representative for $E_1$ and $E_2$.
Suppose that:
- $\eqclass f \sim = \eqclass F \sim = E_1$
and:
- $\eqclass g \sim = \eqclass G \sim = E_2$
From Equivalence Class Equivalent Statements, we have:
- $f \sim F$
and:
- $g \sim G$
So, from Pointwise Multiplication preserves A.E. Equality, we have:
- $f \cdot g \sim F \cdot G$
That is, from Equivalence Class Equivalent Statements:
- $\eqclass {f \cdot g} \sim = \eqclass {F \cdot G} \sim$
which is what we aimed to show.
$\blacksquare$