Pointwise Multiplication on Space of Real-Valued Measurable Functions Identified by A.E. Equality is Well-Defined

From ProofWiki
Jump to navigation Jump to search

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.

$\blacksquare$

Retrieved from "https://proofwiki.org/w/index.php?title=Pointwise_Multiplication_on_Space_of_Real-Valued_Measurable_Functions_Identified_by_A.E._Equality_is_Well-Defined&oldid=579506"