Function Measurable iff Positive and Negative Parts Measurable

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Theorem

Let $\left({X, \Sigma}\right)$ be a measurable space.

Let $f: X \to \overline{\R}$ be an extended real-valued function.

Let $f^+, f^-: X \to \overline{\R}$ be the positive and negative parts of $f$.


Then $f$ is $\Sigma$-measurable iff both $f^+$ and $f^-$ are $\Sigma$-measurable.


Proof

Necessary Condition

Suppose $f$ is measurable.

By definition, its positive part $f^+$ equals the pointwise maximum:

$f^+ = \max \left\{{f, 0}\right\}$

where $0$ denotes the zero function.

By Constant Function is Measurable, $0$ is a measurable function.


Thus, by Pointwise Maximum of Measurable Functions is Measurable, $f^+$ is measurable.


Subsequently, the negative part $f^-$ of $f$ is defined by means of a pointwise minimum:

$f^- = - \min \left\{{f, 0}\right\}$

By Pointwise Maximum of Measurable Functions is Measurable and Negative of Measurable Function is Measurable, $f^-$ is measurable as well.

$\Box$


Sufficient Condition

Suppose $f^+$ and $f^-$ are both measurable.

By Difference of Positive and Negative Parts:

$f = f^+ - f^-$

and hence, by Pointwise Difference of Measurable Functions is Measurable, $f$ is also measurable.

$\blacksquare$


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