Scalar Multiple of Simple Function is Simple Function

Theorem
Let $\left({X, \Sigma}\right)$ be a measurable space.

Let $f: X \to \R$ be a simple function, and let $\lambda \in \R$.

Then the pointwise scalar multiple $\lambda f: X \to \R$ of $f$ is also a simple function.

Proof
Let $\operatorname{Im} \left({f}\right)$ be the image of $f$, and $\operatorname{Im} \left({\lambda f}\right)$ that of $\lambda f$.

Consider the surjection $l_\lambda: \operatorname{Im} \left({f}\right) \to \operatorname{Im} \left({\lambda f}\right)$ defined by:


 * $l_\lambda \left({f \left({x}\right)}\right) := \lambda f \left({x}\right)$

By Real-Valued Function is Simple Function iff Finite Image Set, $\left\vert{\operatorname{Im} \left({f}\right)}\right\vert$ is finite.

Hence Cardinality of Surjection yields that $\left\vert{\operatorname{Im} \left({\lambda f}\right)}\right\vert$ is finite as well.

The result follows from a second application of Real-Valued Function is Simple Function iff Finite Image Set.