Negative Part of Simple Function is Simple Function

Theorem

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

Let $f: X \to \R$ be a simple function.

Then $f^-: X \to \R$, the negative part of $f$ is also a simple function.

Let $f$ have the following standard representation:

$f = \ds \sum_{i \mathop = 0}^n a_i \chi_{E_i}$

Then we see that $f^-$ must satisfy:

$f^- = \ds \sum_{i \mathop = 0}^n \min \set {a_i, 0} \chi_{E_i}$

as the $E_i$ are disjoint, and $\chi_{E_i} \ge 0$ pointwise.

Since all of the $E_i$ are measurable, it follows that $f^+$ is a simple function.

$\blacksquare$

2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $8.7 \ \text{(v)}$