Pointwise Scalar Multiple of Measurable Function is Measurable

Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $f : X \to \overline \R$ be a $\Sigma$-measurable function.

Let $\alpha$ be a real number.

Then the pointwise scalar multiple $\alpha f$ is $\Sigma$-measurable.

Proof
If $\alpha = 0$, then $\alpha f = 0$, and $\alpha f$ is $\Sigma$-measurable from Constant Function is Measurable.

Suppose that $\alpha > 0$.

We want to show that:


 * $\set {x \in X : \alpha \map f x \le t}$ is $\Sigma$-measurable for each $t \in \R$.

Since $f$ is $\Sigma$-measurable, we have that:


 * $\ds \set {x \in X : \map f x \le \frac t \alpha}$ is $\Sigma$-measurable for each $t \in \R$.

Since:


 * $\ds \set {x \in X : \alpha \map f x \le t} = \set {x \in X : \map f x \le \frac t \alpha}$

we have the theorem in the case $\alpha > 0$.

Suppose that $\alpha < 0$.

We again want to show that:


 * $\set {x \in X : \alpha \map f x \le t}$ is $\Sigma$-measurable for each $t \in \R$.

Since $f$ is $\Sigma$-measurable, we have from Characterization of Measurable Functions:


 * $\ds \set {x \in X :\map f x \ge \frac t \alpha}$ is $\Sigma$-measurable for each $t \in \R$.

Since:


 * $\ds \set {x \in X : \alpha \map f x \le t} = \set {x \in X :\map f x \ge \frac t \alpha}$

we have the theorem in the case $\alpha < 0$.