Pointwise Scalar Multiple of Measurable Function is Measurable

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

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

Let $\alpha \in \overline \R$ be an extended real number.

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

Proof
We want to show that:


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

in each of the cases:


 * $(1): \quad$ $\alpha = 0$
 * $(2): \quad$ $0 < \alpha < \infty$
 * $(3): \quad$ $-\infty < \alpha < 0$
 * $(4): \quad$ $\alpha = \infty$
 * $(5): \quad$ $\alpha = -\infty$

If $\alpha = 0$, then $\alpha f = 0$, and $\alpha f$ is $\Sigma$-measurable from Constant Function is Measurable, giving case $(1)$.

Suppose that $0 < \alpha < \infty$.

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 $\alpha > 0$, giving case $(2)$.

Suppose that $-\infty < \alpha < 0$.

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 $\alpha < 0$, giving case $(3)$.

Suppose that $\alpha = \infty$.

Then from the definition of extended real multiplication, we have:


 * $\map {\paren {\alpha f} } x = \begin{cases}\infty & \map f x \ne 0 \\ 0 & \map f x = 0\end{cases}$

If $t < 0$, we have:


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

which is $\Sigma$-measurable.

If $t \ge 0$, we have:


 * $\set {x \in X : \alpha \map f x \le t} = \set {x \in X : \map f x = 0}$

From Measurable Functions Determine Measurable Sets and Constant Function is Measurable, we have that:


 * $\set {x \in X : \map f x = 0} \in \Sigma$

So:


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

giving case $(4)$.

Suppose that $\alpha = -\infty$.

From the rules of extended real multiplication, we have:


 * $\paren {-\infty} \times f = -\paren {\infty \times f}$

From case $(4)$, we have that:


 * $\infty \times f$ is $\Sigma$-measurable.

From case $(2)$, we then have:


 * $\paren {-1} \times \paren {\infty \times f} = \paren {-\infty} \times f$ is $\Sigma$-measurable

hence showing case $(5)$.