Pointwise Difference of Measurable Functions is Measurable
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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): $8.10$