Pointwise Difference of Measurable Functions is Measurable

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

Let $f, g: X \to \overline{\R}$ be $\Sigma$-measurable functions.

Assume that the pointwise difference $f - g: X \to \overline{\R}$ is well-defined.

Then $f - g$ is a $\Sigma$-measurable function.

Proof
We have the apparent identity:


 * $f - g = f + \left({-g}\right)$

By Negative of Measurable Function is Measurable, $-g$ is a measurable function.

Hence so is $f - g$, by Pointwise Sum of Measurable Functions is Measurable.