Pointwise Difference of Measurable Functions is Measurable

Theorem

Let $\struct {X, \Sigma}$ 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:

$f - g = f + \paren {-g}$

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

• 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $8.10$