Negative Part of Simple Function is Simple Function

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Theorem

Let $\left({X, \Sigma}\right)$ 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.


Proof

Let $f$ have the following standard representation:

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


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

$f^- = \displaystyle \sum_{i \mathop = 0}^n \min \left\{{a_i, 0}\right\} \chi_{E_i}$

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.

$\blacksquare$


Also see


Sources