Pointwise Scalar Multiplication preserves A.E. Equality

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.