Pointwise Scalar Multiplication preserves A.E. Equality
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Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g : X \to \overline \R$ be extended real-valued functions such that:
- $f = g$ $\mu$-almost everywhere.
Let $\lambda \in \overline \R$.
Then:
- $\lambda \cdot f = \lambda \cdot g$ $\mu$-almost everywhere
where $\lambda \cdot f$ denotes pointwise scalar multiplication.
Proof
Since:
- $f = g$ $\mu$-almost everywhere
there exists a $\mu$-null set $N \subseteq X$ such that:
- if $\map f x \ne \map g x$ then $x \in N$.
Note that if $\map f x = \map g x$ then $\lambda \map f x = \lambda \map g x$.
So, from the Rule of Transposition we have:
- if $\lambda \map f x \ne \lambda \map g x$ then $\map f x \ne \map g x$
so:
- if $\lambda \map f x \ne \lambda \map g x$ then $x \in N$.
So, we have:
- $\lambda \cdot f = \lambda \cdot g$ $\mu$-almost everywhere.
$\blacksquare$