Pointwise Scalar 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 $\mu$-almost-everywhere equality 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 $\lambda \in \R$.
Let $E \in \map {\mathcal M} {X, \Sigma, \R}/\sim$.
First, we show that if $E = \eqclass f \sim$, then $\eqclass {\lambda f} \sim$ is well-understood.
This follows from Pointwise Scalar Multiple of Measurable Function is Measurable.
We need to show that $\lambda \cdot E$ is independent of the choice of representative for $E$.
Suppose that:
- $\eqclass f \sim = \eqclass g \sim = E$
From Equivalence Class Equivalent Statements, we have:
- $f \sim g$
So, from Pointwise Scalar Multiplication preserves A.E. Equality, we have:
- $\lambda \cdot f \sim \lambda \cdot g$
That is, from Equivalence Class Equivalent Statements:
- $\eqclass {\lambda \cdot f} \sim = \eqclass {\lambda \cdot g} \sim$
which is what we aimed to show.
$\blacksquare$