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.