Pointwise Difference of Measurable Functions is Measurable

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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.

$\blacksquare$


Sources