# Function Measurable iff Positive and Negative Parts Measurable

## 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.

$\blacksquare$

## Sources

- 2005: René L. Schilling:
*Measures, Integrals and Martingales*... (previous) ... (next): $8.11$, $\S 8$: Problem $11$